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© 2023 Mathspace Pty Ltd Copyright Notice This Work is copyright. All rights are reserved. Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. For permission to use material from this text or product, please email hello@mathspace.com.au For our full digital offering, visit mathspace.co Mathspace Pty Ltd Level 11, 477 Pitt St Sydney 2000 The Victorian Curriculum F-10 content elements are © VCAA, reproduced by permission. The VCAA does not endorse or make any warranties regarding this resource. The Victorian Curriculum F-10 and related content can be accessed directly at the VCAA website. Title: Mathspace Victorian Curriculum 2.0 - Year 9 - Edition 2 Print Run: 1 ISBN: 978-1-963022-44-5 Editors and lead authors: Erin Gallagher, Elisha Templin, Sarah Butterworth, Julia Reid, Anjelica Sieger, Jacqueline Cichello, Stephen George, Grace Peake, Jonathan Kim Sing, Toni Smythe, Dot Yung Further reviewers: Melissa Archey, Warren Bardsley, Lianna Beeching, Kathryn Bennett, Benjamin Bui, Eric Chan, Cory Duker, Bahija Elmir, Wayne Gillie, Jeremy Hunter, Shahad Odeh, Helen Soar, Patricia Toutzaris, Paul Unkles, Jie Yang Writing and development team: Adam Humphries, Neil Lopez, Ma. Lourdes Ona, Bernadette Barillo, Sheramie Castillo, Abegail Fontanilla, Shaira Llanita, Lesley Lopez, Kerby Mendoza, Regine Jara, Albert Jara, Rolly Apdo, Michaela Belen, Valerie Baja, Marielle Belen, Allen Calabia, Adrian Doctolero, Dhave Fernandez, Janine Francisco, Rose Francisco, Rona Josue, Crystel Lontoc Images and Design team: Alex Duckmanton, Chastine Marquez, Keith Gimeno

Contents 1

Real numbers

1.01

Rational and irrational numbers

1.02

Operations with integers

Investigation: Construct irrational numbers

Chapter 1 review

Indices 24 2.01

Multiplication law

2.02

Division law

2.03

Power of a power law

2.04

Zero and negative indices

2.05

Scientific notation

Chapter 2 review

Algebra 60 3.01

Algebra review

3.02

Distributive law

3.03

Binomial expansion

3.04

Algebraic factors

3.05

Factorise by grouping

Investigation: Calendar expressions

Chapter 3 review

Linear relationships

4.01

Midpoint

4.02

Gradient

106

4.03

Distance between two points

121

Investigation: Transformations of linear graphs

132

4.04

Gradient-intercept form

134

4.05

Linear graphs

144

4.06

Equation of a straight line

155

4.07

Parallel and perpendicular lines

160

4.08

Direct proportion

170

Contents mathspace.co

iii

4.09

Model linear equations

178

4.10

Linear inequalities

185

Chapter 4 review

194

Quadratics 198 5.01

Quadratic relationships

200

5.02

Graphs and features of quadratics

209

Investigation: Transformations of quadratics

221

5.03

Basic quadratics

223

5.04

Vertical translations

230

5.05

Horizontal translations

236

5.06

Turning point form

242

5.07

Factorised form

254

5.08

Standard form

263

5.09

Solutions to quadratics

271

5.10

Solve algebraically

280

5.11

Solve by factorisation

286

Investigation: Other non-linear graphs

293

Chapter 5 review

295

Finance 300 Investigation: Finances

302

6.01

Cost, profit, revenue

304

6.02

Income

314

6.03

Simple interest

325

6.04

Re-investing with simple interest

334

Chapter 6 review

347

Scale and similarity

350

Investigation: Algorithms in geometry

352

7.01

Similar right-angled triangles

355

7.02

Scale factors

364

Investigation: Guessing distance

375

7.03

Scale and area

377

7.04

Scale and volume

382

Mathspace Victorian Curriculum – Year 9 mathspace.co

389

Chapter 7 review

392

Trigonometry 396 Pythagoras’ theorem

398

Investigation: The triangle inequality

406

Investigation: Ratios in triangles

409

8.02

Trigonometric ratios

411

8.03

Side lengths

423

8.04

Angles

433

8.05

Applications of trigonometry

443

Chapter 8 review

450

8.01

Investigation: Scale diagrams

Measurement 456 9.01

Area

458

9.02

Area of composite shapes

465

9.03

Nets of prisms

475

9.04

Surface area of prisms

482

9.05

Volume of prisms

490

Investigation: Volume versus area

500

9.06

Accuracy of measurement

503

9.07

Surface area of cylinders

513

9.08

Volume of cylinders

521

Chapter 9 review

530

Probability 536 10.01

Experimental probability

538

10.02

Venn diagrams and two-way tables

548

10.03

Arrays and simple tree diagrams

564

10.04

With and without replacement

575

10.05

Multi-stage probability

587

Investigation: Simulation of compound events

603

Chapter 10 review

606

Contents mathspace.co

Statistics 610 11.01

Types of data

612

11.02

Data displays

619

Investigation: Opinion Polls

634

11.03

Back-to-back stem and leaf

636

11.04

Grouped histograms

644

11.05

Shape of data

658

Investigation: Statistical reports in the media

668

Compare data sets

670

Investigation: Plan and conduct statistical investigations

685

Chapter 11 review

689

Answers

694

11.06

Mathspace Victorian Curriculum – Year 9 mathspace.co

“It is impossible to be a mathematician without being a poet in soul.” –Sofia Kovalevskaya

Big ideas Using the correct order of operations on real numbers (rational and irrational numbers) ensures accurate calculations and meaningful estimates.

1 Real numbers Chapter outline 1.01 1.02

Rational and irrational numbers Operations with integers Investigation: Construct irrational numbers Chapter 1 review

4 12 18 22

The world record for reciting π by a person was over 70,000 decimal places! How many can you memorise?

1.01 Rational and irrational numbers After this lesson, you will be able to… • identify and classify numbers as rational or irrational. • understand what surds are and how they are different from other irrational numbers. • estimate the value of irrational numbers using methods involving square numbers.

Rational and irrational numbers A rational number is a number that can be written as a fraction where both the numerator and denominator are integers. Rational numbers consist of: • Integers (positive and negative whole numbers, including zero) • Terminating decimals (decimals with a limited number of digits) • Recurring decimals (decimals with an inﬁnite number of digits, which have a repeating pattern) For example:

An irrational number cannot be expressed as a fraction where both the numerator and the denominator are integers. When expressed as a decimal, irrational numbers have inﬁnite digits, with no repeating pattern. For example:

π = 3.141 592 653 589 793… = 1.414 213 562 373 095… We often come across irrational numbers when working with square roots or other roots. Irrational numbers can take the form of: • The square root of a number that is not a perfect square, such as or • The cube root of a number that is not a perfect cube, such as or , and so on. Rational and irrational numbers together make up the set of numbers known as real numbers, as shown in the diagram. Real Numbers Rational numbers Integers 4

Whole numbers 1 0 1 2 3 4

Irrational numbers

Mathspace Victorian Curriculum – Year 9 mathspace.co

Example 1 Is each number rational or irrational? a

Create a strategy Can the number be expressed as a fraction where both the numerator and denominator are integers? If so, it is rational. Otherwise, it is irrational.

Apply the idea is a fraction that cannot be simpliﬁed. The denominator is an integer, but the numerator is not. This is an irrational number.

Create a strategy Is the number underneath the square root a perfect square? If so, it is rational. Otherwise, it is irrational.

Apply the idea 13 is not a perfect square, so

is irrational.

Reﬂect and check We can explore square root numbers using a calculator. Evaluate the

and investigate the

decimal representation, 3.60555127. We can be fairly certain that it isn’t rational because there appear to be no repeating patterns in the decimal digits.

Create a strategy This number is a decimal, so if it is a repeating or terminating decimal, then it is rational. If it is nonrepeating and non-terminating, it is irrational.

Apply the idea is a recurring decimal, so it is a rational number.

Reﬂect and check If we evaluate on the calculator, we get . This shows that the decimal can be written as a fraction, where both the numerator and denominator are integers, meaning it is rational.

1.01 Rational and irrational numbers mathspace.co

Create a strategy Evaluate the number and examine it to see if it is rational.

Apply the idea If we evaluate

on the calculator, we get 55.901 699 437 5 …. This is a non-terminating and

non-repeating decimal. is an irrational number.

Reflect and check If we used the definition of the power of 5, we can see if this can be simplified to a rational number or not. Write in expanded form Group pairs Use that Evaluate and simplify Since 5 is not a perfect square, we know that

is irrational. So

is irrational.

Idea summary A rational number is a number that can be written as a fraction where both the numerator and denominator are integers. An irrational number is a number that cannot be written as a fraction of two integers. A surd is a type of irrational number. It is the square root of a number that is not a perfect square (or the cube root of a number that is not a perfect cube, etc.).

Estimate irrational numbers We can place irrational numbers on the number line if we know their (approximate) decimal equivalents. For example, π ≈ 3.14, so we can place π on the number line between 3.1 and 3.2. We can place surds approximately on a number line using perfect squares on either side of the number we are taking the square root of. For example, let’s look at

. The perfect squares on either side of 90 are 81 and 100. 81 < 90 < 100

Mathspace Victorian Curriculum – Year 9 mathspace.co

This means that and so So we can place

on the number line between 9 and 10.

In general, if a<b<c then

Example 2 Represent these values on the number line:

π2,

, −1.21,

Create a strategy Estimate the values of the numbers using decimals and plot the values on the number line.

Apply the idea Note that π is estimated as 3.14 π2 ≈ (3.14)2

Square 3.14

≈ 9.9

Evaluate

Find the value of

Estimate the value of than .

The value of

by comparing to the closest square numbers that are bigger and smaller

is closer to

than

Plotting the values on the number line, we have: −2

−1

The numbers in order from smallest to largest are:

Reﬂect and check Comparing the numbers, we can see that:

1.01 Rational and irrational numbers mathspace.co

Idea summary We can estimate the values of surds by ﬁnding square numbers on either side of the number we are taking the square root of. If we want to estimate b , we can ﬁnd square numbers a and c so that a<b<c then a< b< c

1.01 Practice questions Understanding 1

1–4

Adventurer 1–4

Trailblazer 1–4

Match each number with its number type. a

Fraction

−7

Terminating decimal

4.65

Integer

iii

Surd

Which of these are integers?

Which of these are irrational?

Classify these as either rational or irrational: a

Explorer

−4

−0.7

Mathspace Victorian Curriculum – Year 9 mathspace.co

4.435

π2

5, 6, 7a–c, 8a–d, 9ab, 10ac, 11, 12, 14a–d, 15–17

Fluency 5

, we can say that

and

5, 6, 7a–d, 8a–d, 9cd, 10bc, 11, 12, 14c–f, 15–17

and

5, 7, 8e–h, 9cd, 10cd, 11, 13, 14e–h, 16–18

are numbers between 2 and 3.

Are these statements true or false? a

, therefore

and

, therefore

and

, therefore

and

Simplify each of these and classify them as either rational or irrational: a

Ex 1

, where n is odd

−3

e 9

−2

−1

−2.2

0.5

Which integer is each number closest to? a

, where n is even

Plot these sets of numbers on a number line: a

Plot these numbers on the number line: −4

Ex 2

Classify these as either rational or irrational:

g 8

are between 6 and 7.

and

are irrational numbers as they are not perfect squares.

and

e 6

, therefore

Select which surd is: a

Between 3 and 4 A

Between 6 and 7 A

1.01 Rational and irrational numbers mathspace.co

Between 8 and 9 A

For a

: Complete the inequality with two consecutive perfect square numbers that 59 lies between: ⬚ < 59 < ⬚ Complete the inequality with two consecutive integers that

For a

Between 10 and 11 A

⬚<

<⬚

Complete the inequality with two consecutive perfect square numbers that 22 lies between: ⬚ < 22 < ⬚ Complete the inequality with two consecutive integers that ⬚<

<⬚

⬚<

<⬚

⬚<

<⬚

⬚<

<⬚

⬚<

<⬚

For each pair of numbers, select the number with the larger value: or

⬚<

Which integer lies between

and

<⬚

or 5

d h

⬚<

<⬚

⬚<

<⬚

or 3

or 8

For each pair of numbers, select the number with the smaller value: a

lies between:

What two consecutive integers will complete these inequalities? a

lies between:

4π or 12

π2 or 144

Order the numbers

Reasoning

or 11

, from smallest to largest.

21–23

19, 20, 22

Is it possible to ﬁnd two irrational numbers that multiply together to give a rational number? If so, give an example. If not, explain why not.

Is it possible to ﬁnd two irrational numbers that add together to give a rational number? If so, give an example. If not, explain why not.

A rectangle has length

and width

Is the perimeter of the rectangle rational or irrational?

Is the area of the rectangle rational or irrational?

Mathspace Victorian Curriculum – Year 9 mathspace.co

When performing calculations involving π, there are a few common ways to do these, such as: • Using the π button on the calculator • Using 3.14 • Using Order the options from most to least accurate. Justify your answer.

Naomi was asked to estimate . She gave the answer 5.5, and her reason was that she knew the answer would be between 5 and 6. a

Explain how she knew the answer was between 5 and 6.

Is her estimate reasonable? Explain.

Problem-solving 24ab, 26 24

24bc, 25, 26

24–27

Pythagoras’ theorem says that c2 = a2 + b2, where c is the length of the hypotenuse and a and b are the lengths of the two shorter sides. Using Pythagoras’ theorem with the given triangle, answer these questions: a

Calculate the exact value of c if a = 12 and b = 5.

Calculate the exact value of b if c =

Calculate the exact value of c if a = 1 and b = 2.

Calculate the exact value of a if c = 7 and b = 6.

and a =

b cm

c cm

a cm

ISO paper (for example, A4 paper) always has a length that is times longer than the shorter side. How many times longer is the diagonal of a piece of ISO paper than its shorter side?

Fiona has a sheet of card that is

long and

wide.

Fiona estimates the area of the card by using the closest integer value to each square root. What answer did she get?

Fiona then calculates the area using her calculator. What answer did she get, rounded to four decimal places?

What was the percentage error in her estimate, rounded to two decimal places?

1.01 Rational and irrational numbers mathspace.co

The spiral of Theodorus can be drawn to estimate the values of surds. The ﬁrst right triangle in the spiral has short sides of length 1. The length of the hypotenuse of the ﬁrst triangle is

The hypotenuse of the ﬁrst right triangle is then used as one of the short sides for the next right triangle (the other short side has a length of 1). The second right triangle will have a hypotenuse of length

, and so on.

Draw the ﬁrst six triangles of the spiral, starting with a triangle with short sides of 1 cm. Use a protractor to make sure your right angles are accurate. a

Measure the hypotenuse of the sixth right triangle to estimate the value of to one decimal place. Why might your answer to (a) be different from the actual value of rounded to one decimal place?

How might we draw the spiral differently to improve the accuracy of our estimate for ?

1 1 4

2 1

1 17

8 15

9 10

Try drawing the spiral as discussed in part (c) to improve your estimate of .

14 11

1.02 Operations with integers After this lesson, you will be able to… • understand and apply the order of operations in evaluating expressions involving integers. • interpret and solve problems involving mixed operations with integers. • extend the understanding of the order of operations to fractions. • apply the rules of operations to solve problems involving fractions.

Mixed operations with integers We know how to follow the order of operations when evaluating expressions involving more than one operation. The order of operations is as follows: 1. Evaluate whatever is contained in brackets. 2. Evaluate any multiplication or division, reading from left to right. 3. Evaluate any addition or subtraction, reading from left to right.

Mathspace Victorian Curriculum – Year 9 mathspace.co

If we have negative integers in our expression, we still follow the exact same rules; we need to be careful when evaluating the terms involving negatives.

Example 1 Evaluate: a 5 + 9 × (−6)

Create a strategy We ﬁrst need to evaluate the multiplication and then the addition.

Apply the idea 5 + 9 × (−6) = 5 + (−54) Evaluate the multiplication = −49 Evaluate

Reﬂect and check We can check with a calculator. Evaluating 5 + 9 × (−6) = −49 Since scientiﬁc calculators apply the order of operations, this is a good indication that we have applied the order of operations correctly.

b (7 − 8) × (−5)

Create a strategy We ﬁrst need to evaluate the brackets and then the multiplication.

Apply the idea (7 − 8) × (−5) = (−1) × (−5) = 5

Evaluate 7 − 8 Evaluate

Example 2 Evaluate 21 ÷ (−7) × 2

Create a strategy Evaluate the multiplication and division from left to right.

Apply the idea 21 ÷ (−7) × 2 = −3 × 2

Evaluate 21 ÷ (−7)

= −6 Evaluate

1.02 Operations with integers mathspace.co

Idea summary The order of operations is as follows: 1. Evaluate anything inside the brackets. 2. Evaluate any multiplication or division, reading from left to right. 3. Evaluate any addition or subtraction, reading from left to right.

Mixed operations with fractions We can apply the rules for mixed operations to fractions.

Example 3 Evaluate: a

Create a strategy The multiplication needs to be done before the addition.

Apply the idea Simplify by removing common factors Perform the multiplication Use a common denominator for the addition Perform the addition

Create a strategy The indices will need to be done ﬁrst. To do this, we need to express We can then do the addition and subtraction.

Mathspace Victorian Curriculum – Year 9 mathspace.co

as an improper fraction.

Apply the idea Write

as an improper fraction

Evaluate the index: Express 4 as

so we can subtract the ﬁrst two terms

Evaluate the ﬁrst two terms Use a common denominator for the remaining terms Evaluate We may write the answer as a mixed fraction.

Reﬂect and check We can check with a calculator. Evaluating Since scientiﬁc calculators apply the order of operations, this is a good indication that we have applied the order of operations correctly.

Idea summary The order of operations applies when doing arithmetic with fractions.

1.02 Practice questions Understanding 1

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

Are these statements true or false? a

When evaluating an expression with mixed operations, you should always perform addition and subtraction before multiplication and division.

If an expression contains only multiplication and division operations, you should evaluate them from left to right.

Surds can be added in any order but not subtracted.

Two irrational numbers can be multiplied in any order and also divided in any order.

1.02 Operations with integers mathspace.co

Which of these expressions is equivalent to 7 − 4 + (−3)? A

−10 + 4

Ex 2

Ex 3

2×3+3

Fluency

Ex 1

7+4−3

4 − 10 2×

−(5 − 2) + 4

C −3

5a–d, 6a–d, 7a–d, 8a–d, 9ab

− (−3)? C

2×9−3

5b–f, 6c–f, 7b–f, 8b–f, 9ac

5d–h, 6e–h, 7d–h, 8d–h, 9cd

Evaluate: a

−7 + 8 − 2

−10 − 4 + 6

15 − (−11) − 5

(−24 + 18) − 13

−16 − (2 − (−7))

−9 − (−12) − (−3)

1 − (11 + (−14))

(34 − (−52)) − 27

Evaluate: a

−10 + 8 × (−3)

21 − 6 × (−7)

(−9 − 12) × (−2)

−11 × (−3) + 20

18 × (13 + (−15))

−30 × (−1 − 3)

(−4 × 14) − 4 × 3 − 17

−25 × (−3) + 2 × (−17 + 23)

Evaluate: a

−18 + 21 ÷ 7

72 ÷ 6 ÷ 2

(120 ÷ 5) × 3

54 ÷ 3 ÷ (−2)

6 − 25 ÷ 5 + 4

(−36) ÷ (−4) − 14

(−390) ÷ (−19 + 6)

(−100) ÷ (−4) ÷ (−5)

Evaluate: a

−7 + 8 × 6 ÷ 2

(−14) ÷ 2 − 10 ÷ (−2)

2 × (−25 − 20) ÷ (−7 + 2)

20 + (−14) − (−2) − 18

9 × (−6) + 28 ÷ (−7)

(28 × 5) ÷ (14 − 24) + 5

[7 × (6 − 3)] × [(8 + 2) ÷ 5]

[1 + (6 × 3 − 9)] ÷ [20 − (100 ÷ 10)]

Evaluate: a

Reasoning 10

−7 − 4 − 3

Which of these expressions is equivalent to A

Which of these is not equivalent to −(5 × 2) + 4? A

7 + (−4) + (−3)

b 10, 11

c 11–13

d 12–14

Kaylah and Orion are evaluating the expression 10 ÷ 2 × 5. Here is their work: Orion: Kaylah: 10 ÷ 2 × 5 = 10 ÷ 10 =1 Who has the correct solution? Explain your answer.

Mathspace Victorian Curriculum – Year 9 mathspace.co

10 ÷ 2 × 5 = 5 × 5 = 25

An incorrect series of steps for solving an expression with mixed operations and integers is shown below. Identify the error and correct it. Expression: (−2) + 12 × 3 − (−4) (−2) + 12 × 3 − (−4) = 10 × 3 − (−4) = 30 − (−4) = 30 + 4 = 34

Determine the accuracy of this statement and provide the correct result if it is incorrect: 12 + (−63) ÷ (3) = −17

Do the expressions (−7 + 5) × (−3) and − 7 × (−3) + 5 × (−3) yield the same result? Explain why.

A square enclosure has an area of 20 m2. Explain why the side lengths cannot be rational.

Problem-solving 15–17 15

15–17

16–18

Tara is waiting for the next ﬂight to Los Angeles, which was scheduled to be in 36 minutes, but there is a 48-minute delay. She takes a nap and wakes up 24 minutes later. Calculate how much longer Tara has to wait before the plane departs.

A bus departs the depot with 8 passengers on board. At the ﬁrst stop, 4 people got off and 6 people got on. At the second stop, a further 3 people got off, and 7 more got on. Calculate the number of people on the bus when it leaves the second stop.

Luigi enters an elevator at the 7th floor (the ground floor being floor 0). The elevator goes down 3 floors, then up 9 floors and finally it goes down 2 floors, where Luigi gets out.

Which ﬂoor does Luigi get out on?

When Luigi gets off the elevator, how many ﬂoors from his starting point is he?

At the start of the week, Giovanni had $250 in his bank account. On Monday, he withdrew $39. On Wednesday, he deposited $15, and on Thursday, he deposited a further $45. Finally, he withdrew $115 on Friday. Find Giovanni’s ﬁnal bank balance at the end of the week.

1.02 Operations with integers mathspace.co

INVESTIGATION

Construct irrational numbers Investigate rational and irrational numbers on the number line.

Objectives To use geometric constructions to plot an irrational number on the number line

Materials • Pencil

• Compass

Constructing irrational numbers Plotting numbers on a number line All real numbers can be plotted on a number line. The trick is that sometimes we need to construct it, and sometimes we need something with a higher degree of accuracy than the human hand. If you were given a number line with whole numbers and were asked to plot the number , how conﬁdent would you be in your approximation? What if we could use constructions to plot this number? We will learn how to do that in this investigation. 1. Draw a number line. Plot point O, representing zero on the number line. At a length of one unit, plot point A. One unit means it’s just some distance. 1 cm would probably be too small to work with, and 1 m will be too large, so just use any length that fits on your page, and we will call this 1 unit. O 0

2. Construct a line perpendicular to the number line through point A. Do this by ﬁrst drawing an arc centred at point A that intersects the number line twice, labelling these intersection points B and C.

O 0

Mathspace Victorian Curriculum – Year 9 mathspace.co

3. Set the compass so that it is slightly longer than the distance from one point of intersection on the arc to point A. Now, draw a second arc centred at point B through the ﬁrst arc.

O 0

4. Draw a third arc centred at point C through the ﬁrst and second arcs.

O 0

5. Set the compass the length of OA, one unit, and then draw a line from point A through the intersection of the second and third arcs. Mark the end of the line as point D. This line from point A to point D should be the same length as OA, one unit. D

O 0

Investigation: Construct irrational numbers mathspace.co

6. Draw a segment from point D to point O, creating a right-angled triangle. The hypotenuse of the right-angled triangle is (OA)2 + ( AD)2 = (DO)2, and we know the lengths of OA and AD are both one unit. Therefore, DO must be

units. D

1 unit

B 1 unit

7. Set the compass the length of DO.

1 unit

B 1 unit

8. Plot a point on the number line the length of DO. This point must be located at the value D

Mathspace Victorian Curriculum – Year 9 mathspace.co

Investigate 1.

Given the process you’ve just executed for position of

, how might you adapt this to locate the

on the number line?

If you were to plot the square roots of whole numbers from one to ten on the number line, what pattern or relationship do you think you would observe between consecutive values?

Discussion 1.

Using your number line, can you place markers for a few numbers from each category: irrationals, rationals, integers, and natural numbers (positive integers)? Discuss how you determined where each number should go.

Beyond

, there are other irrational numbers like π. How do you think the challenges

of representing these numbers on the number line might differ from the process we’ve used for

Did you know?

Real number system is continuous, meaning there are no gaps between numbers. This continuous nature of real numbers is essential in studying processes that change gradually, such as the growth of plants. For example, when tracking how tall a plant grows each day, the height doesn’t jump from one number to the next. Instead, it changes smoothly over time, and real numbers allow us to represent even the tiniest changes in height, down to the millimetre or smaller.

Investigation: Construct irrational numbers mathspace.co

1 Chapter review 1

Classify these as either rational or irrational: a

−0.175

e 2

Which integer is each number closest to? a

What two consecutive integers will complete these inequalities? a

Between 7 and 8 A

Between 4 and 5 A

Select which surd is: a

Plot these sets of numbers on a number line: a

π+4

A square has side lengths of

c units.

Is the perimeter of the square rational or irrational?

Is the area of the square rational or irrational?

Abeny has a rectangular pool that is

long by

wide.

Abeny estimates the area of the pool using the closest integer value to each square root. What is her answer?

Abeny then calculates the area using her calculator. What answer did she get, rounded to four decimal places?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Evaluate: a

−5 − 4 + 7

13 − (−3) − 7

−1 − (14 + (−3))

(10 − 20) + 30

Evaluate: a

−7 + 16 × (−1)

20 − 5 × (−9)

18 × (13 + (−15))

−20 × (−4) + 2 × (−6 + 6)

Evaluate: a

−8 + 30 ÷ 3

120 ÷ 4 ÷ (−5)

(−36) ÷ (−9) + 3

5 − 12 ÷ 6 + 14

Evaluate: a

−7 + 5 × 15 ÷ 3

11 × 9 ÷ (1 − 2) + 3

6 × (−4) + 35 ÷ (−7)

[5 + (10 × 15 − 15)] ÷ [10 − (30 ÷ 10)]

Evaluate: a

An incorrect series of steps for solving an expression with mixed operations and integers is shown. Identify the error and correct it. Expression: 15 − (−7) × 2 + (−9)

15 − (−7) × 2 + (−9) = 15 + 7 × 2 − 9

= 22 × 2 − 9

= 44 − 9

= 35

Braxton has a debt of $500. He repays $200, then borrows another $300. What is his new debt?

Doreen invested $15 000 in stocks. In the ﬁrst week, her investment tripled. In the second week, she gained $5 000. In the third week, she lost half of the total amount. What is the value of her investment after the third week?

Chapter 1 review mathspace.co

Big ideas Applying index laws simplifies expressions and makes it easier to work with very large or very small numbers.

Curriculum Outcomes VC2M9A01 apply the exponent laws to numerical expressions with integer exponents and the zero exponent, and extend to variables VC2M9M02 solve problems involving very small and very large measurements, timescales and intervals expressed in scientific notation

2 Indices Chapter outline 2.01 2.02 2.03 2.04 2.05

Multiplication law Division law Power of a power law Zero and negative indices Scientific notation Chapter 2 review

26 30 35 40 48 58

A grain of sand is about 10–3 metres across, proving that even the tiniest things can be understood through indices!

2.01 Multiplication law After this lesson, you will be able to… • use the multiplication law to simplify expressions with both numeric and algebraic bases.

Multiply terms with indices Interactive exploration Discover this concept in action online

mathspace.co

Consider the product of a3 × a5. Notice that the terms share the same base, a. We can see that there are eight a’s being multiplied together. a3 × a5 = a × a × a × a × a × a × a × a We can avoid having to write each expression in expanded form by using the multiplication law.

am × an = am + n where a, m and n are integers and a ≠ 0. In this example, the a is called the base, and the m and n are both called indices, powers or exponents. So, when we multiply terms with the same base: • The base remains the same. • We add the indices.

Example 1 Simplify: a 35 × 39

Create a strategy Use the multiplication law to simplify the expression.

Apply the idea 35 × 39 = 35 + 9 14

Apply the multiplication law Simplify

b c7 × c6

Create a strategy Use the multiplication law to simplify the expression.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Apply the idea c7 × c6 = c7 + 6 13

Apply the multiplication law Simplify

c 5d5 × 3d3

Create a strategy Multiply the coeﬃcients, then use the multiplication law to simplify the parts of the terms with indices.

Apply the idea 5d5 × 3d3 = 5 × d5 × 3 × d3 5

=5×3×d ×d 5+3

= 15 × d 8

= 15d

Expand Group the numeric values and algebraic components Multiply the numeric values and apply the multiplication law Simplify

d −3m2 n5 × 7m3 n

Create a strategy Multiply the coeﬃcients, then use the multiplication law to simplify the parts of the terms with indices. Work with the different bases in turn.

Apply the idea −3m2 n5 × 7m3 n = −3 × m2 × n5 × 7 × m3 × n 2

Expand

= −3 × 7 × m × m × n × n

roup the numeric values and algebraic G components

= −21 × m2 + 3 × n5 + 1

ultiply the numeric values and apply M the multiplication law

= −21m5 n6

Simplify

 a e   × a2  3

Create a strategy When a fraction is raised to a power, it means both the numerator and denominator are raised to 3

 a a3 the same power. So   = .  3 33

Apply the idea 3

 a a3   × a 2 = × a2 3  3 3

a3 × a2 27 1 3 = a × a2 27 1 3 +2 = a 27 1 5 = a 27

Apply the power of 3 to both numerator and denominator Evaluate 33 Group the numeric values and algebraic components Apply the multiplication law Simplify

2.01 Multiplication law mathspace.co

Idea summary When multiplying terms with the same base, add the indices: a m × an = am + n If the terms also have coefficients, then multiply the coefficient first, keep the base the same and add the indices: bam × can = (b × c) am + n When a fraction is raised to a power, both the numerator and the denominator are raised to the same power:  a  b 

am bm

2.01 Practice questions Explorer

Understanding 1

a b c

Adventurer

1–3

Trailblazer 2, 3

1–3

Complete the expanded form: 35 = ⬚ × ⬚ × ⬚ × ⬚ × ⬚ Complete the expanded form: 34 = ⬚ × ⬚ × ⬚ × ⬚ Using (a) and (b), complete the expanded form:

35 × 34 = (⬚ × ⬚ × ⬚ × ⬚ × ⬚) × (⬚ × ⬚ × ⬚ × ⬚)

= 3⬚

Complete the statement: When you multiply terms that have the same bases you ⬚ the indices.

Write 53 in expanded form.

Write 2 × 2 × 2 × 2 in index form.

Write a5 in expanded form.

Write b × b × b × b × b × b in index form.

Trent stated that 22 + 32 is equal to 52. a

Use expanded form for each term to explain why Trent is incorrect.

Can 22 + 32 be simplified using the multiplication index law? 4a–i, 5a–d, 6a–c, 7ab, 8a–d

Fluency Ex 1a

4b–j, 5b–h, 6b–f, 7a–d, 8c–e

4c–k, 5e–l, 6c–h, 7c–h, 8d–f

Simplify: a

23 × 24

32 × 32

(−2) × (−2)

35 × 33

2 × 25

210 × 25

34 × 35

2 ×2 ×2

Mathspace Victorian Curriculum – Year 9 mathspace.co

3 ×3×3

54 × 5

73 × 77

Ex 1b

Simplify: a

Ex 1c

Ex 1d

b ×b

a6 × a − 3

x6 × x3

a4 × a

y ×y

u ×u

b6 × b7

x − 5 × x8

d7 × d −4

e −10 × e8

y3 × 3y

2x5 × 2x4

3a4 × 2a6

5y2 × 3y2

a7 × a5 3

Simplify: a

5y2 × y3

−b × − b

7a × 3a

9u × 8u

5c5 d4 × 4c2 d3

Simplify: a

Ex 1e

y3 × y2

3x2 y2 × 2xy3 3 4

7a b × 2ab

−2u4 v2 × 3u6 v9

−3t6 u3 × − 4t5 u3

5a3 b2 c4 × − 5a2 b4 c6

5w5 x3 y9 × − 3wx5 y

3abc × − 3a2 b3 c4 × − 3a2 bc3

 b2  3  3  ×b  

Simplify: 2

 a  2 × a

 v 4  5  × v

 c 2  2 × c

 y2  2  ×y  4

 x2  2  5  × x× x  

Reasoning

9, 10

10, 11

Explain why the expression a4 × b4 cannot be simpliﬁed using the multiplication law.

Marvin is asked to solve the expression (b2) × (b3) × (b7). He claims the result is b12. Is Marvin’s answer correct? Justify your answer.

Are (−2)3 and − 23 equal? Explain.

Are (−2)4 and − 24 equal? Explain.

Generalise a rule for when a negative is raised to an odd or even power.

Problem-solving 12, 13 12

12d–f, 13, 14

Fill in the missing number: a c e

12, 13

a2 × a⬚ = a11

3mc⬚ × ⬚bm⬚ × 2b⬚c3 = −6b4 c5 m7

⬚5 × ⬚6 = b11

c⬚ × c7 = c8

2a2 b⬚ × ⬚a⬚b = 6a4 b3

a⬚b2 c4 × ⬚a⬚b⬚c⬚ =

1 2 3 6 a b c 4

A rectangle has a length a2 cm and width a3 cm. Write an expression for the area of the rectangle.

2.01 Multiplication law mathspace.co

If 2x × 2y = 210, x ≥ y. a

How many positive integer values of x and y are possible for this equation?

Why can we not list the possible solutions if we allow negative integers?

2.02 Division law After this lesson, you will be able to… • apply the division law to simplify expressions involving terms with indices.

Division law Interactive exploration Discover this concept in action online

mathspace.co

Consider the division expression a3 ÷ a5. Notice that the terms share the same base. Writing out the terms in expanded form, it can be seen that we can remove three a‘s as a common factor. a3 5

a× a× a 1 = a× a× a× a× a a× a

We can avoid having to write each expression in expanded form by using the division law.

am ÷ an = am − n where a, m and n are integers and a ≠ 0. When dividing terms with a common base: • The base remains the same. • Subtract the indices. We can also write the division law as:

Example 1 Simplify: a 29 ÷ 24

Create a strategy Use the division law to simplify the expression.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Apply the idea 29 ÷ 24 = 29 − 4 5

Apply the division law Simplify

Reﬂect and check If asked for a basic numeral, we could evaluate 25 = 32.

Create a strategy Use the division law to simplify the expression.

Apply the idea Apply the division law Simplify

Create a strategy Handle the numeric and algebraic components separately. Simplify the numeric component, then use the division law to simplify the algebraic component.

Apply the idea Simplify the numeric component, 15 ÷ 5 = 3

Apply the division law

Simplify

Create a strategy Use the multiplication law to simplify the numerator. Then use the division law to simplify the expression.

2.02 Division law mathspace.co

Apply the idea Apply the multiplication law Simplify Apply the division law Simplify

Reﬂect and check We can also use the division law ﬁrst and then the multiplication law. Use the division law Evaluate the subtraction Use the multiplication law Evaluate the addition

Create a strategy Use the division law twice, once for each base.

Apply the idea Apply the division law for base x Evaluate the subtraction Apply the division law for base y Evaluate the subtraction

Idea summary When dividing terms with the same base, subtract the indices:

If the terms also have coefficients, then divide the coefficient first, keep the base the same and subtract the indices:

Mathspace Victorian Curriculum – Year 9 mathspace.co

2.02 Practice questions Understanding 1

Complete the expression:

Is each statement true or false? a

am ÷ an = am ÷ n

310 ÷ 33

(−8)10 ÷ (−8)3

513 ÷ 512

e 5

a m ÷ bn = a m − n

am ÷ an = am − n

3a–g, 4a–e, 5a–f, 6a–d, 7a–d, 8ab, 9ab

Trailblazer 2

3b–h, 4b–f, 5c–h, 6d–h, 7c–f, 8a–d, 9a–d

3c–l, 4c–h, 5e–l, 6f–l, 7d–h, 8c–h, 9d–h

920 ÷ 99

(−5)8 ÷ (−5)2

77 ÷ 7

(−2)6 ÷ (−2)4

Simplify and evaluate as a basic numeral: a

Ex 1b

Adventurer 2

Simplify: a

Fluency Ex 1a

Explorer 1, 2abef

29 ÷ 25

48 ÷ 45

(−2)11 ÷ (−2)10

b6 ÷ b2

s13 ÷ s4

d20 ÷ d14

Simplify: a

a9 ÷ a5

x7 ÷ x4

c7 ÷ c2

2.02 Division law mathspace.co

Ex 1c

Simplify: a

7 Ex 1d

Ex 1e

8b10 ÷ 2b6

b 30y11 ÷ 6y8

36c8 ÷ 9c3

32t13 ÷ 4t11

15l8 ÷ 3l2

6p13 ÷ 2p10

Simplify:

Simplify: a

f 4 g10 ÷ f 2 g2

m8 p13 ÷ mp3

40h11 j9 ÷ 8h7 j6

10–12

11–13

Simplify:

Reasoning

10, 12

and gets 5d2 as the result. Is Jane’s solution correct?

Jane simpliﬁes the expression Explain your reasoning.

Explain why m5 ÷ z3 is not equal to

Aidan says we can’t evaluate

If m6 = m × m × m × m × m × m then show using the division law why m0 = 1.

Mathspace Victorian Curriculum – Year 9 mathspace.co

. Becky says we can. Who is correct? Explain.

Problem-solving 14a–e, 16, 17 14

14b–f, 16, 17, 18a

14d–h, 15–18

Complete these statements: a

b9 ÷ b⬚ = b5

x⬚ ÷ x3 = x2

Using any of the digits from 1 to 9 at most once, complete the equation.

A rectangle has an area of 8a6 cm2. If one of the sides is of length 2a2 cm, ﬁnd an expression for the other side.

Simplify

Write 3 different expressions using the division law that are all equivalent to: a

4g4

16a2 b12 c3

2.03 Power of a power law After this lesson, you will be able to… • apply the power of a power index law to algebraic expressions. • solve problems that involve multiple index laws, including multiplication and division index laws.

Power of a power law Interactive exploration Discover this concept in action online

mathspace.co

2.03 Power of a power law mathspace.co

Consider (a3)4. 4 times 3 3 (a ) = a × a × a3 × a3 34

= (a × a × a) × (a × a × a) × (a × a × a) × (a × a × a) = a3 × 4 (4 groups of 3 a’s) = a12 We can avoid having to write the expression in expanded form using the power of a power law.

(am)n = amn where a, m and n are integers and a ≠ 0: When simplifying a term with a power that is raised to another power: • The base does not change. • Find the product of the power. Since “powers of powers” involve expressions with brackets, it’s important to remember that every factor inside the brackets is raised to the outside power.

Example 1 Express in simpliﬁed index form: a (25)10

Create a strategy Use the power of a power law to simplify the expression.

Apply the idea (25)10 = 25 × 10 50

Apply the power of a power law Simplify

b (w7)4

Create a strategy Use the power of a power law to simplify the expression.

Apply the idea (w7)4 = w7 × 4 28

Apply the power of a power law Simplify

Mathspace Victorian Curriculum – Year 9 mathspace.co

c (2x2)10

Create a strategy Use the power of a power law to simplify the expression. Remember: the power outside the brackets applies to every factor inside the brackets.

Apply the idea (2x2)10 = 210 × (x2)10 2 × 10

= 1024 × x 20

= 1024x

Raise every factor inside the brackets to the power outside Apply the power of a power law Simplify

d (−3x3)3

Create a strategy Use the power of a power law to expand the brackets. Remember: the power outside the brackets applies to every factor inside the brackets. When dealing with negatives, remember that negatives raised to a positive power will be positive, and negatives raised to a negative power will be negative.

Apply the idea (−3x3)3 = (−3)3 × (x3)3 3×3

= −27 × x

Raise every factor inside the brackets to the power outside Apply the power of a power law

= −27x Simplify

( ) 2

2a  2a2  e  =   5  53 =

23 × a2 × 3

Create a strategy 53

6 a power law to simplify the expression. Use the power 8aof = 125 Remember: the power outside the brackets applies to every factor inside the brackets.

Apply the idea 3

( ) 2

2a  2a2   5  =   53

Raise every factor inside the brackets to the power outside

23 × a2 × 3 Apply the power of a power law = 53 8a6 = Simplify 125

2.03 Power of a power law mathspace.co

Create a strategy Use: • The power of a power law • The multiplication law • The division law

Apply the idea Apply the power of a power law Simplify Apply the multiplication law Apply the division law

Idea summary When a term with a power is raised to another power, multiply the powers: (am)n = amn If the terms also have coeﬃcients, raise the coeﬃcient to the power, keep the base and multiply the indices: (bam)n = bnamn

2.03 Practice questions Understanding 1

( p × p × p) × ( p × p × p)

p 3 × p3

( p × p × p)2

p3 × p2

( p × p × p) × ( p × p)

( p3)2 = p3 × 2

( pm)n = p⬚

True or false: ( p3)2 = p3 + 2

Complete the equation: ( p3)2 = p⬚

Mathspace Victorian Curriculum – Year 9 mathspace.co

Trailblazer 1–3

Adventurer 1–3

Are the expressions equivalent to ( p3)2?

a 3

Explorer 1–3

Fluency Ex 1a

Ex 1b

6 Ex 1c-e

Ex 1f

4a–d, 5a–d, 6a–d, 7abe

4d–h, 5e–h, 6e–l, 7b–e

4g–l, 5i–l, 6i–p, 7d–h

Simplify: a

(23)7

(34)4

(72)9

(53)2

(9 )

(11 )

10 3

(2 )

(512)5

(77)6

((−3)9)3

((−2)7)4

(136)5

Simplify: a

(j3)6

(w2)4

(t4)3

(a7)3

(h7)6

( p2)9

(a5)5

(b4)7

(c10)3

(e11)4

(gx)y

(wa)b

Simplify: a

(ab2)3

(wx3)4

(t2 u3)5

(a7 b2)4

(5c7)2

(ab4 c3)5

m (−2a3)5

(−3a2)3

(32)4 × (33)2

d (2h5)2

(3a2)3

Simplify: a

(25)3 × 27

Reasoning

8, 10

(a3)5 × (a7)4

8–10

Does (w2)3 = (w3)2? Explain.

Find a value of x so that ax × ax = (ax)x. Explain.

A student simpliﬁes the expression

(x2)10 ÷ (x4)2

and gets z4 as the result. Explain the student’s

mistake. What should the correct simpliﬁcation be?

Problem-solving 11a–d, 12 11

11b–e, 12, 13

11c–h, 13, 14

Complete each statement: a

(abc)⬚ = a3 b3 c3

c e g

(⬚m⬚n⬚)3 = 8m6 n9

d (0.1a⬚)2 × (10b4)⬚ = 0.1a6 b4

(4m2)⬚ × (⬚b⬚)2 = 256m6b4

2.03 Power of a power law mathspace.co

Write (16p)4 in the form ab, where a is a prime number.

Write

If the side length of a cube is 5a4, what is the volume?

as a product of prime numbers in index form.

2.04 Zero and negative indices After this lesson, you will be able to… • use index laws to simplify algebraic expressions involving zero and negative indices. • apply zero and negative index laws in real-world situations and problems.

The zero index Consider 25 ÷ 25. Using the division law, we can see that: 25 ÷ 25 = 25−5 = 20 However, we know that if we divide any number by itself, the result is 1. 25 ÷ 25 =

25 25

It follows that: 20 = 1 This is known as the zero index law.

a0 = 1 (provided a ≠ 0)

Example 1 Evaluate (6 × 19)0

Create a strategy Use the zero index law to evaluate the expression.

Apply the idea Everything is to the power of 0 so we can apply the rule to the whole expression. (6 × 19)0 = 1

Mathspace Victorian Curriculum – Year 9 mathspace.co

Example 2 Evaluate 5 × 20 + 3 × 70

Create a strategy Use the zero index law to evaluate the expression. Remember the order of operations for this question: 1. Indices 2. Multiplication 3. Addition In an expression like 5 × 20, the zero index applies to the base of 2, not to the 5.

Apply the idea 5 × 20 + 3 × 70 = 5 × 1 + 3 × 1

Apply the zero index law

=5+3

Evaluate the multiplication

Simplify

Idea summary For any non-zero base a, a0 = 1 That is, raising any non-zero number to the power of 0 will always result in 1.

Negative indices Interactive exploration Discover this concept in action online

mathspace.co

We can use the division law to see that: 25 ÷ 23 = 25 − 3 = 22 When we divide, we subtract the indices. This is true even if subtracting the powers results in a negative. For example: 23 ÷ 25 = 23 − 5 = 2 − 2 Let’s generalise this to a rule. Consider a3 ÷ a5. We can write this in expanded form and remove the common factors.

2.04 Zero and negative indices mathspace.co

a×a×a 1 = a × a × a × a× a a× a Writing this in index form:

This is the negative index law:

A reciprocal is a number that, when multiplied by another number, gives 1. For example the reciprocal 1 1 −n n 1 1 of is 3 because × 3 = 1. Similarly, the reciprocal of a is n , and the reciprocal of − n is a . 3 3 a a We can use this property of reciprocals to help with negative indices. When raising a base to a negative power: • Take the reciprocal of the expression. • Turn the power into a positive. Negative powers are used to represent fractional expressions. Using the equivalence we have just developed, we can also say that:

and in general,

Example 3 Express 6 − 10 with a positive index.

Create a strategy Use the negative index law.

Apply the idea Apply the negative index law

Example 4 Simplify

, giving your answer as a single power of 5.

Create a strategy Use index laws to simplify the expression.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Apply the idea Use the power of a power law

Use the multiplication law

Use the division law

Simplify

Example 5 Simplify, giving your answers with positive indices: a 3q2 × 4q − 5

Create a strategy Apply algebraic conventions and index laws to simplify the expression.

Apply the idea Write in expanded form

Group the numeric and algebraic components together

Multiply the numbers and use the multiplication index law

Use the negative index law

Create a strategy Apply algebraic conventions and index laws to simplify the expression.

Apply the idea Write in expanded form

Divide the numbers by the common factor of 7

Apply the negative index law

Apply the multiplication law

Simplify

2.04 Zero and negative indices mathspace.co

Create a strategy Apply algebraic conventions and index laws to simplify the expression.

Apply the idea Apply the negative index law for fractions Apply the power of a power law Simplify

Idea summary For any base a,

That is, when raising a base to a negative power: • Take the reciprocal of the expression. • Turn the power into a positive. A negative power is used to represent a fractional expression.

2.04 Practice questions Understanding 1

Explorer 1, 2ad, 3, 4, 6

Adventurer 1, 2bd, 3–6

Look at the pattern: 23 = 8

2 =4 1

2 =2

20 = ⬚

2−1 = ⬚

Complete the sentence: Each time the power of 2 decreases by 1, the number on the right is divided by ⬚.

Complete the pattern.

13 − 10

Identify the base and the power. a

Trailblazer 1, 2cd, 3–6

10 − 7

2−4

Mathspace Victorian Curriculum – Year 9 mathspace.co

(−5) − 8

, where a and b are non-zero numbers?

What is the value of A

It’s a mathematical convention without a reason.

When you divide any number by itself, you get 1, and x0 can be thought of as

Zero has no value, so it makes the number disappear.

Because 1 is the multiplicative identity.

The fraction gets inverted.

The fraction becomes a decimal.

The numerator and denominator both become negative.

The fraction becomes zero.

Is 2 − 3 less than or greater than 1? 7a–h, 8, 9ab, 10a–c, 11a–d, 12a–h, 13a–f, 14a–d, 15a–e, 16a–d, 17abij, 18abe, 19ae

7e–l, 8, 9bc, 10c–e, 11c–f, 12e–l, 13b–h, 14c–f, 15e–j, 16c–h, 17c–l, 18b–g, 19a–e, 20abe

7i–p, 8c, 9c, 10e–h, 11d–h, 12i–p, 13e–l, 14e–h, 15e–l, 16e–l, 17e–p, 18e–l, 19c–h, 20

Evaluate: a

33 × 30

(−3)0

−40

−1000

−9 (12 × 15)0

72 ÷ 72

25 × 110

(6 × 19)0

7 (14 × 18)0

Complete: a

When you have a fraction raised to a negative index, what happens?

m 8

Which statement explains why any non-zero number to the power of 0 is 1?

Fluency

Ex 1

7⬚ = 1

Complete the table: a

25 32

105 100 000

33 27

24 16

32 9

2−1

104 10 000

103

102

101

100

3−1

3−2

3−3

10 − 1

2.04 Zero and negative indices mathspace.co

Ex 2

Ex 3

Evaluate: a

22 + 2 + 20

7 × 30 + 32

(2 + 20) × 20

4 × 3 − 2 + 80

6 ÷ 2−1 + 4

(2 + 3 − 2) × 2 − 1

17 − 6

9−1

−8 − 11

Express with a positive index: a e

Ex 4

8 − 15 (−9)

−7

a−9

7 −6 (−12)

−8

−5

−45

p−2

3x − 4

7x − 9

k − 2 q3

2x − 8 y3 m − 5 r − 4 p4

8p − 3

m 3a4 b − 5 c − 7

4 − 2 k − 3 l7

Express with a negative index: a

Complete:

Simplify, leaving your answer in index form: a

511 ÷ 5 − 3

102 ÷ 103

2−3 × 2−4

3−4 ÷ 3−9

90 × 9 − 12

813 ÷ 8 − 9

7 − 7 ÷ 75

10025 ÷ 10026

(63 )5 × 62

i 16

Express with positive indices: a

5 × 50 − 3 × 20

640

Evaluate: a

67 × 6 − 7

48 × 4 − 6

3 − 8 × 311

55 × 5 − 7

4−4 × 2−4

1213 ÷ 127 ÷ 128

5−2 × 3−2

(−4)3 × (−4) − 7

712 ÷ 7 − 7 ÷ 717

Mathspace Victorian Curriculum – Year 9 mathspace.co

17 Ex 5a-b

Simplify, giving your answers with positive indices: a e

Ex 5c

× 4h

7a4 × 4a − 2 3y

−2

× 2y

5x4 × (−3x − 8)

−5

−4y × (−4y 12x5 ÷ 4x − 3

−5

)

3y − 2 × 4y − 3

(5mp)2 × mp − 2

9c6 ÷ 6c8

Simplify, giving your answers with positive indices: (2m) − 3

(4m − 6)4

(3p − 4) − 2

(3y2) − 2

Simplify: a

2y6 × 4y7 × 4y − 5

6y7 × 2y − 5 × 5y3

2x − 2 × 5x5 × 4x − 5

12x − 9 × 2x4 × 3x − 2

4y3 × 3y8 ÷ 2y − 1

−8y10 ÷ 2y − 4 ÷ y3

14x − 8 ÷ 2x × 3x6

(4x3) − 1 × ( y − 2) − 3

g 20

−4

5y9 × 4y − 3

40x

−2

÷ 5x ÷ 4x

−9

Simplify: a

(3x − 4 × 2x2)2

(10x3 y − 4)2

(10x3)3 × (2x4 y3) − 2

Reasoning

21, 22, 24

21–24

What is 04 equal to?

21, 23–25

Explain why 0 − 4 is undeﬁned.

Explain the difference between 2x0 and (2x)0.

When asked to ﬁnd an expression that is equivalent to −20, a student responded 1. Is this answer correct? Explain why or why not.

A student was writing 5a − 1 without negative indices and wrote working is incorrect, and write the correct answer.

Explain why the expression x − n is the same as

. Explain why their

. Does this hold true for negative values of x?

2.04 Zero and negative indices mathspace.co

Problem-solving 26, 27 26

26–28

26–30

Complete each statement: a c e

a⬚b3 × ⬚a2 b⬚ = 2ab − 1

b d

(gh2) ⬚g⬚ = g3 h − 8

Simplify, leaving your answers with positive indices:

(5–1 + 3–1)–1

If n = 2, determine the value of

If the surface area of a cube is 6s2, where s is the length of a side in centimetres, ﬁnd the surface area when s = 3 − 1 cm (without using a calculator).

Without using a calculator, what is the simpliﬁed form of (4 − 1 × 23)2?

2.05 Scientific notation After this lesson, you will be able to… • write numbers in scientific notation form. • perform mathematical operations using numbers in scientific notation. • understand the use of metric prefixes and convert measurements using scientific notation.

Scientific notation Scientific notation, sometimes referred to as standard form, is a compact way of writing very large or very small numbers. For example, • The sun has a mass of approximately 1.988 × 1030 kg, which is much easier to write than 1988 000 000 000 000 000 000 000 000 000 kg. • The mass of an atom of Uranium (one of the heaviest elements) is approximately 3.95 × 10 − 22 g. That is 0.000 000 000 000 000 000 000 395 g. It is called scientific notation because it is often used in scientific contexts. 48

Mathspace Victorian Curriculum – Year 9 mathspace.co

Interactive exploration Discover this concept in action online

mathspace.co

In scientiﬁc notation, numbers are written in the form: a × 10n where a is a decimal number between 1 and 10 and n is an integer (positive, negative, or zero). Note that 1 ≤ a < 10, so a may be equal to 1, but cannot be 10. • A negative index indicates how many factors of ten smaller than a the value is. • A positive index indicates how many factors of ten larger than a the value is. • A index of zero indicates that the value is a because 100 = 1. There are lots of ways to express numbers using powers of ten. The decimal 0.032 can be written as: • • •

(scientiﬁc notation)

Example 1 Express each number in scientiﬁc notation: a 0.000 347

Create a strategy Use the scientiﬁc notation form a × 10n, where a is a decimal number between 1 and 10 and n is a negative integer.

Apply the idea To find the first part of our scientific notation we place the decimal point after the first non-zero digit, so a = 3.47 0.000 347 is 10 000 or 104 times smaller than 3.47. So: 0.000 347 = 3.47 × 10 − 4

b 84 245 000

Create a strategy Use the scientiﬁc notation form a × 10n, where a is a decimal number between 1 and 10 and n is a positive integer.

2.05 Scientific notation mathspace.co

Apply the idea To find the first part of our scientific notation we place the decimal point after the first non-zero digit, so: a = 8.4245 84 245 000 is 10 000 000 or 107 times bigger than 8.4245. So: 84 245 000 = 8.4245 × 107

Example 2 Convert the following from scientiﬁc notation to ordinary numbers: a 2 × 107

Create a strategy Use the fact that 107 = 10 000 000.

Apply the idea 2 × 107 = 2 × 10 000 000 = 20 000 000

Rewrite 107 Evaluate

b 3.62 × 10−4

Create a strategy Use the fact that 10−4 =

= 0.000 1.

Apply the idea 3.62 × 10−4 = 3.62 × 0.000 1 = 0.000 362

Evaluate 10−4 Evaluate

Idea summary In scientiﬁc notation, numbers are written in the form: a × 10n where a is a decimal number (1 ≤ a < 10) and n is an integer (positive, negative or zero). • A negative index indicates how many factors of ten smaller than a the value is. • A positive index indicates how many factors of ten larger than a the value is. • A index of zero indicates that the value is a because 100 = 1.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Operations with scientific notation Just like ordinary numbers, we can add, subtract, multiply, and divide numbers written in scientiﬁc notation. To multiply a × 10m and b × 10n, we multiply a and b and add the powers of 10 (using the index laws): (a × 10m) × (b × 10n) = (a × b) × 10(m+n) To divide a × 10m by b × 10n, we divide a by b, and subtract the second power of 10 from the ﬁrst (following the index law on division):

When multiplying and dividing, it can happen that the answer is not in scientiﬁc notation, so we need to convert it. When adding and subtracting numbers in scientiﬁc notation, it is best to convert the numbers into decimal numbers.

Example 3 Simplify, giving your answers in scientiﬁc notation. a (6 × 107) × (4 × 10−2)

Create a strategy Multiply the number parts, multiply the power parts, then adjust the answer if necessary to ensure that it is in scientiﬁc notation.

Apply the idea (6 × 107) × (4 × 10−2) = 6 × 4 × 107 × 10−2 Remove the brackets and reorder the parts = 24 × 105 1

Multiply the numbers. Use the multiplication law 5

= 2.4 × 10 × 10 6

= 2.4 × 10

Express 24 in scientiﬁc notation onvert to scientiﬁc notation using the C multiplication law

b (2 × 102) ÷ (8 × 105)

Create a strategy When dividing numbers in scientific notation, divide the first number by the second number, divide the first power part by the second power part, then adjust the answer if necessary to ensure that it is in scientific notation.

2.05 Scientific notation mathspace.co

Apply the idea (2 × 102) ÷ (8 × 105) = (2 ÷ 8) × (102 ÷ 105) = 0.25 × 10−3 −1

Use the division law −3

= 2.5 × 10 × 10 = 2.5 × 10

Divide the number parts and divide the power parts

−4

Express 0.25 in scientiﬁc notation Convert to scientiﬁc notation using the multiplication law

c (3.2 × 10−2) + (2.3 × 10−3)

Create a strategy When adding or subtracting numbers given in scientiﬁc notation, it is often best to convert them to ordinary numbers.

Apply the idea (3.2 × 10−2) + (2.3 × 10−3) = 0.032 + 0.0023 = 0.0343

Express both numbers as ordinary decimals Add the decimals

−2

= 3.43 × 10

Express 0.0343 as scientiﬁc notation

Idea summary When multiplying and dividing numbers in scientiﬁc notation:

When adding and subtracting numbers in scientiﬁc notation, it is often best to convert the numbers into ordinary numbers.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Metric prefixes Another way to write very large or very small measurements is using prefixes. We add prefixes to a unit to make the unit bigger or smaller by a power of ten. The metric prefixes kilo-, centi- and milli- are often used for lengths and weights. For example, one kilometre means 1 × 103 metres. Five milligrams means 5 × 10−3 grams. You may also be familiar with the prefixes tera-, giga- and mega- which are commonly used when discussing digital data. For example, one terabyte means 1 × 1012 bytes. Below are some commonly used metric prefixes: Prefix teragigamegakilohectodecadecicentimillimicronanopico-

Letter T G M k h da d c m μ n p

Size 1012 109 106 103 102 101 10−1 10−2 10−3 10−6 10−9 10−12

Example 4 Convert the measurements to the indicated unit, expressing your answer in scientiﬁc notation. a 52 nanoseconds to seconds.

Create a strategy Use the table showing the sizes of different metric preﬁxes.

Apply the idea 52 nanoseconds = 52 × 10−9 seconds Multiply by the size of a nanosecond to convert to seconds = 5.2 × 101 × 10−9 seconds −8

= 5.2 × 10

seconds

Write 52 in scientiﬁc notation Simplify

2.05 Scientific notation mathspace.co

b 0.07 Mb to bytes.

Create a strategy Use the table showing the sizes of different metric preﬁxes.

Apply the idea Mb means megabytes. 0.07 megabytes = 0.07 × 106 bytes

Multiply by the size of a megabyte to convert to bytes

= 7 × 10−2 × 106 bytes

Write 0.07 in scientiﬁc notation

= 7 × 10 bytes

Simplify

Idea summary We can use metric preﬁxes to write very large and very small measurements. Preﬁx teragigamegakilohectodeca-

Letter T G M k h da

Size 1012 109 106 103 102 101

Preﬁx decicentimillimicronanopico-

Letter d c m μ n p

Size 10−1 10−2 10 − 3 10 − 6 10 − 9 10 − 12

2.05 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

Choose the correct statement. In scientiﬁc notation, numbers are written in the form a × 10n where: A

a is a decimal number between 1 and 10 (not equal to 10) and n is a positive integer

a is a decimal number between 1 and 10 (not equal to 10) and n is any integer

a is any decimal number and n is any integer

Which is written correctly using scientiﬁc notation? A

a is an integer between 1 and 10 (not equal to 10) and n is any decimal

8.56 × 1014

0.658 × 103

Mathspace Victorian Curriculum – Year 9 mathspace.co

608.56 × 1014

465.53 × 1021

Simplify using index laws: a

106 × 105

4ace, 5a–e, 6a–e, 7a–e, 8a–e, 9, 10ab, 12–14, 15a–c

Fluency

Ex 2

4bdf, 5c–h, 6c–h, 7c–h, 8c–h, 10c–f, 11, 15c–h

−4 d • 9.37 × 10 • 6 × 10 − 4 • 6 × 10 − 9

• 4.7 × 102 × 103 • 4.7 × 10 − 8 × 108 • 4.7 × 109 × 108

128

2000

884 000

84 626 000

23 456

6.14

9 700 000 000

0.000 74

0.000 059 32

Express in scientiﬁc notation: a

0.6

0.000 676

0.002

Express as ordinary numbers: a

3 × 109

4.13 × 104

9 × 103

8.97 × 105

5.03 × 105

3.014 × 103

8.0045 × 103

1.3008 × 107

4.7 × 10 − 6

2.13 × 10 − 7

6.4 × 10 − 5

Express as ordinary numbers: 5.28 × 10 − 2 0

8.29 × 10

7 × 10 − 3

3.47 × 10

−5

−2

9.72 × 10

⬚ = 7.3 × 10 − 4

Complete: a

0.002 = ⬚ × 10 − 3

⬚

4000 = ⬚ × 103

89 300 000 = 8.93 × 10

⬚ = 7.34 × 105

(3 × 104) × (2 × 105)

(4 × 10 − 3) × (2 × 10 − 2)

(6 × 108) × (3 × 10 − 3)

(8 × 107) ÷ (2 × 102)

(6 × 10 − 1) ÷ (2 × 10 − 4)

(3 × 10 − 1) ÷ (6 × 103)

c e 10

4bde, 5b–f, 6b–f, 7b–f, 8b–f, 10a–d, 11, 12, 14, 15a–f

Ex 3

Express in scientiﬁc notation:

For each set, indicate the largest and the smallest number. −7 −9 −2 a • 7.27 × 10 b • 7.31 × 10 c • 1.25 × 10 • 2.22 × 10 − 7 • 5.13 × 10 − 9 • 1.25 × 10 − 6 −7 −9 • 4.76 × 10 • 8.31 × 10 • 1.25 × 102 −2 e • 2.93 × 10 2 • 7.88 × 10 • 7.88 × 10 − 2

Ex 1

10 − 3 × 10 − 5

0.000 067 = 6.7 × 10⬚

Calculate, expressing your answers in scientiﬁc notation:

2.05 Scientific notation mathspace.co

Calculate, expressing your answers in scientiﬁc notation: a c

Ex 4

3 × 104 + 8 × 105 8

5.74 × 10 − 3.5 × 10

2.1 × 10 − 3 + 4 × 10 − 2

8.2 × 10 − 4 − 7.1 × 10 − 5

The thickness of coating on a camera lens is 0.000 002 metres. Express the thickness in scientiﬁc notation.

The distance between two stars is approximately 5 × 106 metres. Express this distance as a whole number.

The world’s oceans hold approximately 712 000 000 000 000 000 000 litres of water. Express this volume of water in scientiﬁc notation.

Use the table of preﬁxes to convert the measurements to the indicated unit, giving your answer in scientiﬁc notation. a

3.8 milliseconds to seconds

1.67 kL to L

340 pm to m

0.5 decibels to bels

Reasoning

16, 18

Preﬁx Letter Size teraT 1012 gigaG 109 0.022 µg to g megaM 106 63 ML to L kilok 103 0.8 terabytes to bytes h 102 1200 hectopascals to pascals hectodecada 101 decid 10 − 1 centic 10 − 2 millim 10 − 3 μ micro10 − 6 nanon 10 − 9 picop 10 − 12 16, 18

17, 18

Explain why a scientist would want to express a very large or small number in scientiﬁc notation.

Evaluate, rounding to 2 decimal places: i

4.75 × 107 + 1.35 × 103

4.75 × 107 − 1.35 × 103

What do you notice about your answers to part (a)? Why did this happen?

Explain why each of these numbers is not correctly expressed in scientiﬁc notation. a

23.6 × 10 − 3

3.45 × 100.5

Problem-solving 19, 20 19

0.0006 × 102

20–22

9.73 × 82

19–22

Bone marrow in a person’s body produces 2.3 × 106 red blood cells each second. a

Express the amount of red blood cells as an ordinary number.

How many red blood cells are produced each minute? Express your answer in scientiﬁc notation.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Sound travels at a speed of approximately 0.3 kilometres per second, while light travels at a speed of approximately 300 000 kilometres per second. a

Express the speed of sound in kilometres per second in scientiﬁc notation.

Express the speed of light in kilometres per second in scientiﬁc notation.

How many times faster does light travel than sound?

The mass of one helium atom is 6.7 × 10 − 24 g, and the mass of one iron atom is 9.3 × 10 − 23 g. a

Find the difference between the masses of the two atoms. Express your answer in scientiﬁc notation.

What percentage of the mass of an iron atom is a helium atom? Give your answer to one decimal place.

An experiment produces 3.14 × 10 − 1 g of a substance. Express this mass in kilograms.

The track left by a particle in an experiment is 2.76 × 10 − 5 m long. Express this distance in centimetres.

Did you know?

Bacteria can multiply incredibly fast due to the power of indices! In ideal conditions, some bacteria can double their population every 20 minutes. This means that if you start with just one bacterium, after 3 hours, you’d have over 500 bacteria! This exponential growth, modeled by indices, explains why food can spoil so quickly if left out. It’s the same principle scientists use to predict the spread of infections or to grow bacteria for research. Understanding indices helps us track these rapid changes and manage bacterial growth in everything from medicine to food safety!

2.05 Scientific notation mathspace.co

2 Chapter review 1

Simplify, leaving your answer in index form: a

73 × 75

(−2)2 × (−2)4

38 × 3h

b6 × b × b4

x12 × 5x3

e 2

c ×c

−5

Simplify and evaluate as a basic numeral: a

47 ÷ 44

(13)5

(97)2

52 × 50

a− 5

8a7 ÷ 4a6

((−2)2)2

( j13)2

(xy2 z3)5

−70

8 (13 × 17)0

4y−7

5−3 × 5−3

(3m)−2

2y3 × 4y−7

6x − 9 ÷ 2x × 7x7

(c3)5 ÷ (c8)7

83 ÷ 811

16x−7 ÷ 8x−3

Simplify:

Evaluate: a

Simplify, leaving your answers with positive indices:

(−5)20 ÷ (−5)t

Express with positive indices: a

Evaluate: a

Simplify, leaving your answer in index form: a

Simplify and evaluate as a basic numeral: a

(−6)11 ÷ (−6)9

Simplify, leaving your answer in index form: a

4 × 42 × 4 − 1

Mathspace Victorian Curriculum – Year 9 mathspace.co

3 − 2 + 8 × 70

Express in scientiﬁc notation: a

113 000

0.8

0.000 89

1.4 × 100

4 × 10 − 3

Express as a basic numeral: a

6 × 107

3.85 × 104

3.36 × 10 − 2

3.005 × 10 − 6

Calculate, expressing your answers in scientiﬁc notation: a

(2 × 104) × (4 × 105)

(3 × 10 − 3) × (2 × 10 − 4)

(6 × 107) ÷ (3 × 102)

(9 × 10 − 1) ÷ (3 × 107)

Calculate, expressing your answers in scientiﬁc notation: a

234

3.87 × 106 − 3.6 × 105

3.5 × 10 − 3 + 2 × 10 − 2

Use the table of preﬁxes to convert the measurements to the indicated unit, giving your answer in scientiﬁc notation. a

5.1 milliseconds to seconds

3.04 km to m

0.05 terabytes to bytes

0.089 µg to g

23 decibels to bels

Preﬁx teragigamegakilohectodecadecicentimillimicronanopico-

Letter T G M k h da d c m μ n p

Size 1012 109 106 103 102 101 10 − 1 10 − 2 10 − 3 10 − 6 10 − 9 10 − 12

A swimming pool has a length of a4 m, width of a3 m and depth of a m. What is the volume of the pool in cubic metres?

The population of Melbourne is P6 and the population of Margaret River is P2. How many times more populous, is Melbourne compared to Margaret River?

Write (27m)5 in the form ab, where a is a prime number.

Without using a calculator, what is the simpliﬁed form of: (165 ÷ 642) − 1 + 3 × 30

A right-angled triangle has a hypotenuse of 5 × 104 mm and a short side of 4 × 104 mm. Calculate the length of the third side, leaving your answer in scientiﬁc notation.

Chapter 2 review mathspace.co

Big ideas Simplifying, expanding, and factorising are interconnected processes that provide different ways to understand and manipulate algebraic expressions.

Curriculum Outcomes VC2M9A02 simplify algebraic expressions, apply the distributive law to expand algebraic expressions including binomial products, and factorise monic quadratic expressions VC2M9A06 use mathematical modelling to solve applied problems involving change, including financial contexts involving simple interest; formulate problems, choosing to use either linear or quadratic functions or other simple variations; interpret solutions in terms of the context; evaluate the model and report methods and findings

3 Algebra Chapter outline 3.01 3.02 3.03 3.04 3.05

Algebra review Distributive law Binomial expansion Algebraic factors Factorise by grouping Investigation: Calendar expressions Chapter 3 review

62 71 75 81 87 92 94

Algebra simplifications are just like the abbreviations we use in everyday life like BRB (Be Right Back), LOL (Laughing Out Loud) and MSR (Mathspace Rules).

3.01 Algebra review After this lesson, you will be able to… • review adding and subtracting algebraic terms, focusing on collecting like terms. • review multiplying and dividing algebraic terms, highlighting how indices may change. • apply the order of operations in the context of simplifying algebraic expressions.

Algebra review When working with algebraic terms, it can be helpful to remember some vocabulary. Index (power)

3 x Coefficient

Constant

+ 11

Variable

5x2 y + 3xy − 2xy − 8 Terms:

5x2 y, 3xy, −2xy, −8

Like terms: 3xy, and −2xy Add and subtract algebraic terms When adding and subtracting algebraic terms, we collect like terms by adding or subtracting their coeﬃcients, for example 2a + 3a = 5a Multiply and divide algebraic terms When multiplying and dividing algebraic terms, terms do not have to be like terms. We multiply or divide the coeﬃcients and may use index laws, for example 8a4 ÷ 2a = 4a3 When simplifying algebraic expressions we always follow the order of operations.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Example 1 Simplify: a 9a + 3b + 2a − 4b

Create a strategy There are two pairs of like terms, 9a and 2a, and 3b and −4b. Like terms can be collected together.

Apply the idea 9a + 3b + 2a − 4b = 9a + 2a + 3b − 4b

Group like terms

= 11a – b

Combine like terms

Reﬂect and check We can identify two types of terms, terms in a and terms in b. Our solution has two terms, one in a and one in b.

b 5wy2 − 7w2 y + wy − 2w2 y + wy2

Create a strategy wy2 = w × y × y and w2 y = w × w × y, so terms in wy2 and terms in w2 y are unlike. Here: • 5wy2 and wy2 are like • −7w2 y and −2w2 y are like • wy does not have a like term, so it will not change

Apply the idea 5wy2 − 7w2 y + wy − 2w2 y + wy2 = 5wy2 + wy2 − 7w2 y − 2w2 y + wy 2

Group like terms

= 6wy − 9wy + wy Combine like terms

Reﬂect and check The original statement had three different types of terms: wy2, w2 y, and wy Our solution has three terms, one in wy2, one in w2 y, and one in wy.

3.01 Algebra review mathspace.co

Example 2 Simplify: a 5r × (−7)

Create a strategy Multiply the numeric components and then simplify the algebraic components.

Apply the idea 5r × (−7) = 5 × (−7) × r Group the numeric and algebraic components together = −35 × r

Multiply the numbers

= −35r

Simplify

b 3c2 × 5d3

Create a strategy Multiply the numeric components and then simplify the algebraic components.

Apply the idea 3c2 × 5d3 = 3 × 5 × c2 × d3 2

= 15 × c × d 2 3

= 15c d

Group the numeric and algebraic components together Multiply the numbers Simplify

Reﬂect and check Since c2 and d3 have different bases, we cannot use the index laws to simplify them.

c 8ab2 × 3abc

Create a strategy Multiply the numeric components and then simplify the algebraic components.

Apply the idea 8ab2 × 3abc = 8 × 3 × a × a × b × b × b × c Group the numeric and algebraic components together = 24 × a2 × b3 × c Multiply the numbers, and then the variables in turn = 24a2 b3 c

Mathspace Victorian Curriculum – Year 9 mathspace.co

Simplify

Reflect and check Looking at the original statement: • The coefficients are 8 and 3 and 8 × 3 = 24 • The first term has an a and the second term has an a, and a × a = a2 • The first term has a b2 and the second term has a b, and b2 × b = b3 • There is only one c

Example 3 Simplify: a 30m ÷ 5m

Create a strategy Remove common factors from the numeric components and then simplify the algebraic components. Often it is easier to work with algebraic division if it is written as a fraction.

Apply the idea Write as a fraction

Remove the common factor of 5

Simplify the algebraic terms

Reﬂect and check • Considering the coeﬃcients, 30 ÷ 5 = 6 • Considering the algebraic part, m ÷ m = 1

Create a strategy Remove common factors from the numeric components and then simplify the algebraic components. Often it is easier to work with algebraic division if it is written as a fraction.

3.01 Algebra review mathspace.co

Apply the idea Introduce multiplication signs and bring out highest common factor (HCF) of numbers

Remove the common factor, 5

Remove the common factor, w

Remove the common factor, x

Remove the unnecessary multiplication signs

Reﬂect and check We can think about working on each factor separately: , then

, and then

Create a strategy Remove common factors from the numeric components and then simplify the algebraic components. Often it is easier to work with algebraic division if it is written as a fraction.

Apply the idea Reintroduce the multiplication signs Remove the common factors of 3, a and b

Group the numeric and algebraic components together

Simplify

Reﬂect and check We could use the approach we use with index laws:

Use multiplication law in numerator

Divide coeﬃcients and use division law

Simplify

Mathspace Victorian Curriculum – Year 9 mathspace.co

Example 4 Simplify: a 3a2 + 5a × 7a

Create a strategy Use the order of operations and what we know about algebraic terms.

Apply the idea 3a2 + 5a × 7a = 3a2 + 35a2 = 38a

Perform the multiplication Add the like terms

Reﬂect and check The multiplication produces like terms that can be collected together, otherwise, we would not be able to simplify further.

b (7a − 2a) × 3ab

Create a strategy Use the order of operations and what we know about algebraic terms.

Apply the idea (7a − 2a) × 3ab = 5a × 3ab 2

= 15a b

Collect the like terms in the brackets Perform the multiplication

Create a strategy Use the order of operations and what we know about algebraic terms.

Apply the idea Collect the like terms in the numerator and perform the multiplication in the denominator

Divide out the common factor of 15

Introduce multiplication signs

Divide out the common x and simplify.

3.01 Algebra review mathspace.co

Reﬂect and check We could also use the division and negative index laws: Collect the like terms in the numerator and perform the multiplication in the denominator Divide out the common factor of 15 Division index law Negative index law

Idea summary Simplifying algebraic expressions makes use of: • Adding and subtracting algebraic terms, which may use like terms • Multiplying and dividing algebraic terms, which may use index laws • Applying the order of operations

3.01 Practice questions Understanding 1

Explorer

Adventurer

1, 2

Trailblazer 1, 2

For the expression: 3x2 − xy + 5x − 3y − 9 + 5y

How many terms are there?

What are the variables?

What is the coeﬃcient of x?

What is the coeﬃcient of xy?

What is the constant term?

Identify any like terms.

Are these statements true or false? a

9xyz = 9zxy

−4 × 5 × a × b = −9ab

p5 = p × p × p × p × p

2x2 × 5xy cannot be simpliﬁed because they are not like terms.

c8 ÷ c2 = c4

x÷2=

2x + 3y = 5xy

Mathspace Victorian Curriculum – Year 9 mathspace.co

3a–cij, 4a–j, 5a–j, 6a–cgh, 7a–g

Fluency Ex 1

5x + 8x

10y − 7y

7c + 3c − c

3x − 4x − x

6c + 3d + 3c − d

7x + 4y − 2x + y

2x + 7y + 3x + z

3x + 7 − 2y + x

x2 + 6x + 5x + 10

5ab + 2bc − 2ab + 3cb

5a2 b + 7ab2 + 2a2 b − 3ab2

x − 4x + 4x − 16

m 5x − (−3x) + 7 + (−2)

xy + 2x2 y − 7 − 2xy2 + 3xy

Ex 3c

3ab + 9cd − (−2ab) − (−cd) x − (−3x) − 8x + x2

9 × 5u

7r × 4

2 × x × y × 3

4r × 6s

3u3 × 5v5

12p × 5q2

(−3c) × (−5d)

5e × (−5f )

(−3x) × (−10y)

6x2 × 2y2 × 3z2

9a × 3a × (−2a)

5a × 2b × a × 2b

3w3 x2 y × (−3)wx2 y3 × 2xy

−3e7 f 2 × 5e6 f 3 × e × f 2 15x6 ÷ 5x6

Ex 3a, 3b

n p

Simplify:

3 2

4 3

5 3

m 5a b × 4a b

3f–p, 4h–p, 5h–p, 6c–l, 7e–p

Simplify:

o Ex 2

3d–l, 4e–m, 5c–j, 6c–j, 7e–l

7c d × 3c × 2d

Simplify: a

6m ÷ 36

10ab ÷ 50ab

(9cd) ÷ 12de

(−12fg) ÷ (−3g2)

(−32p7) ÷ 8p7

Simplify:

5u × 8v ÷ 4w

(−3a) × (−5b) ÷ 10a2

11xy × 5z ÷ x2 y2

25g ÷ 5 × 7f

(8y ÷ 4x) × (12y ÷ 3x)

(9a × 4b) ÷ (5 × 2c)

3.01 Algebra review mathspace.co

Ex 4

Simplify: b

(19t + 5t) ÷ 4u

(12a − 4a) ÷ (8 × 4a )

5f − 3 × 4f + 7f

2 (4x2 + x × 2x) ÷ 16x

5 (3a × 2b) + 4a × (−3b)

(7c − 3c) × 2c

3x5 × 7y4 − 2x3 y2 × 3x2 y2

Reasoning 8

8, 9

p 9–11

10–12

Explain why we cannot combine 3a + 4b as 7ab.

Is it ever possible that 3a + 4b = 7ab? Explain your reasoning.

Explain why x2 and x are not like terms.

Is there any value of a for which 3a × 2a = 5a2? If so, give an example. If not, explain why.

Simplify the expression 7a + 5a × 2 in two different ways: • Perform the multiplication ﬁrst and simplify. • Repeat the problem and do the addition ﬁrst. Explain why an order of operations is needed when simplifying algebraic expressions.

Consider the problem

Identify and correct the error in each student’s work: Lawrence:

Marika:

Problem-solving 13–15, 17

14–16, 18

15–19

An isosceles triangle has 2 sides of length 3b cm. If the third side has a length of 2c cm, write a simpliﬁed expression for the perimeter of the triangle.

A rectangle has a length of (3x + y) cm and a width of (2x − 2y) cm. Find the perimeter of the rectangle in terms of x and y.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Addition pyramids work by the value in a box being equal to the sum of the two boxes below. The example shows how this works with numbers: 13 9 4

4 −1

Complete the addition pyramids. a

3x + 2y

4a + b a + 2b

3a − b

5x − 2y

4a + 3b

3x − y

A rectangle has a length 4 times its width. The width of the rectangle is 3a cm. Find an expression for the area of the rectangle.

A purse contains x $10 notes, y $5 notes, and z $1 coins. Write an expression for the number of dollars in the purse.

Emma bakes x batches of chocolate chip cookies, with y cookies in each batch.

How many chocolate chip cookies does Emma bake in total?

If there are an average of y chocolate chips in each cookie, and she bakes the same number of cookies on 3 consecutive days, how many chocolate chips does she use?

A piece of paper with dimensions 8.5 centimetres by 11 centimetres will have a square with length x cut out from each corner. Write and simplify an expression that can be used to represent the perimeter of the shape formed once the corners have been removed.

3.02 Distributive law After this lesson, you will be able to… • review and apply the distributive law for both numerical and algebraic expressions. • use algebra tiles to illustrate the distributive law and create equivalent mathematical statements. • expand and simplify real-world algebraic expressions using the distributive law. • apply the distributive law to expand algebraic terms with brackets.

3.02 Distributive law mathspace.co

The distributive law Interactive exploration

Discover this concept in action online

mathspace.co

We use the distributive law to expand algebraic terms with brackets. The term on the outside of the brackets is multiplied by each term inside the brackets. A(B + C) = AB + AC When multiplying terms with indices, we may need to use the multiplication index law: am × an = am+ n

Example 1 Expand: a 6s (9t + 8)

Create a strategy Use the distributive law A (B + C) = AB + AC to expand the expression.

Apply the idea 6s (9t + 8) = 6s × 9t + 6s × 8 = 54st + 48s

Expand the expression Simplify

Reﬂect and check We can use arcs to help us remember to multiply each term inside the brackets by the factor outside of the brackets.

b 4u7 (4u7 + 2u6)

Create a strategy Use the distributive law A (B + C) = AB + AC to expand the expression. We will also use the multiplication index law am × an = am + n.

Apply the idea 4u7 (4u7 + 2u6) = 4u7 × 4u7 + 4u7 × 2u6 14

= 16u + 8u

Expand the expression Simplify

Reﬂect and check Remember that we multiply the coeﬃcients and add the indices for each product.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Example 2 Expand and simplify: x (x − 6) − 8 (x + 7).

Create a strategy Use the distributive law A(B + C) = AB + AC to expand the expression.

Apply the idea x (x − 6) − 8 (x + 7) = x2 − 6x − 8x – 56

Expand both brackets

= x − 14x – 56

Simplify

Reflect and check When the factor outside of the brackets is negative, we need to multiply each term in the brackets by the negative.

Idea summary Expand means to write an algebraic expression without brackets. We can use the distributive law to expand an algebraic expression with brackets: A(B + C) = AB + AC and if the second term in the brackets is negative: A (B − C) = AB − AC

3.02 Practice questions Understanding

Explorer 1–3a

Adventurer 1–3

Trailblazer 1–3

Ludivina states that (a3)(a3) = a9. Dortha states that (a3)(a3) = a6. Who is correct and explain why.

Explain why 2(a − 3) is the same as (a − 3) × 2.

Insert (=) or (≠) in each statement. a b c

3a(2a + b) ⬚ 3a × 2a + 3a × b

6m2 v(3 − 2m2 + v) ⬚ 6m2 v × 3 − 6m2 v × 2m2 + 6m2 v × v 4ab(3b + 2c − 6a) ⬚ 4ab × 3 + 4ab × c − 4ab × 6

3.02 Distributive law mathspace.co

4, 5a–i, 6a–ce, 7ac, 8a–df

Fluency 4

c 5

3 (2x + 4) = ⬚x + 12

a(b + 7)

m(m + 7)

y(1 + y)

4 (m2 + 10)

m(m − 3)

y(4y − 10)

m(8 − m)

5t(t + 14)

2b(2b + 16)

−5a(3a − 4) 3

−w(w + 1)

−m(11n + 19p)

7s (8t − 9)

6wy( y + w)

(5y − 6) (−3y2)

6a4 (3a6 + 6a4)

3u4 (8u + 10u3)

Expand: a

7y( y2 + 4)

−3x2 (5x − 2x4)

Expand: 2f ( f − 8g − 7h)

6a(a − 5b + 3c)

(5x + 9y − 2) (4x)

2a2 (a2 + 4a − 7)

Expand and simplify: a

8 ( y + 6) + 3 ( y + 9)

h(h + 5) + 9 (h + 9)

x(x + 4) + 2 (x + 5)

3c(9c2 + 2c + 7)

4c(8c2 − 3c − 5)

8y2 − 7y + 9y( y − 5)

4y2 − 9y + 6y( y − 8)

5 (9x − 6y) − 3(8y + 2x)

9x(6x + 8y) − 7x(2y + 4x)

2 (4 (x − 3) + 5)

3(2(x – 4) + 5)

6u(5u + 8v) − 7u(4v + 2u)

9, 11

n 9, 11

9, 10

Four students gave incorrect answers when expanding −6(4x − 25). a

Explain the errors made by each student. i

Leon: −24x − 25

Jill: −2x − 25

iii

Claire: 24x − 150

Beryl: −24x − 150

What is the correct answer?

There are three consecutive integers. a

If x is the ﬁrst and smallest of the integers, write an expression for the: i

⬚ (3g − ⬚) = −6g + 14

Reasoning

2 (−4s + 5) = ⬚s + ⬚

Expand:

a Ex 2

5 (⬚p + ⬚) = 15p − 30

m −4t(4t + 3) Ex 1b

4cd, 5h–p, 6d–h, 7cd, 8g–n

Complete the statements: a

Ex 1a

4bc, 5d–l, 6abef, 7bc, 8d–jm

Second integer

Third integer

The sum of the ﬁrst, and two times the second is 15 more than two times the third. Write an equation for this.

What are the three integers?

Are there any other sets of three consecutive integers that satisfy the equation in part (b)? Explain how you know.

Mathspace Victorian Curriculum – Year 9 mathspace.co

The perimeter of a circle is 3π + 2. Find and expand an expression that represents the perimeter of 5 of these circles.

Problem-solving 13–15

12–15

Complete the equation so that x = −7 is the only solution:

2 (x − 12) + 9 = 3x − ⬚

Write two or more expressions for the area:

By expanding each of the expressions given in part (a), show that they are equivalent and hence come up with a single expression for the area.

c 14

x+7

x+1

2 x

Explain why x cannot be negative.

A rectangular garden has a length that is one metre less than twice the width (w). a

Write and expand an expression for the area of the rectangle.

When a border 2 metres wide is added around the garden, the total area (garden and border) is 2w2 + 7w + 12. What is the area of the border only?

The landscaper designing the garden has 100 square metres of pavers to use for the border. Determine the largest garden the landscaper can border before running out of pavers.

Expand and simplify −2a(a3 − 4a + 2) + 5a

The rule for the distributive law is that A(B + C) = AB + AC.

13–17

Write A(B + C) if A = x + 1, B = 2x and C = 3.

What is the expansion for A(B + C)?

A rectangular prism has dimensions of width = x, length = 2x − 1 and height = 6x. a

Write an expanded expression for the surface area.

If materials needed to make the prism dictate that the width could be between 1 cm and 10 cm, ﬁnd the range of the possible surface area.

3.03 Binomial expansion After this lesson, you will be able to… • apply the distributive law for multiplying binomials, even when variables are involved. • recognise the pattern for squaring a binomial.

3.03 Binomial expansion mathspace.co

Binomial expansion A binomial is an expression with two non-like terms that are being added or subtracted. For example: • 3x + 2 • −x2 + 9 • 10xy2 − 140xyz +

A C

Consider the expression ( A + B) (C + D). We can represent it visually using this area model.

B BC

This diagram shows that

+ D

x x

( A + B) (C + D) = AC + AD + BC + BD. In other words, the area of the whole shape is the sum of the smaller areas.

Let’s look at using this model to expand (x + 3) (x + 5).

5 × x = 5x

x × x = x2

+ 3

Here we can see that: (x + 3) (x + 5) = x2 + 3x + 5x + 15 We can simplify this to:

3 × x = 3x

x2 + 8x + 15

5 × 3 = 15

Exploration Complete the area models for multiplication shown: 7

7x3

3x2

3x2(7x3 + 9)

3(7 + 9) Complete the new area models for multiplication:

x 20

10 −3

(10 − 3) (20 + 1)

−3

(x − 3) (2x + 1)

What do the area models have in common?

What’s different about the area models?

How do you think multiplying binomials can be related to multiplying integers?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Example 1 Expand and simplify (x + 1) (x + 3)

Create a strategy Use the rule ( A + B) (C + D) = AC + AD + BC + BD.

Apply the idea (x + 1) (x + 3) = x × x + x × 3 + 1 × x + 1 × 3

Apply the rule

Simplify the products

Combine like terms

= x + 3x + x + 3 = x + 4x + 3

Idea summary We can expand the product of two binomial expressions using the rule: ( A + B) (C + D) = AC + AD + BC + BD

Square of a binomial Sometimes binomials such as (a + b) (a + b) are written as (a + b)2. If we need to expand binomials like this, we can still follow the same structure. (a + b)2 = (a + b) (a + b) 2

= a + ab + ab + b 2

= a + 2ab + b

Write as binomial product 2

Binomial expansion Simplify

Notice the middle term. It is 2ab, which is twice the product of a and b.

Example 2 Expand (x + 9)2

Create a strategy Use the pattern for squaring a binomial.

Apply the idea (x + 9)2 = (x + 9) (x + 9)

Write as a binomial product

Binomial expansion

Simplify

= x + 9x + 9x + 81 = x + 18x + 81

3.03 Binomial expansion mathspace.co

Example 3 Expand (5x − 2y)2

Create a strategy Use the pattern for squaring a binomial.

Apply the idea (5x − 2y)2 = (5x − 2y) (5x − 2y) 2

Write as a binomial product 2

= 25x − 10xy − 10xy + 4y 2

= 25x − 20xy + 4y

Binomial expansion Simplify

Idea summary Squaring a binomial means multiplying a binomial by itself. (a + b)2 = (a + b) (a + b)

(a − b)2 = (a − b) (a − b)

= a2 + ab + ab + b2

= a2 − ab − ab + b2

= a2 + 2ab + b2

= a2 − 2ab + b2

3.03 Practice questions Understanding 1

Adventurer 1cd, 2–4

Trailblazer 1cd, 2–4

Simplify: a

(8)2

(−5)2

(3x)2

(−12y)2

Are the expressions equivalent to ( y − 5) ( y − 8)? a

y( y − 5) − 8 ( y − 5)

5 ( y − 8) − ( y − 5)

y( y − 5) + y − 8

y( y − 8) − 5 ( y − 8)

The area of this ﬁgure is represented by the expression ( y + 3) ( y + 6). a

Explorer 1–4

Find the area of: i

Rectangle A

Rectangle B

iii

Rectangle C

Rectangle D

Write an equivalent, simpliﬁed expression for the area of the ﬁgure.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Is each statement true or false? a

(3x + 5)2 = 9x2 + 25

After completing a binomial expansion, adding products in different orders will always give the same answer.

The product (3x − 8) (3x + 8) will have two terms.

Fluency Ex 1

Ex 2

Ex 3

5a–cg, 6a–d, 7a–d, 8a–c, 9a–d

5c–g, 6c–h, 7b–e, 8c–e, 9b–e

5e–j, 6e–j, 7e–h, 8c–h, 9d–h

Expand and simplify: a

(x + 6) (x − 12)

(x + 2) (x + 6)

(v + 11) (v + 10)

(x − 4) (x − 7)

(m − 5) (m + 6)

(10m + 9) (m − 7)

(3w + 5) (7w + 6)

(3y − 3) (10y + 9)

(7y + 9) (−4y − 4)

Expand and simplify: a

(a + 3) (a − 3)

(x + 2) (x − 2)

( y + 5) ( y − 5)

(2x + 5) (2x − 5)

(4x + 6) (4x − 6)

(5x + 2) (5x − 2)

(3x + 7) (3x − 7)

(3y + 4) (3y − 4)

(9m + 8) (9m − 8)

(12x + 1) (12x − 1) (x − 3)2

(x − 5)2

Expand and simplify: a

(x + 7)2

(x + 4)2

(x + 6)2

(x − 2)2

Expand and simplify: a

(3a + 4b)2

(2a − 3)2

(4m − 5n)2

(x − 2y)2

(2x − 7y)2

(6p + 3q)2

Expand: a

(a + 6) (b − 12)

(k − 2r) (k + 5)

(q2 − 7) (t + 9)

(x + 5y) (x2 + y)

(3m2 − 4p) (m − p)

(4g2 + 3h) (6g + 5h3)

(3a3 − b2) (7a − 6b2)

(5x2 y + 4x) (7y3 − 12x5)

10–13

10–14

Reasoning

10–13

Explain why (x + 2)(x + 2) has three terms when expanded but (x − 2)(x + 2) has only two terms when expanded.

Marvin has a square bedroom. His sister Jane has a room that is 1 m shorter in length than Marvin’s room, but 1 m wider.

Show that Marvin has the larger bedroom.

How much bigger is Marvin’s bedroom compared to Jane’s bedroom?

Are (x − 4)2 and (4 − x)2 equivalent expressions? Explain your answer.

3.03 Binomial expansion mathspace.co

Rahul attempts to expand the expression (x + 9)2 and gets a result of x2 + 81. Is his answer correct? Explain where you think Rahul has gone wrong and identify the correct expansion.

Anette decides to use the area model to expand and simplify (10x + 2) (10x + 3). To warm up, she decides to try a simple integer example of (10 + 2) (10 + 3): 10

100

100 + 20 + 30 + 6 = 156 She then tried this method on (10x + 2) (10x + 3) 10x

10x

100x

20x

30x

100x + 20x + 30x + 6 = 150x + 6 Find the error in Anette’s answer. Correct her work and solution.

Problem-solving 15a–d, 16a–d, 17ab, 19 15

15c–f, 16c–f, 17bc, 18, 20, 21

15e–h, 16e–h, 17bd, 18, 20, 21

Expand and simplify: a

3 ( y + 3) ( y + 5)

8 (x − 2) (x − 6)

−(x + 5) (x + 6)

−9 ( y − 7) ( y + 6)

3 (−4x + 3) (4x − 3) g

4 (6x − 2) (x + 2)

−4 ( y + 1) ( y + 9)

Complete by filling in the blank: a

(t + 8) (⬚) = t2 + 10t + 16

(t + 6) (⬚) = t2 + 17t + 66

(t − 2) (⬚) = t2 − 7t + 10

(t − 2) (⬚) = t2 + 4t − 12 (t + 2) (t + 9) = t(t + 9) + ⬚ (⬚)

(t − 5) (⬚) = t + 7t − 60

(v − 2) (v − 3) = v(v − 3) − ⬚ (⬚)

(x + 3)3

(2x − 4)3

(2x + 3) (x + 5y − 4z)

(t + 3) (2t − 1) (3t + 5)

Expand:

(t − 2) (t + 8) = t(t + 8) − ⬚ (⬚)

A square has a perimeter of 8x + 20. What is its area?

Find the area of a rectangle which has a length of (x + 4) cm and a width of (x + 3) cm.

What expression, when multiplied by n + 6, gives n2 + 14n + 48?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Consider the rectangular prism shown with its dimensions labelled. Write an expression for the volume, and then expand and fully simplify the result.

(x + 7) m (x + 5) m 4m

3.04 Algebraic factors After this lesson, you will be able to… • identify common factors in algebraic expressions. • rewrite algebraic expressions in their factorised form. • apply factorisation techniques to simplify algebraic expressions.

Factorise algebraic expressions Factorising can be thought of as being the reverse process to expanding. Expanding would be: 6(x + 3) = 6x + 18 Factorised

Expanded

Factorising would be putting the expanded form into brackets by taking out the highest common factor (HCF): 6x + 18 = 6(x + 3) We can view this as the opposite of the distributive law: AB + AC = A(B + C) A is the highest common factor, which is taken out. This extends to three or more terms: AB + AC + AD = A(B + C + D) In any algebraic term, such as 12ab or 9ac, there can be numeric and algebraic components. When factorising we need to ﬁnd the highest common numeric factor and the highest common algebraic factor. In the case of 12ab and 9ac: • the highest common numeric factor is 3 • the highest common algebraic factor is a We can see this when we write the terms out in an expanded way: 12ab = 2 × 2 × 3 × a × b 9ac = 3 × 3 × a × c By identifying all the factors that are common to both terms we can see that the highest common factor of 12ab and 9ac is 3a.

3.04 Algebraic factors mathspace.co

Exploration The expanded form for 5x3 y4 z is 5 × x × x × x × y × y × y × y × z. The expanded form for 2y3 z5 is 2 × y × y × y × z × z × z × z × z. 1.

How many y’s are common in both expressions?

How many z’s are common in both expressions?

What is 5x3 y4 z + 2y3 z5 in factorised form?

What can we do to check our answer?

We can write 5x3 y4 z + 2y3 z5 = y3 z(5x3 y + 2z4) Expanded Factorised

To verify if you have factorised correctly, you can expand the result and check if it matches the original equation.

Example 1 Find the highest common factor for the following pairs of terms: a 10xy and 15yz

Create a strategy

Apply the idea

Break down the terms into factors.

10xy = 2 × 5 × x × y Rewrite 10xy 15yz = 3 × 5 × y × z Rewrite 15yz HCF = 5 × y = 5y

b 6a2 b and 21ab2

Create a strategy Break down the terms into factors.

Apply the idea 6a2 b = 2 × 3 × a × a × b

Rewrite 6a2 b

21ab2 = 3 × 7 × a × b × b

Rewrite 21ab2

HCF = 3 × a × b

Select the common factors

= 3ab

Mathspace Victorian Curriculum – Year 9 mathspace.co

Simplify

Select the common factors Simplify

Example 2 Factorise −2s − 10

Create a strategy Check all the terms for common numeric and algebraic factors. Factorise using AB + AC = A (B + C).

Apply the idea There are no common algebraic factors in −2s and −10, but there is a common numeric factor, −2 −2s − 10 = −2 × s + (−2) × 5 = −2 (s + 5)

Write the terms as products of −2 Factorise

Reﬂect and check It is important to be careful about factorising a negative because the sign of each term will change. If the leading term in an expression is negative, then we should factorise the negative out.

Example 3 Factorise 42x − x2

Create a strategy Check all the terms for common numeric and algebraic factors. Factorise using AB + AC = A (B + C).

Apply the idea There are no common numeric factors in 42x and − x2 but there is a common algebraic factor, x. 42x − x2 = x × 42 − x × x = x (42 − x)

Write the terms as products of x Factorise

Reﬂect and check Let’s check our factorisation of x(42 − x), by expanding it. x(42 − x) = x × 42 − x × x x(42 − x) = 42x − x2 This is the same as the original expression, so we know that we have factorised this correctly.

3.04 Algebraic factors mathspace.co

Example 4 Factorise 2ab2 + 16ab

Create a strategy Check all the terms for common numeric and algebraic factors. Factorise using AB + AC = A (B + C).

Apply the idea 2ab2 and 16ab have a common numeric factor of 2 and a common algebraic factor of ab. 2ab2 + 16ab = 2ab × b + 2ab × 8

Write the terms as products of 2ab

= 2ab (b + 8)

Factorise

Idea summary We can factorise expressions by writing the highest common factor of the terms outside of the brackets: AB + AC = A(B + C) Factorising is the reverse of expanding.

3.04 Practice questions Understanding 1

Trailblazer 1e–h, 2, 3

8ab and 6cd

10x2 and 25yz

12j2 k and 16x2 y

18xyz2 and 6ab2 c

24a2 n3 y6 and 8b4 m2 h6

36x7 y and 6a7 z

100z10 a8 k5 and 20j10 b8 n5

27ax z and 18pr k

Expand each expression to see which ones are a factorised form of −12x + 20x2: a

Adventurer 1d–f, 2, 3b–d

Find the highest common numeric factor of each pair of terms:

g 2

Explorer 1a–d, 2, 3a–c

−4 × 3x + 5

−2 (6x − 10)

4x (5 − 3x)

x(−12 + 20x)

All of these terms have a common factor of 5b. Write out the terms as a product of their factors, cross out the 5b that is common and determine what is remaining for each.

10ab

5ab

b f

15a2 b 2 2

120a b

Mathspace Victorian Curriculum – Year 9 mathspace.co

5ab

5a2 b

4a–c, 5, 6a–e, 7a–f, 8a–f, 4ace, 5d–f, 6e–h, 7b–g, 9a–c 8d–j, 9d–f

Fluency Ex 1

Find the highest common algebraic factor of each term: a

13y2 an and 17my2 b

8my and 12am

9bxm and 5bxa

2t2 + 2t = 2t (⬚ + ⬚)

Complete: 2

3y + 6y = ⬚ ( y + 2) −y2 − 2y = ⬚ ( y + 2)

8v − v2 = v (⬚ − ⬚)

5a − 10

15 + 20a

−18a + 16

4ab + 6a

11 km − mn

−12s + 10

8hj − 9gh

16jkt − 5kt

−35m + 49n

4 − 6a

−5c − 30

−8 − 12d

−w − 6w

−45mnq − 72mq

3x − x2

z2 + 6z4

Factorise: −3a − 6 −4b − 5ab

−abc − 5a

k2 − 9k

5u − 15u

5k2 t + 40k2 t2

Factorise: a

y2 + 4y

z + 8z

44uv − 8u2 v

m −30w − 25w y

−8w + 3w y

−63r2 + 28

49p2 q − 28pq2

−16a2 − 18a2 b

3x + 9 + 12yx

−10u v + 9uv

Factorise: a

4x + 12 + 16yx

4x2 y + 8xy2 + 12xz2

5ab c + 25bc + 100abc

30b2 c + 10abc + 20c2

5a2 b2 + 2ab − 3a2 b2 − 4ab

5x2 y2 + 15yz2 + 25y2 + 60y

2x + 10x y + 8yx 2

Reasoning 10

−m2 + 19m = ⬚ (m − 19)

4a + 12

e 8

y2 + 5y = y (⬚ + ⬚)

Factorise:

Ex 4

9xn and 4ax

Ex 3

18b n and 7b m

9x and 8x

Ex 2

4bdf, 5ef, 6g–j, 7c–h, 8f–n, 9f–h

10, 11

10–12

A student incorrectly factorised and wrote −5x2 − 15x = −5x (x − 3). Explain how to correct the error, and provide the correct answer.

Alex and Beth are both asked to factorise −5x + 10y. Alex wrote down −5(x − 2y) and Beth wrote down 5(2y − x). Who is correct? Explain.

Louis was asked to fully factorise 2x3 + 8x2 − 4x and gave as his answer 2(x3 + 4x2 − 2x). What has he done wrong?

3.04 Algebraic factors mathspace.co

Problem-solving 13–15 13

14, 16–18

The rectangle shown has an area of 4x2 − 16x square units. What expression describes the length of the rectangle?

? 4x

A property management company has an area of land, 20n + 5n2 available for parking. When factorised, this represents the length and width of the block of land. a

Factorise the expression to determine the dimensions of the block.

Given that the length is longer than the width for all n > 1, which term is the length and which is the width?

The length is further divided into 5 equal sections, and one of these sections is fenced off for storage. What is the length of fencing needed?

You are to design a photo collage made of two large square photos with a side length of x and four smaller rectangular photos that have a height of x and a width of 4 units. a

Draw an example of what this arrangement would look like.

Find an algebraic expression for the area of the rectangle formed if the photos are all placed in a single row.

Fully factorise your answer from part (a) and then draw a different photo arrangement that would match these dimensions.

A rectangle has a width of 2xy cm and a length of 2(x − y) cm. a

Write an expression for the area of the rectangle, and expand it.

Write an expression for the length of a rectangle with the same area if its width is 4xy cm.

Write a factorised expression for the length of a rectangle with the same area if its width is: i

13, 14, 16, 18

4 cm

xy cm

A container in the shape of a rectangular prism has a length of (2a − b) metres, a width of 6ab metres, and a height of 5ab metres. a

Write an expression for the volume of the prism, and expand it.

Write a factorised expression for the volume of a container that has twice the volume of the rectangular prism.

What percentage of the rectangular prism has been ﬁlled if water is poured in to a depth of 2ab m?

Write an expression for the percentage of the rectangular prism that has been ﬁlled if water is poured in to a depth of 3a metres.

Factorise 3t (t − 5) + 4 (t − 5)

Mathspace Victorian Curriculum – Year 9 mathspace.co

3.05 Factorise by grouping After this lesson, you will be able to… • apply factorising by grouping in algebraic expressions. • identify the characteristics of a monic quadratic trinomial and the steps to factorise it. • apply factorising techniques to solve problems involving algebraic expressions with three or more terms.

Factorise by grouping Common factors can be more than 1 term. For example, 5 (a + 7) and b (a + 7) have a common factor of (a + 7). So, 5 (a + 7) + b(a + 7) = (5 + b) (a + 7) For expressions with 4 terms, sometimes separating them into 2 groups can help fully factorise the expression.

Example 1 Factorise: 3x + 6 + 2x2 + 4x

Create a strategy Using inspection, determine the 2 groups that will likely end up with a common factor.

Apply the idea 3x + 6 + 2x2 + 4x = (3x + 6) + (2x2 + 4x) = 3 (x + 2) + 2x (x + 2)

Separate into two groups Factorise each group

= (x + 2) (3 + 2x) Identify (x + 2) as the common factor and ﬁnish the factorisation

Reﬂect and check There may be multiple ways to separate the original expression into two groups that share a common factor. For example, we could have factorised the expression this way: 3x + 6 + 2x2 + 4x = 2x2 + 3x + 4x + 6 2

Rearrange the terms

= (2x + 3x) + (4x + 6)

Separate into two groups

= x (2x + 3) + 2 (2x + 3)

Factorise each group

= (2x + 3) (x + 2) Identify (2x + 3) as the common factor and ﬁnish the factorisation

Notice that this is the same result we got before, the terms are simply rearranged.

3.05 Factorise by grouping mathspace.co

Example 2 Fully factorise: 5(a + b) + v(a + b)

Create a strategy Factorise using the law, AB + AC = A(B + C).

Apply the idea Both terms have a common factor of (a + b) so we can factorise this out. 5 (a + b) + v (a + b) = (a + b) (5 + v)

Factorise

Example 3 Factorise: 2x + xz − 40y − 20yz

Create a strategy Group the expression in pairs and factorise using the law, A(C + D) + B(C + D) = ( A + B) (C + D).

Apply the idea To factorise an expression with 4 terms, we factorise two pairs of terms, then take out the common factor binomial. 2x + xz − 40y − 20yz = (2x + xz) + (−40y − 20yz)

Group terms with common factors

= x (2 + z) − 20y (2 + z)

Factorise each pair

= (2 + z) (x − 20y)

Factorise

Idea summary When we factorise expressions with 4 terms, we may need to break the term into 2 groups and factorise each group individually. If there is a common factor between these two groups, we may need to factorise further.

Monic quadratic trinomials Interactive exploration

Discover this concept in action online

Mathspace Victorian Curriculum – Year 9 mathspace.co

mathspace.co

A monic quadratic trinomial is in the form x2 + bx + c. It has three features: • A trinomial has 3 terms. • A quadratic has a highest power of 2. • A monic quadratic is one where the coefficient of x2 is 1. We factorise monic quadratic trinomials by ﬁnding the factors of the constant, c, that also add up to the coeﬃcient of x, b.

Example 4 Factorise: x2 + 6x + 8

Create a strategy The constant is 8 and the coeﬃcient of x is 6. We need to ﬁnd two factors of 8 that add to 6.

Apply the idea The factors of 8 that add to 6 are 2 and 4. x2 + 6x + 8 = x2 + 4x + 2x + 8

Rewrite the 6x term as 4x + 2x

= (x + 4x) + (2x + 8) Separate into two groups = x (x + 4) + 2 (x + 4)

Factorise each group

= (x + 2) (x + 4)

Factorise using the common factor of (x + 4)

Idea summary When factorising a monic quadratic trinomial in the form ax2 + bx + c, we: 1. Find the factors of c ( p and q) that add up to b 2. Rewrite the middle term in terms of the factors: ax2 + px + qx + c, where px + qx = bx 3. Factorise by grouping.

3.05 Practice questions Understanding

Explorer 1–3

Adventurer

Trailblazer

1–3

For the expression 3(x + 7) + x(x + 7), what is the common factor?

Complete this statement: To factorise x2 + 9x + 18, we need to ﬁnd two numbers whose product is ⬚ and whose sum is ⬚.

Given that p < q, ﬁnd the values of p and q in each pair of equations:

p + q = 19 pq = 90

p + q = −2 pq = −63

p + q = −10 pq = 24

3.05 Factorise by grouping mathspace.co

Fluency Ex 1

Ex 3

x2 + 5x + 6

x2 + 7x + 10

x2 + 6x + 9

x2 + 8x + 12

x2 − 3x + 2

x2 − 6x + 8

x2 − 5x + 6

x2 − 6x + 9

x + 11x + 24

x + 17x + 72

x − 4x + 3

x2 − 17x + 60

m x2 − x − 30

x2 + x − 20

x2 − 3x − 54

x2 − 3x − 70

−8 − 6x − x

−12 + 7x − x2

40 + 13x + x

35 − 12x + x

Factorise: a

y( y + 5) + 7 ( y + 5)

a (a − 4) − 3 (a − 4)

p( p − 3) + 6 ( p − 3)

5 (q + 4) − q (q + 4)

5 (r − 3) − r (r − 3)

6r (2r − s) − rs (2r − s)

7t (t + u) + 2u (t + u)

x ( y − z) − w ( y − z)

5y (4w + 3x) − z (4w + 3x)

8y ( y − 4) + 3 (4 − y)

Factorise: a

8x + xz − 16y − 2yz

24 + 3y + 8x + xy

7xy + wx + 7yz + wz

x2 + 2x + 5x + 10

x2 − 3x + 8x − 24

50 + 5y + 10x + xy

12xy + wx + 12yz + wz

x2 + 5x + 8x + 40

x2 − 5x + 10x − 50

2mp + 6 + 3p + 4m

5mp + 6 + 2p + 15m

2x + 18yz + 12xy + 3z

m 4x + 24yz + 32xy + 3z

a3 + 8a2 + a + 8

x2 − 4x + 4

Factorise: a

x2 + 8x + 16

x2 + 10x + 25

64 + 16x + x2

36 − 12x + x2

Reasoning 8

4i–t, 5e–j, 6g–n, 7bdf

Factorise:

q Ex 2

4e–p, 5c–g, 6e–k, 7bdf

4a–h, 5a–d, 6a–f, 7ace

x2 − 20x + 100

8, 9

Identify and explain the error: 5xy − 5xz + 7z − 7y = 5x ( y − z) + 7 (z − y) = (5x + 7) ( y − z)

What is the largest possible integer value of k that will allow m2 + km + 24 to be factorised using whole numbers? Justify your answer.

Problem-solving 10a–d, 13 10

10e–h, 11, 12cd, 13, 14

Factorise each expression by ﬁrst factoring out a numeric factor: a

2x2 + 14x + 20

4x2 + 24x + 32

3x2 − 21x + 30

−2x2 − 18x + 44

−3x2 + 12x − 12

4x2 − 4x − 288

5x2 + 5x − 30

−4x2 + 12x + 40

2x3 + 16x2 + 30x

Factorise: a

10b–e, 11, 12ab, 13

x3 + 7x2 + 12x

Mathspace Victorian Curriculum – Year 9 mathspace.co

Factorise: a c

3f ( g + h) + (g + h)

3x2 (x + 4y) − 5y(x + 4y)

(2c − d) (c + 5d) − 3 (d − 2c)

Find the perimeter of the square.

A square has area t2 + 14t + 49. a

2y(2x2 + 3z) − (2x2 + 3z)

Find the side length of the square.

Find an expression for the shaded area in the ﬁgure in terms of x. Write your answer in factorised form. x

3 x+1 4

Did you know?

Algebra is the backbone of roller coaster design! Engineers use algebra to calculate the perfect curves, drops, and loops that create thrilling, yet safe, rides. By solving equations that involve speed, acceleration, and force, they ensure that the coaster delivers excitement without exceeding safety limits. Every twist and turn is carefully calculated to ensure that the G-forces feel exciting but not overwhelming for riders. Even the timing of the ride, from the initial climb to the final stop, is determined using algebraic models. The result is a smooth, exhilarating ride that feels like flying! This precise math allows roller coasters to provide an adrenaline rush while keeping everyone safe. Without algebra, designing such complex, high-speed rides would be impossible. Algebra ensures the balance between excitement and safety on every thrilling roller coaster ride!

3.05 Factorise by grouping mathspace.co

INVESTIGATION

Calendar expressions Formulate problems with quadratic polynomials using a calendar game.

Objectives To expand, simplify, and factorise algebraic expressions

Materials • Monthly calendar

•

Pencil

• Paper

Calendar expressions How to play the game Firstly, choose dates on a calendar that form a pattern, without your partner knowing. Create an expression that must be expanded, simpliﬁed, or factorised so that your partner can guess which days you’ve chosen. For example, you could choose two consecutive days and create an expression that requires binomial expansion. Give your partner the expression, the shape of the pattern on the calendar and the month the dates are in. They can now try to solve the problem. Here is an example of how Bubaker and Tanesha played the game. Remember that you can be as creative with your expressions as you would like: JUNE 2023

JULY 2023

Sun Mon Tue Wed Thu

Fri

Sat

Sun Mon Tue Wed Thu

Fri

Sat 1

Mathspace Victorian Curriculum – Year 9 mathspace.co

Bubaker chooses 2 June, 9 June, and 16 June without showing Tanesha. The expression they provide to Tanesha is 2x2 + 50x + 288. The instructions they give are to ﬁnd three days in June that fall in a vertical line.

Tanesha factorises the expression and gets 2 (x + 9) (x + 16). The factorised expression gives the three dates shown vertically on the calendar.

Tanesha chooses 10 July and 18 July without showing Bubaker. The expression they provide to Bubaker is 2 (8x + 4 − 3x + 5). The instructions they give are to ﬁnd two days in July that fall on a diagonal.

Bubaker expands the expression and gets 10x + 18. They are able to identify the dates on the calendar.

Tanesha and Bubaker continue taking turns choosing dates, creating patterns and guessing the days.

Find the four days in June 2023 that form a square with four days on the calendar, given the expression a2 b3 (1 − a7 b7).

Investigate 1.

How did you create some of your expressions to challenge your partner?

As the game went on, did you choose to become more creative when creating expressions? Why or why not?

Were there any moments in the game where you or your partner misunderstood or made a mistake with an algebraic expression? What was the mistake, and how did you correct it?

Discussion 1.

Consider other ways we could have designed the game or other rules we could have implemented. How might different rules have changed your strategies or the algebraic expressions you used?

How could we incorporate other mathematical concepts or operations into this game? For example, could we introduce roots, powers, or other algebraic functions to make the game more challenging?

Create a calendar game that involves different mathematical concepts.

Investigation: Calendar expressions mathspace.co

3 Chapter review 1

Simplify: a

11y − 6y

7a + 2a − a

3x + 3y − (−x) + 4y

a2 + 5a − 4a + (−11a2)

6u2 × 7v4

−c8 d2 × 3c6 d × 2d2

Simplify: a c

5x × 8y × 2z

Simplify: a

2r × 3s

5b2 ÷ 25b

−(8xy) ÷ 12yz

4x2 × 7y ÷ 2xy2

Simplify: a

A construction company charges $2 000 for each job. They also charge labour costs of $50 per hour and $75 per hour for overtime. If x is the amount of regular hours worked and y is the amount of overtime hours worked, write an expression to represent the total cost of the job.

Expand: a

6f ( f 2 + 3)

7c3 (3d4 − 9)

5g (g − 4f + 8)

5 (7n − s) − 2 (6n − 3s)

4c2 − 7c + 3c(c − 11)

−a (11b + 9c)

−4a2(4a − 2a4)

−2g (2g + 4)

Expand and simplify:

a c 9

Expand: a

x (2x − 5)

5 (3 ( y − 2) + 7)

Complete the equation so that x = 11 is the only solution: 4(x − 8) + 13 = x +⬚

Expand and simplify: a

(b + 3) (b + 7)

(c − 11) (c − 8)

( f + 9) ( f − 9)

(12g + 10) (12g − 10)

(3d + 5) (d − 6)

(4c + 5d)2

Expand and simplify: a

(h + 3)2

Mathspace Victorian Curriculum – Year 9 mathspace.co

(3e − 4) (5e + 6)

Expand: a

(r − 8) (s + 7)

( p + 13w) ( p2 + w)

Find the area of a triangle with a base length of (2x − 5) mm and a height of (x − 1) mm.

Find the highest common algebraic factor of each term: a

16ab − ac

−8x − 4

−35xy − 21x

7c2 − 28c

−3f 2g + f 2g2

5k + 35jk + 15

3st2 + 9s2t + 30t

Factorise the expression to determine the dimensions of the playground.

Given that the length is longer than the width for all x > 1, which term is the length and which is the width?

If one of the longer sides of the playground is along the existing community hall, how much fencing is needed to fence the rest of the playground?

Factorise: x (x + 5) + 7 (x + 5)

7x (x − 3) + 2 (3 − x)

3a + ab − 6c − 2bc

x2 − 4x − 32

7a (a2 − 3b) − 13b (a2 − 3b)

Factorise: a

g2 + 2g + 6g + 12

3xy + 10 + 15x + 2y

Factorise: a

3x + 9

The area of a rectangular playground is 18x2 + 3x. When factorised, this represents the length and width of the playground.

a 19

24xy2 and 10y2z

Factorise: a

13t3 and 7t

x2 + 11x + 28

x2 − 14x + 48

81 − 18x + x2

Factorise: a

3x2 + 21x + 30

Chapter 3 review mathspace.co

Big ideas • Understanding linear relationships allows us to model, analyse, and solve real-world problems effectively. • Applying transformations, proportional relationships, and inequalities in linear graphs helps us visualise connections, make accurate predictions, and find a range of possible solutions.

Curriculum Outcomes VC2M9N01 recognise that the real number system includes the rational numbers and the irrational numbers, and solve problems involving real numbers with and without using digital tools VC2M9A03 sketch linear graphs of equations in various algebraic forms, using the coordinates of 2 points, and solve linear equations VC2M9A04 find the gradient of a line segment, the midpoint of the line interval and the distance between 2 distinct points on the Cartesian plane VC2M9A06 use mathematical modelling to solve applied problems involving change, including financial contexts involving simple interest; formulate problems, choosing to use either linear or quadratic functions or other simple variations; interpret solutions in terms of the context; evaluate the model and report methods and findings VC2M9A07 experiment with the effects of the variation of parameters on graphs of related functions, using digital tools, making connections between graphical and algebraic representations, and generalising emerging patterns VC2M9M03 solve spatial problems, applying angle properties, scale, similarity, ratio, Pythagoras’ theorem and trigonometry in right-angled triangles VC2M9M05 use mathematical modelling to solve practical problems involving direct proportion, rates, ratio and scale, including financial contexts; formulate the problems and interpret solutions in terms of the situation; evaluate the model and report methods and findings

Linear relationships Chapter outline 4.01 4.02 4.03

Midpoint Gradient Distance between two points Investigation: Transformations of linear graphs 4.04 Gradient-intercept form 4.05 Linear graphs 4.06 Equation of a straight line 4.07 Parallel and perpendicular lines 4.08 Direct proportion 4.09 Model linear equations 4.10 Linear inequalities Chapter 4 review

98 106 121 132 134 144 155 160 170 178 185 194

The longest distance between any two places on Earth is the distance between Chincha Alta in Peru and Siem Reap in Cambodia, which is over 20,000 kilometres!

4.01

Midpoint After this lesson, you will be able to… • use counting squares to find the midpoint of a line segment. • apply the midpoint formula to find the coordinates of a midpoint. • solve real-world problems involving the midpoint of a line segment.

Midpoint from horizontal and vertical segments The midpoint of a line segment is the point exactly halfway along the segment. A

That is, the distance from the midpoint to both endpoints is the same. AB = BC : means that the distance from A to B is the same as the distance from B to C.

Midpoint

Interactive exploration Discover this concept in action online

mathspace.co

To find the midpoint of a line segment AB, we can count squares to find the lengths of the legs of the right-angled triangle with hypotenuse AB. Then we find the midpoint, the point halfway along, each leg. 12 11 10 9 8 7 6 5 4 3 2 1

length = 10

length = 6

1 2 3 4 5 6 7 8 9 10 11 12

Triangle with hypotenuse AB

12 11 10 9 8 7 6 5 4 3 2 1

length = 5

length = 3

1 2 3 4 5 6 7 8 9 10 11 12

The midpoint of each leg and the midpoint of AB

The coordinates of the midpoint of each leg helps us ﬁnd the midpoint of the hypotenuse. • The x-value of the horizontal leg’s midpoint is the x-value of the midpoint of AB. • The y-value of the vertical leg’s midpoint is the y-value of the midpoint of AB. The midpoint between two coordinates such as (0, 4) and (0, 9) can be found by finding the 4+9 average of the two numbers 4 and 9. The midpoint of the y-coordinate is = 6.5 and the 2 x-coordinate is 0. So, the midpoint is (0, 6.5).

Mathspace Victorian Curriculum – Year 9 mathspace.co

Example 1 M is the midpoint of A(−8, −4) and B(2, 8). Find the coordinates of M.

8 7 6 5 4 3 2 1 −8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 A

x 1 2

Create a strategy Draw a right-angled triangle with hypotenuse AB, then ﬁnd the midpoint of each leg.

Apply the idea 10

12 M

8 7 6 5 4 3 2 1

−9 −8 8− −7 7 −6 6 −5 −4 −3 −2 −1−1

The length of the horizontal leg is 10 units, so the midpoint will be 5 units to the left of B.

B (2, 8)

The length of the vertical leg is 12 units, so the midpoint will be 6 units above A. The coordinates of M are (−3, 2). x 1 2

−2 −3 −4

A (−8, –4)

Idea summary 12 11 10 9 8 7 6 5 4 3 2 1

To ﬁnd the midpoint of a line segment AB, draw a right-angled triangle with hypotenuse AB, and ﬁnd the midpoint of each leg.

• The x-value of the horizontal leg’s midpoint is the x-value of the midpoint of AB. • The y-value of the vertical leg’s midpoint is the y-value of the midpoint of AB. A

1 2 3 4 5 6 7 8 9 10 11 12

4.01 Midpoint mathspace.co

Midpoint formula The midpoint (M) of any two points has coordinates that are exactly halfway between the x-values and halfway between the y-values. This means we can find the average of the two x-coordinates to find the x-coordinate of the midpoint and the average of the two y-coordinates to find the y-coordinate of the midpoint.

M M==

xx1 1++xx22

yy1 1++yy22

22 22 average average ofofthe thexx values values

average average ofofthe theyy values values

(x1, y1)

ﬁrst endpoint of the line segment

(x2, y2)

second endpoint of the line segment

Example 2 M is the midpoint of A (4, −6) and B (8, −12). What are the coordinates of M ?

Create a strategy To ﬁnd the coordinates of M, we want to ﬁnd the average of the two x-coordinates and the average of the two y-coordinates. We can use the midpoint formula:

Apply the idea Write the midpoint formula

Substitute the coordinates (4, −6) as (x1, y1) and (8, −12) as (x2, y2)

Simplify each part

Simplify

The coordinates of the midpoint M are (6, −9).

Reﬂect and check We can check our answer by plotting the segment and the midpoint on a Cartesian plane.

100

Mathspace Victorian Curriculum – Year 9 mathspace.co

The length of the horizontal leg is 4 units, so the midpoint will be 2 units to the right of A.

x 2

The length of the vertical leg is 6 units, so the midpoint will be 3 units above B.

−2 −4 −6

This shows our answer is correct. The midpoint is at (6, −9).

−8 −10 −12

Idea summary For points A(x1, y1) and B(x2, y2), the midpoint (M) will be:

(x1, y1) (x2, y2)

ﬁrst endpoint of the line segment second endpoint of the line segment

4.01 Practice questions Explorer

Understanding

Adventurer

1–3

How many midpoints does a segment have?

Find the coordinates of the midpoint M. a

A(2, 8) and B(2, 2) 9 8 7 6 5 4 3 2 1 –4–3–2–1 –1 –2 –3 –4

Trailblazer

1–3

A(6, 5) and B(14, 7) 9

6 5

B x 1 2 3 4 5 6 7 8 9

1–3

M A

3 2 1

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14

4.01 Midpoint mathspace.co

101

Segment AB and point M are plotted on the Cartesian plane. Is M the midpoint of AB? a

y B

2 −4−3−2 −1

9 8 7

x 1 2 3 4 5 6 7 8 9

−2

−8

−6

Ex 1

−4

Fluency

4, 5a–e, 6ace, 7–9

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14

4, 5c–g, 6bdf, 7–10

4, 5f–j, 6gh, 7–10

Find the coordinates of the midpoint M of segment AB. a

A (5, 4) and B (5, 10)

A (−9, −8) and B (−2, −8)

y −10−9 −8 −7 −6 −5 −4 −3 −2 −1 11 10 9 8 7 6 5 4 3 2 1

x 1

2 3 4 5 6 7 8 9

A (−7, −2) and B (1, 6)

y 8 7 6 5 4 3 2 1

−9−8−7−6−5−4−3−2 −1 −1 −2 A −3 −4

102

−1 −2 −3 −4 −5 −6 −7 B −8 −9 −10

x 1 2 3

Mathspace Victorian Curriculum – Year 9 mathspace.co

A (−7, 3) and B (1, −7) A

3 2 1

−8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8

x 1 2

A (−8, 5) and B (4, −3)

A (−6, 2) and B (10, −10)

y A

4 3

−8 −6 −4 −2 −2

2 1 −8−7−6−5−4−3−2 −1 −1

−6

1 2 3 4

−8

−2

Ex 2

2 4 6 8 10

−4

−3

−10

Find the midpoint of segment AB. a

A (7, 3) and B (9, 9)

A (5, 7) and B (2, 3)

A (−9, 8) and B (−1, 10)

A (3, −2) and B (5, −11)

A (−11, −8) and B (−7, 0)

A (−14, −4) and B (4, 8)

A (−7, 2) and B (11, −14)

A (−13, −12) and B (−5, −2)

and

For the pair of points A and B: i

Find the coordinates of the midpoint, M.

Plot the segment AB and the point M on a Cartesian plane.

A (5, 5) and B (13, 9)

A (−12, −1) and B (6, 9)

A (−9, 1) and B (5, −7)

A (1, −11) and B (9, 7)

A (4, 5) and B (6, 9)

A (−10, −1) and B (4, 5)

A (−7, 5) and B (1, −9)

A (−12, −6) and B (4, −4)

The graph shows the annual net proﬁt (in millions) of a company over the last few years. It shows that its proﬁt has been growing approximately linearly from $19 million in 2008 to $39 million in 2014.

Profit (millions) (2014, 39)

What was the company’s net proﬁt in 2011? (2008, 19)

Year

4.01 Midpoint mathspace.co

103

The graph shows a straight line that approximates the global life expectancy of a child over a period of 100 years. The graph shows the life expectancy to be 55 years in 1910, and 71 years in 2010.

Age

(2010, 71)

Use the midpoint formula to estimate the life expectancy of a child in 1960. (1910, 55)

Year

Lines of latitude and longitude measure position on the earth’s surface and work like coordinates. The first coordinate represents how far above or below the equator you are, and the second coordinate measures how far from Greenwich Mean Time you are. A plane starts its flight at (20° N, 62° E). It is bound for its destination at (49° N, 144° E). Find its position halfway through the flight.

The table shows the number of smartphone users in a particular city: a

Using the midpoint formula, estimate the number of smartphone users in 2009.

Using the midpoint formula, estimate the number of smartphone users in 2013.

Reasoning 11

12ab

Year 2007 2011 2015

Number of smartphone users 175 904 238 502 298 080

11, 12

11–13

What are the properties for the diagonals of a parallelogram?

ABCD is a quadrilateral with vertices A (−2, −1), B (−10, 3), C (−15, 7) and D (−7, 3). Find the midpoints of the diagonals. i

c 12

Is ABCD a parallelogram? Explain your answer.

For the points P (−3, −2), Q (19, 18) and R (10, 15).

Find the coordinates of T, the midpoint of PQ.

If T is the midpoint of RS, ﬁnd the coordinates of S.

What type of quadrilateral is PRQS? Explain your answer.

Q R

5 −15 −10 −5 P−5 −10 −15

Explain how the midpoint formula is related to the average of two numbers.

104

Mathspace Victorian Curriculum – Year 9 mathspace.co

x 5

Problem-solving 14, 15a–d, 16 14

15c–h, 16–19

15e–j, 16, 18–23

For points A and M, where M is the midpoint of A and B: i

Find the coordinates of point B.

Plot the segment AB and the point M on a Cartesian plane.

A (1, −1) and M (4, 3)

A (−3, 1) and M (−4, −3)

If the midpoint of points A and B is M, what are the coordinates of A? a

B (9, 6) and M (7, 6)

B (10, −5) and M (8, −5)

B (9, 12) and M (7, 9)

B (11, 2) is M (8, −2)

B (−1, 0) and M (−3, 3)

B (12, 5) and M (8, 1)

B (2, −9) and M (−1, −7)

B (−4, −6) and M (1, 8)

B (−7, −9) and M (−5, −3)

B (−1, 0) and M (−4, −2)

A circle with centre at Point C (5, 3) has AB as its diameter. The Point A has coordinates (2, 7) as shown:

9 8 7 6 5 4 3 2 1

Find the coordinates of B and justify your answer.

−1 −1 −2 −3

y A C x 1 2 3 4 5 6 7 8 9 10 11

Find the coordinates of the midpoint of A (2x, 5y) and B (6x, y).

The points P, Q, R, S, T lie on a straight line such that PQ, QR, RS and ST have the same length. Find the coordinates of the points Q, R and S given P (−4, −2) and T (−12, 10).

M is the midpoint of AB. The coordinates of point A are (a, 12), the coordinates of point M are (c, 4), and the coordinates of B are (6, −4). If the length of AB = 20 units, ﬁnd the x-coordinates of A and M.

For the points A (−10, 10) and B (16, 4): a

Find the coordinates of the midpoint M of AB.

For what value of k will the line y = 6x + k bisect (cut in half) the line segment AB?

Find the coordinates of a point that is three-quarters of the way from point A (−3, 1) to point B (1, 7).

A segment AB has endpoints midpoint of the segment.

M (4a + 2, 5b − 3) is the midpoint of S (20, −12) and T (−18, 6). a

Find the value of a.

and B (−7r, 5s). Find the coordinates of the

Find the value of b.

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4.02

Gradient

After this lesson, you will be able to… • identify increasing and decreasing lines. • calculate the gradient using the “rise over run” method. • apply the gradient formula using the coordinates of two points on a line. • examine the special case of horizontal and vertical lines and their gradients.

Increasing and decreasing lines The gradient of a line is the measure of its slope. The gradient value describes: • If a line is increasing left to right (positive gradient) • If a line is decreasing left to right (negative gradient) • How much the line moves vertically (change in the y-value) compared to how much it moves horizontally (change in the x-value)

Interactive exploration Discover this concept in action online

Positive gradient:

Negative gradient: 4

1 −4 −3 −2 −1

−1

−4 −3 −2 −1

−1

−3

−3 −4

Mathspace Victorian Curriculum – Year 9 mathspace.co

x 1

−2

−4

Line with increasing gradient, going up from left to right

−2

106

mathspace.co

Line with decreasing gradient, going down from left to right

Example 1 Describe the gradient of the line as increasing or decreasing. 4

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

Create a strategy

Apply the idea

Identify if the line is increasing or decreasing from left to right.

The line is decreasing from left to right, which means that the line has a negative gradient.

Idea summary If the gradient is positive, the line is increasing from left to right. If the gradient is negative, the line is decreasing from left to right.

Gradient as rise over run The gradient can be calculated by using the formula: m= Note: we use the variable m to denote the value of the gradient. The rise and run are found by forming a right-angled triangle using any two points on the line. rise 4 y = This line has a gradient of run 3 2 In this case, the gradient is positive because, over 1 4 the 3-unit increase in the x-values, the y-value has rise x increased. −3

−2

−1 −1

3 run

−2

Remember that increasing lines have positive gradients, and decreasing lines have negative gradients.

Interactive exploration Discover this concept in action online

mathspace.co

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107

7 6 5

y y = 4x + 1 m=4 y = 2x + 1 m=2

4 3 2 1 −1

−1

x 1

Graphs of y = 2x + 1 and y = 4x + 1 The steeper the graph, the larger the value of m.

Example 2 Find the gradient of the line shown using points A and B, by creating a right-angled triangle and identifying the rise and run.

8 7 6 5

4 3

2 1

x 1

2 3 4

5 6 7 8 9

Create a strategy We can draw a triangle to ﬁnd the rise and run, and then use the gradient formula.

Apply the idea 9 8 7 6 5 4 3 2 1

B 2 rise A

3 run x

108

We can now see that this line has a run of 3 and a rise of 2. rise m= run 2 = 3

2 3 4 5 6 7 8 9

Mathspace Victorian Curriculum – Year 9 mathspace.co

Reﬂect and check The line is increasing so it should have a positive gradient.

Idea summary We calculate the gradient by using the formula:

Gradient

rise is the vertical measurement run is the horizontal measurement

Gradient formula 6

We have just been using

A (3, 6)

Gradient =

5 4 3 2 1 −1

x 1

−2

B (7, −2)

to ﬁnd the gradient of a line. This works when it’s easy to see the graph, and construct a rightangled triangle with easy to measure lengths of rise and run. We can extend this thinking to use the coordinates of points A and B and construct a general formula.

Let’s look at an example.

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The line is decreasing so should have a negative gradient, and it does. 7

For any two points (x1, y1) and (x2, y2), we can ﬁnd the gradient using the formula:

Gradient =

5 4

(x2, y2)

Gradient =

2 1 −1

−1

(x1, y1) 1

x 3

Example 3 What is the gradient of a line that passes through the points A(3, 5) and B(−2, 10)?

Create a strategy Use the gradient formula.

Apply the idea Use the gradient formula

Substitute the coordinates

Evaluate

Reﬂect and check By doing a quick sketch of the line we can see that the line is decreasing, so it should have a negative gradient.

y 10 8 6 4 2 −2 −2

110

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x 2

Idea summary For any two points (x1, y1) and (x2, y2), we can ﬁnd the gradient using the formula:

Gradient = (x1, y1) (x2, y2)

are the coordinates of the ﬁrst point are the coordinates of the second point

Gradient of horizontal and vertical lines On horizontal lines, the y-value is the same for every point on the line. 4 A (−4, 4) B (−1, 4) 3

y C (3, 4)

The equation of this line is y = 4.

1 −4 −3 −2 −1

Coordinates A, B, and C all have the same y-coordinate. Every point on the line has a y-value equal to 4.

x 1

−1

Since gradient is calculated by no rise, then:

and there is

rise run 0 = run =0

−2 −3 −4

The gradient of a horizontal line is always 0. On vertical lines, the x-value is always the same for every point on the line.

A (−3, 8)

Coordinates for A, B, and C all have the same x-coordinate, x = −3.

The equation of this line is x = −3.

Since gradient is calculated by

no run, then:

4 B (−3, 3)

3 2

−2 −3 C (−3, −3)

rise run

rise 0 = undefined =

1 −4 −3 −2 −1 −1

and there is

x 1

Division by 0 is undefined. The gradient of a vertical line is always undefined.

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Example 4 What is the gradient of the line passing through A(−3, −1) and B(−3, 5)?

Create a strategy Observe that the x-coordinates have the same value, so the points lie on the same vertical line.

Apply the idea The gradient of any vertical line is undeﬁned, thus the gradient of the line passing through A(−3, −1) and (−3, 5) is undeﬁned.

Reﬂect and check Use the gradient formula in an attempt to calculate the gradient. Gradient = Write the gradient formula Simplify Since there is a 0 on the denominator, the gradient is undeﬁned.

Idea summary Horizontal lines have a gradient 0. Vertical lines have a gradient that is undeﬁned.

4.02 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

A line is drawn on the Cartesian plane.

Which of these describe the relationship between the gradients of the segments?

Gradient AB is equal to gradient AC.

Gradient AB is twice gradient AC.

Gradient AC is twice gradient AB.

1 −4 −3 −2 −1

Mathspace Victorian Curriculum – Year 9 mathspace.co

B 1

−1

112

2 A

x 3

Identify whether each graph has a positive gradient or negative gradient. y

1 −4 −3 −2 −1

x 1

−1

−4 −3 −2 −1

−2

−3

−4

Identify whether each graph has a zero gradient or undefined. b

2 1 −4 −3 −2 −1

−1

1 −2 −1

−3 −4

Fluency

−2

−1

−2

Ex 1

x 1

−1

x 1

−2

−5

−3

−6

−4

4–9, 10ab, 11a–el, 12, 15a–d

4–9, 10cd, 11e–l, 12–14, 15b–e

6, 7, 8ef, 9, 10d, 11e–n, 12–14, 15c–f, 16

Identify if the lines are increasing or decreasing, from left to right. a

4 3

1 −4 −3 −2 −1

−1

−2

x 1

−4 −3 −2 −1

−1

x 1

−2

−3

−4

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113

Identify if the line segments have a positive or negative gradient. a

y 4

1 −4 −3 −2 −1

−1

x 1

−4 −3 −2 −1

−1

−2

−3

−4

y 9 8 7 6 5 4 3 2 1

−4 −3 −2 −1−1

3 2 1 −4 −3 −2 −1 x 1

−1

−2

−2 −3 −4

−3 −4

Order the gradients of these graphs from smallest to largest. A

4 3

1 −4 −3 −2 −1

114

x 1

Describe the gradient of these lines: a

−1

x 1

−4 −3 −2 −1

−1

−2

−3

−4

Mathspace Victorian Curriculum – Year 9 mathspace.co

x 1

y 4

1 −4 −3 −2 −1

x 1

−1

−4 −3 −2 −1

−2

Ex 2

x 1

−1

−2

−3

−4

For each of these intervals between points A and B: i

Find the rise.

iii

Find the gradient.

Find the run.

6 A

2 1

x 1

2 3 4

y 10

8 6

−9 −8 −7 −6 −5 −4 −3 −2 −1

14 12

−4 −3 −2 −1

5 6 7 8 9

4 2

x 1

−6 −5 −4 −3 −2 −1 −2 A

x 1 2 3 4

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−4 −3 −2 −1 A −2

2 1

−4 −6

−4 −3 −2 −1

−1

−10

−2

4 B

−3 −2

−1

x 1

−1

−2

−1

−3

y B

y 6 5 4 A

−8 −7 −6 −5 −4 −3 −2 −1 −1

−4 −3 −2 −1

−1

−3 −4

Sketch the line that passes through the given points.

Find the gradient.

(1, 3) and (0, 6)

(2, 1) and (8, −2)

(−1, 0) and (0, −5)

(−6, −6) and (6, 2)

Find the gradient of the line that passes through the given points: a

(6, 4) and (3, 4)

(1, −4) and (0, 0)

(2, −6) and (0, 0)

(−1, 0) and (0, 3)

(−4, 7) and (1, 10)

(−6, 5) and (4, 5)

Mathspace Victorian Curriculum – Year 9 mathspace.co

−2

1 2 3

−2

116

−4

Find the gradient of each line passing through points A and B: a

Ex 3

−3

−14

x 1

−8 −12

(10, 5) and (15, −5)

(−8, 20) and (−4, 10)

(−2, −5) and (−9, −12)

(−3, −1) and (−5, 1)

(−15, −35) and (−30, −25)

and

m Ex 4

and

 33 1 1 33 55 −−22, ,44 and88, ,−−88

What is the gradient of the line passing through A(5, −2) and (5, −3)?

A line is graphed where A (2, 0) and B (0, 4) both lie on the line.

y 4 B

As x increases, what happens to the value of y?

3 2 1 x

A −1

−1

A line is graphed where A (−5, 0) and B (0, 2) both lie on the line.

y 4

As x increases, what happens to the value of y?

3 2

1 x

A −5 −4 −3 −2

−1

1 −1

Find the gradient of each line. a

1 −4 −3 −2 −1

−1

−2 −3 −4

x 1

1 −4 −3 −2 −1

−1

x 1

−2

4.02 Gradient mathspace.co

117

1 −4 −3 −2 −1

−1

x 1

−2

−7 −6 −5 −4 −3 −2 −1

−3

x 1

3 2

−2

−4 −3 −2 −1

−3

−1

x 1

−2

−4

1 −1

4 2

−4 −3 −2 −1

−1

−2

−4

x 1

The line y = 5x + 2 is graphed: a

Find the y-value of the point on the line where x = 5.

Find the gradient of the line.

3 2 1 −4 −3 −2 −1

−1

−2 −3 −4

Reasoning 17

17, 19, 20

17–20

The points A (−11, −9), B (−5, 1) and C (−2, 6) are plotted on the Cartesian plane. a

Find the gradient of AB.

Find the gradient of BC.

Do the points A, B and C lie in a straight line? Explain your answer.

Explain why a vertical line has an undeﬁned gradient.

118

Mathspace Victorian Curriculum – Year 9 mathspace.co

x 1

A line is drawn on the Cartesian plane. Tom uses points A and C to ﬁnd the gradient of the line, and Johannes uses points B and C to ﬁnd the gradient of the line. Will their results be the same or different? Explain your answer.

3 2 B 1

−4 −3 −2 −1

−1

−2 −3

x 1

−4

Two gradients are calculated. • Line 1 gradient = 3 • Line 2 gradient = 5 Which line is steeper, (increases faster)? Explain why.

Problem-solving 21ab, 23–26 21

21cd, 22–24, 26, 27

Complete the coordinates using the points and gradients provided. a c

21bc, 23–26

(4, −3) and (1, ⬚), gradient = −2 (5, 3) and (⬚, 63), gradient = 4

b d

(5, 3) and (2, ⬚), gradient = −4

(11, ⬚) and (−20, 16), gradient =

The points A (26, m − 24), B (−1, m) and C (−10, 9) are plotted on the Cartesian plane. Find m, given that A, B and C all lie on the same line.

Pair up these points to create 3 lines, one with a positive gradient, one with a negative gradient and one with a gradient of 0. A (3, 5), B (4, 6), C (−1, 8), D (7, 5), E (−3, 10), F (4, 12)

Two lines L1 and L2 have equations y = x − 3 and y = −x + 5 respectively. The lines and their point of intersection have been graphed: a

When x = 4, ﬁnd the y-coordinate of the corresponding point on L1.

When x = 4, ﬁnd the y-coordinate of the corresponding point on L2.

Find the gradient of the two lines: i

Find the product of their gradients.

6 5 4 3 2 1 −3−2 −1 −1 −2 −3 −4 −5 −6

y L2

x 1 2 3 4 5 6 7 8 9 L1

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119

The graph illustrates Jay’s walk to school along a straight track: a b

Distance (m)

The line has a positive gradient, what does this mean in the context of the problem? What does the gradient context?

represent in this

250 200 150 100 50 Time (min) 1

The lines plotted here show the distance of an object in metres over time measured in seconds. i

10 9 8 7 6 5 4 3 2 1

Time (seconds) 1

iii

Distance (metres)

Time (seconds) 1

Distance (metres)

Time (seconds) 1

Distance (metres) 10 9 8 7 6 5 4 3 2 1

10 9 8 7 6 5 4 3 2 1

Distance (metres)

Time (seconds) 1

For each graph:

120

Is the gradient positive, negative or neither?

What is the value of the gradient?

What does this value mean in the context of the graph?

A car begins at the starting line of a drag race and travels 100 metres down the track in 5 seconds. a

Create a graph that represents this context.

Is the gradient of the line positive or negative?

Find the gradient of the line.

What does the gradient of the line represent?

Mathspace Victorian Curriculum – Year 9 mathspace.co

4.03

Distance between two points

After this lesson, you will be able to… • recognise the relationship between Pythagoras’ theorem and the distance formula. • calculate the distance between two points using Pythagoras’ theorem. • solve problems requiring the calculation of distances between points on the Cartesian plane. • apply the distance formula in special cases, such as horizontal or vertical lines.

Distances on the Cartesian plane Exploration B (−1, 8) 9

Triangle ABC is shown.

C (3, 8)

6 3 −4 −3 −2 −1 −3

x 1 2 3 4 5 6

−6 −9

2. Explain how to ﬁnd the distance from B to C. 3. How can we use the distance from A to C and the distance from B to C to ﬁnd the distance from A to B?

−12 −15 −18

1. Explain how to ﬁnd the distance from A to C.

A (3, −18)

We can use Pythagoras’ theorem to help us ﬁnd distances and lengths on the Cartesian plane. We already know that Pythagoras’ theorem can help us calculate the side lengths in a right-angled triangle. a2 + b 2 shorter side lengths

= c2 hypotenuse

The value c is used to represent the hypotenuse which is the longest side of the triangle. The other two lengths are represented by a and b.

4.03 Distance between two points mathspace.co

121

We can also use Pythagoras’ theorem to ﬁnd the distance between two points on the Cartesian plane.

3 2 1 −3 −2

−1

x 1

−1 A

−2

Draw the right-angled triangle where the interval is the hypotenuse. Then, we can use the horizontal and vertical distances as the lengths of the sides.

AC2 = AB2 + BC2

−3

Example 1 The points A(−3, −2), B(−3, −4), and C(1, −4) are the vertices of a right-angled triangle, as shown on the Cartesian plane.

y 2 1 −3

−2

−1

x 1

−1 A

−2 −3

−4 4

a What are the lengths of AB and BC?

Apply the idea Using the graph, AB is 2 units and BC is 4 units.

b Use Pythagoras’ theorem to ﬁnd the length of the interval AC to three decimal places.

Create a strategy Use Pythagoras’ theorem AC2 = AB2 + BC2 with the values we identified in part (a).

Apply the idea Let the length of AC be denoted by c. Use Pythagoras’ theorem Substitute a = 2 and b = 4 Evaluate the indices Evaluate the addition

122

Mathspace Victorian Curriculum – Year 9 mathspace.co

Take the square root of both sides Evaluate Since AC is denoted by c, the length of AC is 4.472 units.

Reﬂect and check When we want to know the exact distance, we leave the answer in square root form, so the exact distance between points A and C is units. Recall that there are two answers to c2 = 20: a positive result and a negative result.

Since lengths and distances are always positive, we only state the positive result.

Idea summary If the line is horizontal, the difference of the x-coordinates of the points is its distance. If the line is vertical, the difference of the y-coordinates of the points is its distance. c2 = a2 + b2

We can use Pythagoras’ theorem to ﬁnd the distance between any two points, by drawing a right-angled triangle where the interval is the hypotenuse. Then we ﬁnd the shorter side lengths and use Pythagoras’ theorem to solve for the hypotenuse. The result will be the distance between the two points.

5 4 b

3 c 2 1

a 1

x 4

The distance formula We can generalise the method for ﬁnding the distance between any two points, (x1, y1) and (x2, y2), using Pythagoras’ theorem. Create a right-angled triangle.

y (x2, y2)

We can label the hypotenuse as d, the horizontal side as a, and the vertical side as b. By Pythagoras’ theorem: d2 = a 2 + b 2

d (x1, y1)

4.03 Distance between two points mathspace.co

123

y (x2, y2)

The horizontal distance (a) is the size of the difference between the two x-coordinates. b = y 2 – y1

a = x2 − x1 The vertical distance (b) is the size of the difference between the two y-coordinates.

(x1, y1) a = x2 – x1

b = y2 − y1

Use Pythagoras’ theorem to ﬁnd the length of d:

This equation is known as the distance formula.

horizontal vertical distance distance (width) (height) d distance between the two points (x1, y1) coordinates of the first point (x2, y2) coordinates of the second point

Example 2 Find the distance between Point A (−2, −9) and Point B (2, −14), rounded to two decimal places.

Create a strategy Determine the coordinates, then use the distance formula.

Apply the idea The coordinates are A (−2, −9) and B (2, −14). Use the distance formula

Substitute (x1, y1) = (−2, −9), (x2, y2) = (2, −14)

Evaluate the brackets

Evaluate the indices

Evaluate the addition

Evaluate and round

124

Mathspace Victorian Curriculum – Year 9 mathspace.co

Reﬂect and check y −3 −2

−1

We could have found the answer by plotting the points on a Cartesian plane. Since the distance formula is derived from Pythagoras’ theorem, we will get the same answer.

x 1

−2

−4 −6 −8

−10

5 units

−12 −14 4 units

Idea summary The distance between two points (x1, y1) and (x2, y2) is given by:

d (x1, y1) (x2, y2)

is the distance between two points are the coordinates of the ﬁrst point are the coordinates of the second point

4.03 Practice questions Understanding 1 2

Explorer 1, 2ab, 3ab, 4

Adventurer 1, 2ac, 3ac, 4

Trailblazer 1, 2cd, 3cd, 4

What is the formula for ﬁnding the distance between two points on a Cartesian plane? Find the distance between the two points, A and B. a

5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3 −4 −5

A(2, 5) and B(2, 8)

x 1

5 4 3 A 2 1 −5 −4 −3 −2 −1 −1 −2 −3 B −4 −5

x 1 2 3 4 5

A(−7, −1) and B(−7, −6)

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125

Find the distance between the two points, A and B. a

y −6 −4 −2

x 2

8 10

−1

−2

−6 −5 −4 −3 −2 −1 −1 B −2

−3 A

y 2

−4

Ex 1

−4

A(3, 2) and B(9, 2)

A(−4, 5) and B(−7, 5)

Is each statement true or false? a

If two points have the same y-coordinate, subtract the smaller x-coordinate from the larger x-coordinate to ﬁnd the distance between them.

To ﬁnd the distance between two points on the same vertical line, calculate the positive difference between their x-coordinates.

If the distance between two points is zero, the points lie on the same horizontal line.

The formula for the distance between two points is derived from Pythagoras’ theorem. 5ab, 6ab, 7a, 8a, 9ab, 10, 11a–c, 12ab

Fluency 5

1 2 3 4

−3

−5

5ac, 6bc, 7a, 8b, 9bc, 10cd, 11cd, 12, 14

5cd, 6d, 7b, 8b, 9, 10, 11ef, 13–15

For each triangle: i

Find the length of interval PQ.

iii

Find the length of PR, rounded to two decimal places.

P(−6, 5), Q(−6, 2) and R(−2, 2)

Find the length of interval QR.

P(−1, 9), Q(−1, 2) and R(−5, 2)

6 P

5 3

−7 −6 −5 −4 −3 −2 −1

Mathspace Victorian Curriculum – Year 9 mathspace.co

1 x

126

−6

−5

−4

−3

−2

−1

2 1 x

P(−4, 5), Q(−4, 2) and R(4, 2)

P(−3, −5), Q(−3, −1) and R(1, −1)

−4 −3 −2

−1

−3

−4 −3 −2 −1

2 −2

3 Q

1 −11

−4

x 1

−1

−5

−6

Use Pythagoras’ theorem to ﬁnd the length of side AC, rounded to two decimal places. a

y A

8 7

3 2

x 1

y 4

2 3 4 5 6 7 8 9

y A

1 x

−14 −12 12 −10 10 − −8 8 − −6 6 − −4 4 − −2 C

A 1

−1 B −2

x −3 −2 2 −1 −1 −2 −2 B

−4 4

2 3 4 5 6

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127

Use Pythagoras’ theorem to ﬁnd the length of side AC, as an exact value. a

−7 −6 −5 −4 −3 −2 −1 −1 A B −2 − −3 −4 −5 − −6 − −7 −8 C −9

7 6 5

4 3 2 1

2 3 4 5 6 7 8 9

y x 1 2 3

The vertices of a triangle are listed. i

Sketch the triangle on the Cartesian plane.

Find the length of AC as an exact value using Pythagoras’ theorem.

A(−4, 3), B(−4, −4), C(2, −4)

A(6, 4), B(6, −2), C(−7, −2)

Find the distance of the given point P from the origin. a

P(12, 16)

24 21

P(7, −9)

−1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11

P(−5, −4)

x 1

2 3 4 5 6 7 8 9

y −6 −5 −4 −3 −2

−1 −1 −2 −3

Mathspace Victorian Curriculum – Year 9 mathspace.co

−7 −6 −5 −4 −3 −2 −1

x 2

P (−7, 24)

128

−4 −5

Use the distance formula to ﬁnd the length of the interval AC. Write your answer in exact form. a

y A

y 1

A −1 −1

−5 −4 −3 −2

−3

−4

C 1

y A 9

6 5 4 3 2 1

7 6 5 4

−4−3−2−1 −1 −2 −3 A −4

2 1 −2 −1

x 1

−5

Ex 2

−2

y C

x 1 2 3 4 5 6 7 8 9

Find the distance between Point A and Point B. Write your answer in exact form. a

A (1, 4) and B (7, 12)

A (4, 2) and B (−8, −7)

A (−5, 5) and B (3, −3)

A (−1, 9) and B (−4, 1)

A (1.4, −3.8) and B (6.2, −10.2)

and

Find the length of the segment indicated, rounded to two decimal places. A (−1, 5), B (3, 7), C (0, 6), D (−4, −5), E (−3, 0), F (2, −4)

Find the length of the segment indicated, rounded to two decimal places. A (0, 5), B (−5, 9), C (0, −2.3), D (−1.5, −5), a

, d

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129

Given P (4, 3), M (−3, −4) and N (−7, 1): a

Find the distance from P to M, rounded to two decimal places.

Find the distance from P to N, rounded to two decimal places.

Which point is further from P ?

For the points A (12, 3) and B (14, 0): a

Find the length of AB rounded to two decimal places.

If M is the midpoint of AB, ﬁnd the length of AM rounded to two decimal places.

Reasoning 16

16, 17

For the points M (−9, −1) and N (1, 5): a

Find the exact distance from M to the origin.

Find the exact distance from N to the origin.

Which point is closer to the origin? Explain your answer.

For the points A (−1, −7), B (4, −2) and C (0, −3), ﬁnd: a

The exact length of AC.

The exact length of BC.

What do you notice about the distance between the points?

Is C the midpoint of AB? Justify your answer.

Problem-solving 18, 19

20, 22–23, 25

20, 21, 24, 26–28

K is the midpoint of A (3, 1) and C (15, −7). Find the distance from A to K rounded to one decimal place.

A triangle has vertices at A (1, −1), B (−3, −4) and C (5, −4). a

Find the length of sides: i

b 20

iii

Is this triangle equilateral, isosceles or scalene?

Find the length in exact form: i

130

A triangle has vertices at A (9, −2), B (2, −8) and C (3, 5). a

iii

How can we show that the triangle has a 90° angle and two 45° angles?

Does the triangle have a 90° angle and two 45° angles?

Given the points P (10x, −6x) and Q (4x, 2x), where x > 0: a

Find the distance between P and Q in terms of x.

Find the coordinates of the midpoint of the segment PQ in terms of x.

Mathspace Victorian Curriculum – Year 9 mathspace.co

The points M(−3, −5) and R(4, −12) are the endpoints of an interval. N(−1, −7) is a point on this interval. a

Find the exact distance between the points: i

M and N

N and R

What is the ratio MN : NR?

Find the perimeter of the parallelogram whose vertices are A(3, 2), B(−2, 7), C(2, 9) and D(7, 4). Round your answer to one decimal place.

△PQR has vertices P(2, −6), Q(−9, −17) and R(−5, 1). a

Find the exact length of the sides: i

b 25

iii

Is the triangle right-angled?

Find the exact length of the sides: i

b c

iii

Classify the quadrilateral.

ABCD is a rhombus whose vertices are A(1, 2), B(3, 10), C(11, 12) and D(9, 4). Find: a

The exact length of diagonals:

i AC ii BD The exact area of the rhombus.

△PQR is isosceles. a

Find the area of the triangle.

Find the exact length of PR.

Find the exact value of d, the perpendicular from Q to the side PR.

y 11 P 10 9 8 7 6 5 4 3 2 1 Q

Find the exact length of the diagonals: i

P(−4, −4), Q(5, −7), R(9, −11) and S(0, −8) are the vertices of a quadrilateral. a

d R x 1 2 3 4 5 6 7 8 9 10 11

A circle with centre at point C(3, 4) has point A(−12, 12) lying on its circumference. Find: a

The radius of the circle.

The exact circumference of the circle.

The distance between point (20, 4) and the centre.

Does the circle also pass through the point (20, 4)? Explain your answer.

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INVESTIGATION

Transformations of linear graphs Investigate the effects of using transformations on the graph of a linear equation.

Objectives • To describe the transformations of a linear graph using transformation vocabulary • To write the equation of a line by comparing its transformations to the equation y = x

Transformations of linear graphs Explore transformations of a linear graph.

Interactive investigation Discover this concept in action online

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Vertical enlargement or compression When m is greater than 1, what type of transformation do you observe? When m is between 0 and 1, how does the graph of y = mx compare to y = x? Set m to a value greater than 1. As you increase the value of m, what happens to the steepness of the line? Reﬂection What happens to the graph when m is negative? Set m to values less than 0 and describe the effect on the graph’s direction with a negative gradient. Vertical translation Set m to 1 (so we have the line y = x + b), describe what happens to the line as you increase and decrease b. Can you create a line that is twice as steep as y = x and is shifted down 3 units? What are the values of m and b?

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Investigate 1.

Given the equation y = −2x + 3, describe all the transformations that have been applied compared to the equation y = x.

The graph of y = x, has been vertically compressed by a factor of 0.75 and shifted 2 units upwards. What is the resulting equation?

Given the equation y = −3x − 1, describe all the transformations that have been applied compared to the equation y = x.

The graph of y = x, has been vertically enlarged by a factor of 4 and then translated 2 units downwards. What is the resulting equation?

The graph of y = x, has been reﬂected in the horizontal axis, vertically enlarged by a factor of k and translated n units upwards. What is the resulting equation?

Discussion 1.

Imagine a linear graph that represents the relationship between hours studied and test scores. If the gradient (or slope) represents the eﬃciency of your study, would a steeper slope indicate more or less eﬃcient study habits?

If b represents your score without any study, what does a positive b value represent? What about a negative b value?

Create another real-world scenario that could match the relationship of a linear graph and describe how transformations applied to the graph could continue to represent the relationship.

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4.04 Gradient-intercept form After this lesson, you will be able to… • understand and apply the gradient-intercept form of a line. • explore the effects of changing the gradient and y-intercept on the equation and graph of a line. • interpret the significance of positive, negative and zero gradients. • determine the y-intercept of a line and understand its role in the equation. • derive the equation of a line from a graph.

The gradient-intercept form of a line Interactive exploration Discover this concept in action online

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All linear equations can be written in the form:

y = mx + c m the gradient c the y-value of the y-intercept For the gradient: • If m < 0, the gradient is negative and the line is decreasing. • If m > 0, the gradient is positive and the line is increasing. • If m = 0, the gradient is 0 and the line is horizontal. • The larger the magnitude of m, the steeper the line. For the y-intercept: • If c is positive, the line crosses the y-axis above the origin. • If c is negative, the line crosses the y-axis below the origin. An equation of the form y = mx + c is known as the gradient-intercept form of a line because we can easily identify both the gradient, m, and the value of the y-intercept, c.

Example 1 Find the equation of a line whose gradient is −8 and crosses the y-axis at −9. Express your answer in the form y = mx + c.

Create a strategy

Apply the idea

Substitute the gradient for m and the value of the y-intercept for c.

From the information, we have m = −8 and c = −9, so the equation is: y = −8x − 9

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Idea summary For all equations in gradient-intercept form, y = mx + c: • The value of m is the gradient. The larger the value of m, the steeper the line is. An increasing line has a positive gradient and a decreasing line has a negative gradient. • The value of c is the y-value of the y-intercept. The line passes through the point (0, c).

The equation of a line from a graph Exploration During a snowstorm, the initial depth of snow is 2 centimetres. The snow falls at a rate of 1.5 centimetres per hour. The graph and equation both represent this situation. Equation:

y, Depth (cm) 22 20 18 16 14 12 10 8 6 4 2

y = 1.5x + 2

x, Time (hours) 1 2 3 4 5 6 7 8 9 10 11 12131415

Choose any point on the line, substitute its coordinates into the equation, and evaluate.

Repeat with two more points on the line. What do you notice?

Choose any point not on the line, substitute its coordinates into the equation, and evaluate.

Repeat with two more points not on the line. What do you notice?

Substitute a value between 1 and 15 into the equation for x and evaluate.

Explain the relationship between the equation and the points on the graph.

To find the equation of a line from a graph we need the gradient and y-intercept. Find the gradient (m) of the line. Using the coordinates of any two points. Identify the y-intercept by finding where the line crosses the y-axis. The coordinates will be in the form (0, c), but we only need the y-value (c) to create the equation. Once we have identified these two features, we can write the equation of the line in the form y = mx + c. 4.04 Gradient-intercept form mathspace.co

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Example 2 Find the equation of the line graphed on the Cartesian plane.

y 7 6 5 4 3 2 1 −4 −3 −2 −1

−1

x 1

a What is the gradient of the line?

Create a strategy Find two points that lie on the line and use the gradient formula.

Apply the idea We can easily identify A (0, 4), and B (1, 1). Use the gradient formula

Substitute the coordinates

Simplify numerator and denominator

Evaluate

b What is the value of the y-intercept?

Create a strategy

Apply the idea

We can identify the y-intercept by ﬁnding where We can see that the line crosses the y-axis at 4. the line crosses the y-axis. The coordinates will be The value of y is 4 when x = 0. So c = 4. in the form (0, c) but we only need the y-value (c).

c Write the equation in gradient-intercept form.

Create a strategy

Apply the idea

Substitute the values found in parts (a) and (b) into the equation y = mx + c.

The equation is y = −3x + 4.

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Idea summary We can ﬁnd the equation of a line of the form y = mx + c by ﬁnding the gradient for the value of m, and the y-value of the y-intercept for the value of c.

4.04 Practice questions Understanding 1 2

Trailblazer 1–4

Describe the effect on the graph of the line, if m is: Positive

Negative

Zero

In each of the equations, identify the value of m and c. a

Adventurer 1–4

In the equation of a line, y = mx + c, what do m and c represent?

a 3

Explorer 1–4

y = 15x − 10

y = 7x

y=6−x

For the graph shown: a

What is the gradient of the line?

By how much does the y-value change as the x-value increases by 1?

10 9 8 7 6 5 4 3 2 1 −1 −1 −2

x 1

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5, 6a–e, 7, 9, 10, 12–14, 17, 19ab, 20a

Fluency 5

What is the y-intercept?

−4 −3 −2 −1

138

What is the gradient?

−4 −3 −2 −1

−2

−3

−4

6 5 4 3 2 1

x 1

−1

−2

−1

−6−5−4 −3−2 −1 −1 −2 −3 −4 −5 −6

5cd, 6d–g, 8, 11, 12c–f, 16, 18, 19cd, 20b

For each graph:

Ex 1

5bd, 6d–g, 7, 8ab, 9, 12c–f, 15–17, 19bc, 20b

4 3 2 1

x 1 2 3 4 5 6

–3 –2

–1

x 1

–2 –3 –4

Find the equation in the form y = mx + c, for: a

A line with a gradient 2 and y-intercept 5

A line with a gradient −6 and y-intercept 9

A line with gradient

and y-intercept 3

A line with gradient

and passes through the point (0, 3)

A line with gradient 8 and passes through the point (0, −4)

A line with gradient 0 and passes through the point

A line that has the same gradient as y = 9 − 8x and the same y-intercept as y = −5x − 3

Mathspace Victorian Curriculum – Year 9 mathspace.co

Ex 2

For each equation: i

What is the gradient, m?

What is the value of the y-intercept, c?

y = −2x + 9

y = −4x − 8

y = 6 + 8x

y = 3 (4x − 3)

By rewriting the equation in the form y = mx + c: i

What is the gradient, m?

What is the value of the y-intercept, c?

9x − y − 8 = 0

3y = 12x − 15

hat are the values of the gradient, m, and the W y-intercept, c?

Write the equation of the line in gradient-intercept form.

By substituting into your equation from part (b), find the value of y when x = 27

y 5 4 3 2 1 −2

−1

x 1

−1

−2

This graph shows the relationship between the time (x) in minutes and distance ( y) in kilometres of a high-speed train ride at a constant speed.

y (kilometres) 5

Find the equation of the line in gradient- intercept form.

Find how far the high-speed train travelled, y, after 29 minutes.

3 2 1 x (minutes) 1

This graph shows the relationship between the time (x) in seconds and distance ( y) in metres for a sprint at a constant speed that Jocel did. a

Find an equation of the line in gradient- intercept form.

By substituting into your equation from part (a), find how far Jocel ran, y, after 50 seconds.

10 9 8 7 6 5 4 3 2 1

y (metres)

x (seconds) 1 2 3 4 5 6 7 8 9 10 4.04 Gradient-intercept form mathspace.co

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Find the equation of the line in gradient-intercept form. a

y 1

−1

−2

−1

2 4 6 8 10 12 14 16 18

−1

−2

−3

−4

−5

y 2

1 x −1

−1

−2

−1

−2

−1

−3

−2

4 3

−2

1 −1

−3 −4

Find the equation in gradient-intercept form of a line that has a gradient of 5 and has each of these y-intercepts: a

y=3

y = −6

Find the equation in gradient-intercept form of a line that has a y-intercept of 3 and has each of these gradients: a

140

1 −2

−2

−1

m=4

Mathspace Victorian Curriculum – Year 9 mathspace.co

m = 0.6

A line has a gradient of −3 and crosses the y-axis at y = −2. a

Find the equation of the line in the form y = mx + c.

Does the point (−2, 4) lie on this line?

A line has a gradient of −3 and a y-intercept at (0, 8). a

Find the equation of the line in the form y = mx + c.

Does the point (8, −31) lie on this line?

For the two points given: i

Find the gradient of the line passing through the points.

Find the equation of the line in the form y = mx + c.

(0, 2) and (2, 6)

(0, −9) and (5, 1)

(0, 2) and (−7, 44)

(0, −3) and (1, 4)

Find the equations of these lines in the form of y = mx + c : a

Passes through the points A (0, 4) and B (3, 0)

Passes through the points A (0, −3) and B (9, 0)

Passes through the points A (0, −1) and B (−1, 0)

Passes through the points A (0, −9) and B (−45, 0)

Sketch on a Cartesian plane: a

The line with a y-intercept of 3 and gradient of

The line with equation y = 2x + 5

The line with a y-intercept of −2 and gradient of −3

The line with equation

Write the equation of the line in gradient-intercept form.

Find the y-coordinate of the y-intercept.

iii

Find the x-coordinate of the x-intercept of the line.

Sketch the line on a Cartesian plane.

A line that has gradient of −2 and passes through the point (3, −8).

A line that has gradient of

and passes through the point (−10, 4).

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Reasoning 21

21, 22, 26

22–23, 25, 26

24–27

Look at the three lines:

• Equation 1: y = x + 1 • Equation 2: y = 2x + 1 • Equation 3: y = 4x + 1 a

What do all the equations have in common?

What do all the graphs have in common?

Describe all lines that have the form y = mx + 1.

y=x+1

2 1

x −2

−1

−1 4x + 1 −2 y = 2x + 1

Look at the three lines: 8

• Equation 1: y = 2x + 4 • Equation 2: y = 2x + 8 • Equation 3: y = 2x − 4

What do all the equations have in common?

What do all the graphs have in common?

Describe all lines that have the form y = 2x + c.

4 y = 2x + 4

−8 −6 −4 −2 −2

x 2

−4 y = 2x − 4 −6 y = 2x + 8 −8

Given: • Line A: 5x + 3y + 5 = 0 • Line B: 7x + 6y − 3 = 0

Express both lines in the form y = mx + c.

Which line is steeper? Justify your answer.

Which of these lines is steeper? Justify your answer. Line A: 2x + 5y − 5 = 0 Line B: 4x + 4y + 1 = 0

The equation y = −3x + 5 represents a straight line. Is the graph of this line increasing, decreasing, vertical or horizontal? Justify your answer.

Explain the difference between the graphs of the equations y = 2x and y = −2x.

Is it possible for a vertical line to be represented by an equation in the gradient-intercept form y = mx + b? Why or why not?

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Problem-solving 28 28

28–30

28–32

A farmer noticed that the growth of his corn plants follows a linear relationship with the number of days since planting. He observed that on day 4, the plants were 8 cm tall, and on day 16, they were 36 cm tall. Find the equation of the line representing the relationship between the height of the corn plants ( y) and the number of days since planting (x).

For each equation: i

What is the gradient, m?

What is the value of the y-intercept, c?

x − y = 13

3x − 9y − 27 = 0

3x − 4y − 28 = 0

3x − 10y = −2

16x + 12y = 10

−9x + 9y = 27

A straight line has gradient −1 and goes through the points (0, 2) and (a, −6). a

Write the equation of the line in the form y = mx + b.

Find the value of a.

A car rental company charges a ﬁxed fee per day for renting a car, plus an additional fee for each kilometre driven. The company found that a customer who drove 30 kilometres in a day was charged $45, while another customer who drove 60 kilometres in a day was charged $75. Find the equation representing the cost ( y) as a function of the number of kilometres driven (x).

A gardener charges a callout fee of $85 plus $28 per hour. Write an equation to represent the total amount charged, y, by the gardener as a function of the number of hours worked, x.

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4.05 Linear graphs After this lesson, you will be able to… • use a table of values to sketch a linear graph. • identify x and y-intercepts and how to use them to sketch a linear graph. • use the gradient and a single point to sketch a linear graph. • apply knowledge of sketching linear graphs to solve problems.

Sketch a graph using a table of values We can use a table of values to sketch a graph. x y

This is a table of values. The ﬁrst row contains the independent variable (in this case, x). The second row contains the dependent variable, which is the variable that depends on the value of x.

To create a table of values, we will need to ﬁnd the y-value for each x-value in the table. We can do this by substituting each x-value into the equation.

Example 1 For the equation y = 2x − 4: a Complete the table of values. x y

Create a strategy Substitute each x-value into the given equation.

Apply the idea For x = 0, y=2×0–4 = −4

Substitute x = 0 Evaluate

y=2×1–4 = −2

Substitute x = 1 Evaluate

y=2×2–4 =0

Substitute x = 2 Evaluate

y=2×3–4 =2

Substitute x = 3 Evaluate

For x = 1,

For x = 2,

For x = 3,

x y

144

0 −4

1 −2

2 0

3 2

Mathspace Victorian Curriculum – Year 9 mathspace.co

b Using the table of values, plot the points onto a Cartesian plane.

Apply the idea 4

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

c Using the points plotted above, sketch the line that passes through two points.

Create a strategy Connect the plotted points.

Apply the idea 4

Reﬂect and check Notice that the line passes through all of the plotted points.

y = 2x − 4

2 1 −2 −1 −1

x 1

−2 −3 −4

Idea summary We can sketch a line by using a table of values. We do this by: • Completing a table of values by substituting the x-values, to ﬁnd the corresponding y-value. • Plotting the points onto a Cartesian plane. • Drawing a line through these points.

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Sketch a graph using its intercepts x and y-intercepts are often some of the easiest points to find from an equation to use to plot the line. • The x-intercept is where the line crosses the x-axis. It occurs when the y-value is 0. To find the x-intercept, set y = 0 in the equation and solve for x. • The y-intercept is where the line crosses the y-axis. It occurs when the x-value is 0. To find the y-intercept, set x = 0 in the equation and solve for y.

Interactive exploration Discover this concept in action online

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Example 2 For the linear equation y = 2x − 2: a What are the coordinates of the y-intercept?

Create a strategy

Apply the idea

Substitute x = 0 into the given equation.

y=2×0–2 = −2 The coordinates are (0, −2).

b What are the coordinates of the x-intercept?

Create a strategy Substitute y = 0 into the given equation.

Apply the idea Write the equation

Substitute y = 0

Add 2 to both sides

Evaluate

Divide both sides by 2

Evaluate

The coordinates are (1, 0).

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Substitute x = 0 Evaluate

c Sketch the line y = 2x − 2.

Create a strategy Plot the points found and draw a line through them.

Apply the idea 4

3 2

y = 2x − 2

1 −3 −2 −1

−1

x 1

−2 −3 −4

Example 3 For the equation 2x − 2y + 4 = 0: a Complete the table of values. x y

Create a strategy Because we need to substitute multiple values of x into the equation, start by rearranging the equation and then substituting each value.

Apply the idea Start by rearranging for the equation: 2x − 2y + 4 = 0

Write the equation

2x + 4 − 2y + 2y = 0 + 2y Add 2y to both sides of the equation 2x + 4 = 2y

Simplify

2x + 4 2 y = 2 2

Divide both sides by 2

x+2=y

Simplify

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Now we can use this equation to evaluate y for the x-values in the table. For x = 0, y=x+2 =0+2 =2 For x = 1, y=x+2 =1+2 =3 For x = 2, y=x+2 =2+2 =4 For x = 3, y=x+2 =3+2 =5 x y

0 2

1 3

2 4

3 5

b Using the table of values, plot the points onto a Cartesian plane.

Apply the idea 7

6 5 4 3 2 1 −4 −3 −2 −1 −1 −2

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x 1 2 3 4 5 6 7

c Using the points plotted above, sketch the line that passes through two points.

Create a strategy Connect the plotted points.

Apply the idea

Reﬂect and check 7

Notice that the line passes through all of the plotted points.

6 5 4 3 2 1 −4 −3 −2 −1 −1

2x − 2y + 4 = 0 x 1 2 3 4 5 6 7

−2

Idea summary To sketch a line using the x and y-intercepts: • Set y = 0 to ﬁnd the x-intercept. • Set x = 0 to ﬁnd the y-intercept. If the equation is in gradient-intercept form ( y = mx + c), we don’t need to let x = 0, because we know that the y-intercept is c.

Sketch a graph using its gradient and one point We can sketch a graph by identifying one point and its gradient. For example, if a line passes through the point (−2, −1) and has a gradient of 3: 4

First, we plot the point (−2, −1)

3 2 1 −3 −2

−1

x 1

−2 −3 −4

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We then use the gradient to ﬁnd the second point.

y 4 3

We know that the gradient is 3 or

2 run 1

in fraction

form. We also know that the gradient is x

–3 –2 –1 rise –1 –2

So the rise = 3 and the run = 1.

–3 –4

We then connect the two points with a line.

4 3 2 1 −3 −2

−1

x 1

−1

−2 −3 −4

We can also graph a line by identifying the gradient and the y-intercept, from the equation, when it is in the form y = mx + c. We know that the y-intercept occurs at c, and the gradient is m. y 6

For example, if we have the equation , we know the y-intercept is at (0, 3) and as the

gradient is , another point will be at

(0 + 5, 3 + 1) = (5, 4).

3 2 1

x 1

Note: if the gradient is negative, we will move down instead of up, when ﬁnding the second point.

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Example 4 Sketch the line that passes through the point (−1, 2) and has a gradient of

1 . 2

Create a strategy We can draw a line by plotting any two points on the line.

Apply the idea We have been given only one point, so we want to use the gradient to ﬁnd a second point. 1 A gradient of is already written as . The rise is 1 and the 2 run is 2.

y 5 4 3 2 1 −4 −3 −2 −1

−1

Starting with the point (−1, 2), we move two units to the right (run of 2) and then one unit up (rise of 1).

x 1

When we do that, we have the second point (1, 3). Using the two points, we can plot the line.

−2 −3

Example 5 Sketch the line that has a gradient of −3 and an x-intercept of −5.

Create a strategy Use the gradient to ﬁnd another point from the x-intercept.

Apply the idea 4

The x-intercept has coordinates of (−5, 0).

The gradient is

the run is 1. We can use this to ﬁnd another point on the line.

2 1 −5 −4 −3 −2 −1 −3

1 −1

so the rise is −3 and

x 1

By moving right 1 unit and down 3 units, we get to the point (−4, −3).

−2 −3 −4

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151

Now, we can draw our line through these two points.

3 2 1 −5 −4 −3 −2 −1

−1

x 1

−2 −3 −4

Idea summary To sketch a graph using its gradient and one point: • We plot the given point. • Use the gradient to ﬁnd the second point. • If the gradient is positive, we go up for the rise. • If the gradient is negative, we go down for the rise. • Connect the two points with a line. If we are given a line in gradient-intercept form y = mx + c remember that: • The y-intercept is (0, c). • The gradient is m.

4.05 Practice questions Understanding 1

Explorer 1–2

Adventurer 1–3

Trailblazer 1–3

When sketching the graph of a straight line using its intercepts, what are you speciﬁcally finding? A

The point(s) where the line is steepest.

The points where the line crosses the x-axis and the y-axis.

The point(s) where the line is farthest from the origin.

The point where the line crosses the x-axis.

Fill in each blank to make a true statement. To graph

, plot the coordinate (0, ⬚) on the y-axis, then rise ⬚ units up and run ⬚

units to the right which gives the point ⬚, then draw a line passing through the points ⬚ and (3, 6). 3

152

Explain how to find the x-intercept of the line −x − 8y = 8?

Mathspace Victorian Curriculum – Year 9 mathspace.co

4, 6, 7a–d, 10–12

Fluency Ex 1

5, 6, 7e–h, 10, 11cd, 12, 13

For each equation: i

Complete the table of values.

Sketch the graph of the line on a Cartesian plane.

y=x+9

y = 3x + 3

y = 3x − 4

y = −2x + 8

Complete the given table of values.

Sketch the graph of the line on a Cartesian plane.

y = −3x − 1

Ex 3

y = −x + 1

y = 2x − 9 0

−2

x x −3 0 3 6 y y

For each equation:

Ex 2

−1

x y

y = 4x

x x −3 0 3 6 y y

5, 8–10, 11cd, 13

For the equation y = −x − 2: a

Complete the table of values.

Sketch the line on a Cartesian plane.

Find the coordinates of the y-intercept.

Find the coordinates of the x-intercept.

Find the y-value when x = 3.

Find the x-value when y = −4.

x y

−2

−1

iii

Sketch the line.

For each equation: i

Find the x-intercept.

y=x+1

y = 3x − 1

y = 3x − 3

y = −2x + 4

Find the y-intercept.

For the equation 2x − 3y − 18 = 0: a

Complete the table of values.

Sketch the line on a Cartesian plane.

Find the coordinates of the y-intercept.

Find the coordinates of the x-intercept.

Find the y-value when x = 6.

y = −x + 4

y = −5x − 4

y = −3x + 6 x y

−3

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153

Ex 4

Ex 5

For each equation: i

Find the x-intercept.

3x + y − 1 = 0

9x + 2y + 18 = 0

−6x + 2y + 18 = 0

−20x + 5y − 40 = 0

iii

Sketch the line.

A line that passes through the point (−1, 2) and has a gradient of 4

A line that passes through the point (−2, 4) and has a gradient of −3

Sketch: a

A line that has a y-intercept of −2 and whose gradient is 4

A line that has an x-intercept of −2 and whose gradient is 3

A line that has a y-intercept of 2 and whose gradient is

A line that has a y-intercept of 7 and whose gradient is

For each equation: i

Find the gradient of the line.

Find the coordinates of the point on the line where x = 0.

iii

Sketch the line.

y = 3x − 3

y = −2x − 1

y=x−1

Sketch using the gradient and y-intercept: y = 4x − 1

Reasoning

14, 15

b 14, 15

14–16

Elisabeth and Arthur’s teacher asks them to sketch a graph that passes through the point (−1, 3) and has a gradient of

. Both students sketch different graphs.

Elisabeth’s graph 7

Arthur’s graph y

1 −4 −3 −2 −1 −1

x 1

Who was correct? Why?

154

Find the y-intercept.

Sketch these lines:

Mathspace Victorian Curriculum – Year 9 mathspace.co

1 −4 −3 −2 −1 −1

x 1

Does the graph of y = −4x + 1 rise from left to right? Why?

The equations of y = 7x − 1 and y = 7x + 3 have the same gradient but with different y-intercepts. What can you conclude about their graphs? Why?

Problem-solving 17, 18

17, 18

17–19

Mary is saving money to buy a new bike. She currently has $50 saved and plans to save $10 per week until she reaches her goal. Write an equation representing the relationship between the number of weeks (x) and the amount of money saved ( y). Graph the corresponding line and find how many weeks it will take for Mary to save $150.

For each equation:

Find the gradient of the line.

Find the coordinates of the point on the line where x = 0.

iii

Sketch the line.

6x − 3y + 18 = 0

−6x + 3y + 24 = 0

The perimeter of a rectangle is 10 cm. The length (L) is L cm. a

Write an expression for the width (W ) of the rectangle in terms of L.

Use a table of values to show the relationship between the length (L) and width (W ).

Sketch a graph to visually show the relationship.

Usually, when sketching graphs, there shouldn’t be any endpoints. Should this graph have any endpoints? Why?

4.06 Equation of a straight line After this lesson, you will be able to… • apply the point-gradient formula to find the equation of a line.

The point-gradient formula We know that to ﬁnd the gradient given two points, we use:

We can use one of the points, say (x1, y1), with the equation y = mx + c to ﬁnd c. y1 = mx1 + c where c = y1 − mx1

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155

If we put this in the equation y = mx + c, we get y = mx + y1 − mx1

Substitute c = y1 − mx1

y − y1 = mx − mx1

Subtract y1 from both sides

y − y1 = m (x − x1)

Factor out m

This gives us a new generalisation that we can use to ﬁnd the equation of any line with gradient m, and point (x1, y1). We call this the point-gradient formula.

Example 1 A line passes through point A (8, 2) and has a gradient of 2. Write the equation of the line in gradient-intercept form.

Create a strategy Use the point-gradient formula and rearrange the equation into gradient-intercept form.

Apply the idea y − y1 = m (x − x1)

Start with the point-gradient formula

y − 2 = 2 (x − 8)

Substitute the values for m = 2 and (x1, y1) = (8, 2)

y − 2 = 2x – 16

Expand the brackets

y = 2x – 14

Add 2 to both sides

Reﬂect and check It’s also possible to ﬁnd the equation another way, using just the gradient-intercept form. Let’s take a look: y = mx + c

Start with the gradient-intercept form

2 = 2 (8) + c

Substitute the values for m = 2 and (x, y) = (2, 8)

2 = 16 + c

Evaluate the multiplication

−14 = c

Subtract 16 from both sides

Since we now know the gradient and y-intercept, we can write the equation as y = 2x − 14.

Example 2 A line passes through the points (2, −7) and (−5, 7). a Find the gradient of the line.

Create a strategy Use the gradient formula,

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Apply the idea m=

y2 − y1 x2 − x1

7 − ( −7) −5 − 2 14 = −7 = −2 =

Write the gradient formula Substitute (x1, y1) = (2, −7) and (x2, y2) = (−5, 7) Evaluate the subtraction Evaluate

The gradient of the line is m = −2.

b Using the point-gradient formula, express the equation of the line in gradient-intercept form.

Create a strategy Use the point-gradient formula, y − y1 = m (x − x1), substitute the gradient and the coordinates of one point, then rearrange into the form y = mx + c.

Apply the idea y − y1 = m (x − x1)

Use the point-gradient formula

y − (−7) = −2 (x −2)

Substitute (2, −7) and m = −2

y + 7 = −2 (x − 2)

Evaluate the adjacent signs

y + 7 = −2x + 4

Expand the brackets

y + 7−7 = −2x + 4 −7 y = −2x −3

Subtract 7 from both sides Evaluate

Idea summary To ﬁnd the equation of a line given the gradient and a point, use the point-gradient formula:

y − y1 = m (x − x1) m (x1, y1)

is the gradient are the coordinates of the given point

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157

4.06 Practice questions Understanding 1

Explorer

Adventurer

1–4

Trailblazer

1–4

For each linear relationship, complete the statement: As x increases by 1, y changes by ⬚. a

y 4

1 −4 −3 −2 −1

−1

x 1

−4 −3 −2 −1

−2

Ex 1

−3

−3 −4

1 −8

(2, 7)

(12, 0)

y = −x + 6

x y

x+y=9

0 5

2 7

4 9

6 11

8 13

? c

y = −x − 6

Identify the gradient of the line.

Identify a point on the line.

y − 6 = −3 (x − 1)

y=x−6

For each equation:

Fluency

5a–e, 6, 7ac, 9ab

158

y=x+6

Which is the gradient-intercept form of the equation: x − y + 6 = 0? A

−1

−4

Does the point lie on the line a

x 1

−2

x −3 −2 −1 0 y 4 1 −2 −5

y+3=x+8

5b–f, 6, 7bc,9a–c

5e–j,6,8,9bd–f, 10, 11

Find the equation of each line. a

A line that passes through the point (−5, −4) and has a gradient of −4.

A line that passes through the point (−4, 3) and has a gradient of 4.

A line that passes through the point (7, −6) and has a gradient of −3.

A line that passes through the point (−4, 3) and has a gradient of −9.

A line that passes through the point (−6, −3) and has a gradient of −2.

A line that passes through the point

Mathspace Victorian Curriculum – Year 9 mathspace.co

and has a gradient of 2.

Ex 2

A line that passes through the point (4, 3) and has a gradient of

A line that passes through the point

A line that passes through the point (8, 1), and has a gradient of .

A line that passes through the point (−4, 5) and has a gradient of

and has a gradient of 7.

Find the equation.

Sketch the graph.

A line has gradient 2 and passes through the point (−5, −3).

A line has gradient

and passes through the point (−2, 2).

For each line described: i

Find the gradient.

Find the equation.

A line passes through the points (3, −3) and (5, −11).

A line passes through the points (−7, 1) and (5, −2).

A line passes through the points (−6, 7) and (−8, −4).

Find the equation of a line which has a gradient of answer in the form ax + by + c = 0.

For each table of values:

For each line described:

and crosses the x-axis at −10. Write your

Find the equation of the line in the form y = mx + c.

Complete the table.

x 1 2 3 4 y −3 −6 −9 −12

x y

−28

−4 −9

−2 −8

0 −7

2 −6

x −15 1 2 3 y 82 74 66

4 58

x y

−8 21

−4 13

3 −1

6 −7

x −3 −1 1 3 y 18 2 −14 −30

x y

−9 −45

−6 −30

3 15

11 55

Find the equation of the line that: a

Passes through the points (9, −5) and (0, 40).

Passes through the points (−2, −4) and (10, 20).

Passes through the points (10, 6) and (15, −24).

Passes through the points (−3, −13) and (9, −21).

Does the equation have a gradient of 7 and pass through the point (1, 4)? a

d 4.06 Equation of a straight line mathspace.co

159

Reasoning

12, 13

12–15

Is it possible to ﬁnd the equation of a line if you are only given the coordinates of intercepts? Explain.

In the point-gradient form of the equation of a line, y − y1 = m (x − x1), what would the formula become if the line is horizontal?

If an equation is given in point-gradient form, y − y1 = m (x − x1), how could you sketch the graph of the line?

Write down the equations of three lines that pass through the point (1, 3). Explain how your lines are different.

Problem-solving 17, 18

16–18

16–20

Find the equation of the line that passes through (−4, 6) and is parallel to the line 5x − 4y + 2 = 0.

The equation of a line is 2x + y − 8 = 0.

Find the x-intercept of the line.

Find the equation of a line with a gradient of −4 that passes through the x-intercept of the given line.

A line has a gradient of

and passes through the midpoint of A (−6, −6) and B (8, 8).

Find the coordinates of M, the midpoint of AB.

Find the equation of the line.

A circle with centre C (11, 13) has a diameter with endpoints A (5, 14) and B ( p, q).

Find the value of p.

Find the value of q.

Find the equation of the line passing through B with gradient .

The line 4x + y = 12 intersects the x-axis at P and the y-axis at Q. Let R be the origin. Find the area of △PRQ.

4.07 Parallel and perpendicular lines After this lesson, you will be able to… • understand the characteristics of parallel and perpendicular lines. • apply the understanding of parallel and perpendicular lines to find equations of such lines. • interpret and solve problems involving parallel and perpendicular lines.

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Parallel lines Lines that never intersect are called parallel lines.

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Parallel lines are lines that have the same gradient.

When two lines are parallel, they never cross and have no points in common.

3 2 1 −4 −3 −2 −1

mathspace.co

x 1

−1

−2 −3 −4

Let’s look at how we can identify parallel lines given their equations. Equation form

Characteristic of parallel lines

y = mx + c

Parallel lines have the same m value.

Examples y = 2x – 1 y = 4 + 2x

For every straight line, there exists an inﬁnite number of lines parallel to it. 4

Here is the line y = x.

y=x+3

Two more lines in the same family of parallel lines are also graphed:

3 y=x

y=x+3

1 −4 −3 −2 −1

−1

x 1

−2

y=x–4 Notice that they have the same gradient (m-value) but different x and y-intercepts (c-values).

−3 −4 y = x − 4

The gradient of any vertical line is undefined. But by definition, any two vertical lines are parallel.

4.07 Parallel and perpendicular lines mathspace.co

161

Example 1 Consider these questions about parallel lines: a Are the lines y = 3x + 2 and y = 3x – 5 parallel?

Create a strategy Lines are parallel when the gradients are the same. Identify the gradients for each equation. The equations are in the form y = mx + c. In this form, m represents the gradient.

Apply the idea y = 3x + 2, will have a gradient of 3. y = 3x – 5, will have a gradient of 3. So these lines are parallel.

b Which of these three lines are parallel? y = –3x + 2 –3y –x = 5 y + 3x = 7

Create a strategy Lines are parallel when the gradients are the same. The equations will need to be rearranged into the format of y = mx + c to compare the gradients.

Apply the idea y = –3x + 2 is in the form y = mx + c. It has a gradient of –3. –3y –x = 5 This needs to be rearranged. –3y –x = 5 –3y = x + 5 −3 y−3 yx=+ x5+ 5 = −3 −3 −3 −3 1 1 5 5 y =y−= −x3−x − 3 3 3

Write the equation Add x to both sides Divide both sides by –3 Write in gradient-intercept form

y + 3x = 7 This needs to be rearranged. y + 3x = 7 y = –3x + 7

Write the equation Evaluate

The lines that are parallel are y = –3x + 2 and y + 3x = 7 as both have a gradient of –3.

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Example 2 and passes through the point (9, −1).

Find the equation of the line that is parallel to

Create a strategy First, ﬁnd the gradient of the given line. Next, substitute the gradient and the given coordinates into the equation y = mx + c, and solve for c.

Apply the idea The gradient of

. Since the new line is parallel to the given line, the gradient of

the new line is also . Equation of a line in gradient-intercept form Substitute Substitute x = 9 and y = −1 Evaluate the multiplication Subtract 3 from both sides Substituting the values of m and c into the gradient-intercept form of a line, the equation is:

Reﬂect and check y −4 −3 −2 −1

−1

x 1

We can graph the given line with the line we found to see that they are parallel.

−2 −3 −4 −5 −6 −7 −8

Idea summary If two lines are parallel, they have the same gradient. If two lines have the same gradient, they are parallel. All vertical lines are parallel to each other.

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163

Perpendicular lines Lines that meet at right angles (90°) are called perpendicular lines.

Interactive exploration Discover this concept in action online

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Perpendicular lines have gradients whose product is equal to −1. For two perpendicular lines, y = m1 x + c1 and y = m2 x + c2: m1 × m2 = −1 We say that m1 is the negative reciprocal of m2. This means the two gradients have opposite signs and are reciprocals of each other. Here are some examples of negative reciprocals:

If the gradient of the ﬁrst line is m1, then we can ﬁnd the gradient m2 of a perpendicular line using the formula:

Special case: a vertical line and a horizontal line are perpendicular but do not have gradients which have a product of −1. This is because the gradient of a vertical line is undeﬁned.

Example 3 Find the equation of a line that is perpendicular to y = 6x + 10, and has the same y-intercept. Give your answer in the form y = mx + c.

Create a strategy Find the gradient of the perpendicular line and the y-intercept of the given line, then substitute both values into the equation y = mx + c.

Apply the idea The gradient of y = 6x + 10 is 6, and the negative reciprocal of 6 is perpendicular line is

164

, so the gradient of the

. The y-intercept is c = 10. So the equation of the line is:

Mathspace Victorian Curriculum – Year 9 mathspace.co

Reﬂect and check y 15 12 9

We can check that the product of the gradients is equal to −1, and we can graph both lines to see they are perpendicular.

y = 6x + 10

1 y = − x + 10 6

6 3 x −6

−3

Idea summary Two lines are perpendicular if their gradients are negative reciprocals of each other. To test if lines are perpendicular, multiply the gradients together. If the result is −1, then the lines are perpendicular. Vertical and horizontal lines are perpendicular to each other, even though their gradients do not have a product of −1.

4.07 Practice questions Understanding 1

Explorer 1–4

Adventurer 1–5

The equations y = 2x, y = 2x + 5 and y = 2x − 7 have been graphed on the same Cartesian plane: a

What do all of the equations have in common?

What do all the lines have in common?

Trailblazer 2–6 y 10 y = 2x 8 6 y = 2x + 5 4 2 −10 −8 −6 −4 −2 −2 −4 −6 −8 −10

2 4 6 8 10 y = 2x − 7

If two non-vertical lines are perpendicular, what is the product of their gradients?

4.07 Parallel and perpendicular lines mathspace.co

165

For the given graph: a

Which two lines are parallel?

Find two pairs of perpendicular lines.

y = 3x + 15

10 8 6 4 2

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

Complete the statement:

The x-axis and the y-axis are a pair of ⬚ lines. y = −1

166

y = 2x − 1

y=x

Any line parallel to this line will have a negative gradient.

Any line perpendicular to this line will have a positive gradient.

Any line perpendicular to this line will have a gradient of zero.

Any line perpendicular to this line will have a negative gradient.

1 y = − x−9 3

x=5

7–9, 10ab, 11–13, 14a–d, 15, 16, 18a

8–13, 14a–d, 15–18

8–11, 14–18

Which of these lines are parallel to y = 7x + 3? a

y = 7x − 3

y = 6x + 3

y = 7x + 4

y = 7x

y = 6x + 4

y = −7x + 3

y = 3x + 7

y = −3x − 7

Are these pairs of lines parallel? a

y = −3x − 2 and y = −3x + 9

y = −2x − 5 and y = −2x − 8

y = 7x + 8 and y = −5x + 8

y = 4x − 1 and y = 4x − 6

x = 4 and y = 5

y = 7x − 5 and y = −7x + 6

y = 3x − 5 and

y = −8x − 2 and 9x + 7

The line L passes through the point (9, −5) and is parallel to the line y = −5x + 2. a

Ex 2

1 y=− x 3

If the equation of a line is y = mx + c, and m > 0, which of the statements is true?

Fluency Ex 1

x 2 4 6 8 10

Which of these lines is parallel to the y-axis? A

y = 1.9x

Find the gradient of line L.

Find the equation of line L.

Find the equation of the line described: a

Parallel to the line y = −2x + 9 and passes through the point (−3, 1).

Parallel to the line y = 8x − 3 and crosses the y-axis at 5.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Parallel to the line y = −3x − 8 and crosses the y-axis at −4.

Parallel to the x-axis and passes through (−10, 2).

Parallel to the y-axis and passes through (−7, 2).

Find the gradient of the line perpendicular to: a

A line with gradient 6

c 12

y = −3x

Ex 3

y=8−x

3y − x = 5

x + 3y = −10

y = 9x

y = −9x + 5

9y = −x + 2

Are these pairs of lines perpendicular? a

y = x + 1 and y = x − 1

y = 3x − 7 and y = −3x + 6

y = −3x + 6 and

and

and y = −(3x − 27)

y = 3x − 5 and

and

The line L is perpendicular to y = 5x − 4 and crosses the y-axis at 1. a

Are these lines perpendicular to y = 9x + 2? a

A line with gradient

Are these lines perpendicular to y = 3x + 2? a

Find the gradient of line L.

Find the equation of line L.

Given the lines L1: 8x = −7y + 3 and L2: 7y − 8x − 7 = 0 a

Find the gradient of line L1.

Find the gradient of line L2.

Are the two lines perpendicular?

Given the lines L1 : 5x + 8y − 10 = 0 and L2 : a

Find the gradient of line L1.

Are the two lines parallel?

Find the gradient of line L2.

Find the equation of the line described: a

Perpendicular to

Perpendicular to y = −6x + 10 and has the same y-intercept.

Perpendicular to

Perpendicular to the x-axis and passes through (−8, −1).

Perpendicular to the y-axis and passes through (−8, −8).

and passes through (−8, 5).

and goes through the point (0, 6).

4.07 Parallel and perpendicular lines mathspace.co

167

Reasoning 19

19, 20

19, 21

19–21

A farmer wants to create an enclosure for his sheep. He puts fencing along these lines:

What is the shape of the enclosure? Justify your answer. 20

If every point on the line y = 2x + 2 is shifted 2 units up, ﬁnd the equation of the new line.

Every point on a particular line is shifted 6 units down. The resulting line has equation y = 2x − 2. Find the equation of the original line.

Problem-solving 22a, 23–25 22

22, 23, 25, 28, 30

22, 25–33

Each pair of lines are parallel. Find the value of b. a

0 = 5x − 9y + 6

y = bx + 4

0 = bx + 3y − 2 23

The line L goes through (−9, 10) and (2, −1). a

Find the gradient of line L.

Find the equation of the line that passes through (10, 8) and is parallel to line L.

The line L goes through (1, 1) and (3, 13). a

Find the gradient of line L.

Find the equation of the line that has a y-intercept of (0, −4) and is perpendicular to line L.

A construction company is building a new road that runs parallel to an existing road. If the equation of the existing road is y = 0.5x + 3, ﬁnd the equation of the new road that passes through the point (2, 6).

The line L goes through (3, 2) and (−2, 4). Find the equation of the line that has a y-intercept of 1 and is parallel to line L.

Each pair of lines are perpendicular. Find the value of k. a

5x + 4y + 8 = 0

−6x + 6y − 8 = 0

kx + 9y − 8 = 0 kx + 10y + 8 = 0 28

168

The line L is perpendicular to y = 10x + 8 and passes through the point of intersection of the lines y = x + 5 and 8x − 10y + 60 = 0. a

Find the gradient of line L.

Find the point of intersection of y = x + 5 and 8x − 10y + 60 = 0.

Find the equation of the perpendicular line L.

Mathspace Victorian Curriculum – Year 9 mathspace.co

The line y = 2x + 4 is perpendicular to the line cx − 4y + 7 = 0. Find the value of c.

The 4 vertices of square ABCD have been plotted on a Cartesian plane. a

Find the gradient of side AB.

Find the gradient of side BC.

Find the product of the gradients in parts (a) and (b).

Are sides AB and BC perpendicular?

y 7 6 A

5 4 D

2 C

1 1

The points P(0, −2), Q(−3, −2), R(0, 4) and S(2, 2) are graphed on the Cartesian plane:

y 4 R

Find the gradient of PQ.

Find the gradient of RS.

Are PQ and RS parallel?

Are QR and PS parallel?

What type of quadrilateral is PQRS?

x 5

−4 −3 −2 −1 Q

−1

S x 1

−2 P −3 −4

The vertices of a quadrilateral are A(2, 2), B(6, 4), C(8, 10), and D(4, 8). a

Find the gradient of each side.

Is quadrilateral ABCD a parallelogram?

Is quadrilateral ABCD a rectangle?

A quadrilateral ABCD has been graphed on the Cartesian plane. a

Find the gradient of each side: i

iii

What type of quadrilateral is ABCD? Explain your answer.

y x

−1 1 2 3 4 5 6 7 8 9 −1 D −2 −3 −4 A −5 C −6 −7 −8 B −9

4.07 Parallel and perpendicular lines mathspace.co

169

4.08 Direct proportion After this lesson, you will be able to… • define what is directly proportional relationship. • determine the constant of proportionality in a direct proportion. • identify a proportional relationship in a graph or table of values. • solve problems involving directly proportional quantities.

Direct proportion Two quantities are said to be directly proportional if an increase in one leads to an increase in the other, or if a decrease in one leads to a decrease in the other. This means that they change at the same rate. For example, if you earn $18 per hour, your total income depends on the number of hours you work. The more hours you work, the more money you make. We can write this as: Income = $18 × hours worked or y = 18x where y is your income, and x is the number of hours worked. The constant of proportionality is the number that links the two quantities together. In this case, it’s 18, because for every hour you work, you earn $18. Notice that this constant is the same as the slope or gradient of the line when you graph the relationship. For directly proportional relationships, the starting point ( y-intercept) is always zero, meaning if you work 0 hours, you earn 0 dollars. The general equation for any directly proportional relationship is:

y = kx k is the constant of proportionality The constant of proportionality is also called the unit rate. While it can be any number except zero, we usually focus on positive values for k.

Interactive exploration Discover this concept in action online

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Example 1 Consider the values in each table. Which of them could represent a directly proportional relationship between x and y? Table A: x y

170

1 2

Table B: 2 8

3 18

4 32

Mathspace Victorian Curriculum – Year 9 mathspace.co

x y

1 2

2 4

3 6

4 8

Create a strategy Determine which table shows that the y-values are a constant multiple of the x-values.

Apply the idea Table A x

Table B x

2 3 4 ×2 ×4 8 18 32

Multiplier has changed not proportional

2 3 4 ×2 ×2 ×2 ×2 2 4 6 8

Multiplier has stayed the same proportional

Example 2 Consider the equation P = 90t. a State the constant of proportionality.

Create a strategy

Apply the idea

For equations in the form y = kx, k is the constant of proportionality.

The constant of proportionality in P = 90t is 90.

b Find the value of P when t = 2.

Create a strategy

Apply the idea P = 90t

Substitute the given value into the equation.

Write the equation

= 90 × 2

Substitute t = 2

= 180

Evaluate

Example 3 Oliver is making cups of fruit smoothie. The amount of bananas and strawberries he uses is shown in the proportion table. Strawberries Bananas

8 5.5

16 11

24 16.5

32 22

40 27.5

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a Sketch the graph of this proportional relationship.

Create a strategy Use two pairs of values from the table as coordinates of points in the graph.

Apply the idea 11 10 9 8 7 6 5 4 3 2 1

The points (8, 5.5) and (16, 11) have been plotted, and a line drawn through them.

Bananas

Strawberries 2

10 12 14

Reﬂect and check Conﬁrm that the other points (24, 16.5), (32, 22), and (40, 27.5) lie on the same line you have created.

b What is the unit rate of this relationship?

Create a strategy The unit rate is the gradient or constant of proportionality.

Apply the idea 11 10 9 8 7 6 5 4 3 2 1

5.5 units

8 units

Strawberries 2

172

We can see in the graph that rise = 5.5 and run = 8.

Bananas

10 12 14

Mathspace Victorian Curriculum – Year 9 mathspace.co

c Select ALL the statements that describe the proportional relationship. A

For every 5.5 bananas Oliver uses, he adds 8 strawberries.

The unit rate of bananas in respect to strawberries is

For every 8 bananas, William uses 5.5 strawberries.

The unit rate of bananas in respect to strawberries is

Create a strategy Use the table given in the question, and the unit rate from part (b).

Apply the idea From the table, when William uses 5.5 bananas, he adds 8 strawberries. So option A is correct. The unit rate is

bananas per strawberry. So option B is correct.

Reﬂect and check We can also see in the graph that for every 5.5 bananas Oliver uses, he increases or adds 8 strawberries.

Bananas

11 10 9 8 7 6 5 4 3 2 1

5.5 units

This also shows why option A is correct.

8 units

Strawberries 2

10 12 14

Idea summary A directly proportional relationship is a linear relationship with a y-intercept of zero:

y = kx k is the constant of proportionality

4.08 Practice questions Understanding 1

Explorer 1–3

Adventurer

Trailblazer

1–3

Find the gradient of each linear relationship represented in the table. a

x y

1 3

2 6

4 12

8 24

x y

0 14

2 9

4 4

6 −1

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Which graphs indicate that y is directly proportional to x? a

y 4

1 −4 −3 −2 −1

−1

x 1

−4 −3 −2 −1

−2

−3 −4

y 4

174

−4 −3 −2 −1

−1

x 1

−2

−3

−4

Is each statement true or false? a

In a directly proportional relationship, the constant of proportionality represents the gradient of the line.

The graph of any directly proportional relationship will go through the point (0, 0).

The graph of a directly proportional relationship can be curved.

A directly proportional relationship can be increasing or decreasing.

The equation

Fluency 4

−2

−4

−1

x 1

−1

−3

−4 −3 −2 −1

represents a direct proportionality between two quantities y and x.

4, 6, 7ab, 8, 10, 12, 14ac

5, 6, 7bc, 9–12, 14bc

Given that y = 2x, ﬁnd the missing values in the table.

Given that y = 0.4x, ﬁnd the missing values in the table.

Mathspace Victorian Curriculum – Year 9 mathspace.co

5, 6, 7cd, 9–11, 13, 14bd, 15

x y

Ex 1

Which could represent a directly proportional relationship between x and y? a

x 1 3 5 7 y 20 16 12 8

x y

1 16

5 12

6 8

20 4

x 1 2 3 4 y 5 20 45 80

x y

1 5

2 10

3 15

4 20

Find the equation relating the given variables in each table. a

a 0 1 2 3 b 0 9 18 27

x y

0 −10

4 −2

9 8

16 22

s 0 5 10 15

t 8

Ex 2

Ex 3

q is directly proportional to r and q = 10 when r = 20. a

Graph the relationship between q and r.

Find the gradient of the line.

Express q in terms of r.

Find the value of q when r = 30.

Find the value of r when q = 22.

a is directly proportional to b and a = 54 when b = 9. a

Graph the relationship between a and b.

Find the gradient of the line.

Express a in terms of b.

Find the value of a when b = 7.

Find the value of b when a = 60.

Given A = 70w: a

Find the value of A when w = 4.

What is the constant of proportionality?

Given

What is the constant of proportionality?

Find the value of M when n = 50.

William is making rice. The amount of rice and water he uses is in the table. Rice (g) Water (mL)

100 210

110 231

120 252

130 273

140 294

Sketch the graph of the proportional relationship.

Find the unit rate of the relationship between rice and water.

Write two statements that describe the proportional relationship.

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The table shows the proportional relationship between hours spent ﬁshing and the number of ﬁsh caught:

Hours Fish caught

4 9

8 18

12 27

Sketch the graph of the proportional relationship.

Find the unit rate of the relationship.

By considering parts (a) and (b), make a conclusion about the number of ﬁsh caught.

For each table: i

Sketch the graph.

x 5 10 15 y 9.5 19 28.5

x 15 25 35 45 y 36 60 84 108

25 47.5

x y

10 7.5

20 15

30 22.5

40 30

55 132

x y

1 1.25

3 3.75

5 6.25

7 8.75

20 38

Find the constant of proportionality, k.

Reasoning

16, 17

50 37.5 9 11.25

when x = 3:

If y and x are directly proportional, and a

Find the constant of proportionality.

Write the equation for y in terms of x.

16, 17

16, 18

Two quantities are directly proportional. If one of the quantities increases by a factor of 7 (is multiplied by 7), what happens to the other quantity?

Jimmy and Ben are both saving money for a vacation. Jimmy can represent his savings over , where x represents the number of days and y represents the

time using the equation

savings. Ben saves $7 every 4 days. a

Complete the table showing Ben’s savings over time. Days Amount saved

12 21

Graph the amount each of them have saved over time.

Explain how to determine who is saving more per day.

A real estate agent advertised a house for sale online and in the newspaper. The agent received 4 online enquiries each day and 2 enquiries each day through the newspaper advertisement. a

176

8 14

Let y represent the number of online enquiries received x days after initially advertising the house. Write the equation for y based on x. The line representing the number of online enquiries has been graphed. On the same set of axes, graph the line representing the number of newspaper enquiries after x days.

y 40 35 30 25

Online

20 15 10 5

x 2

Why is the line representing the number of online enquiries steeper than the line representing the number of newspaper enquiries?

Why do both lines pass through the origin?

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Problem-solving 19–22

20–24

20, 23–26

Write an equation involving a constant k such that m varies directly with p.

The revenue of a company, r, varies directly with the number of sales, n. Write an equation involving a constant k.

Paul paints 15 plates every 6 hours. a

Complete the proportion table: Hours worked Plates painted

b 22

6 15

12 20

24 40

36 60

48 80

Complete this table for Harry: Number of weeks Number of books read

Harry and Carl love reading. Harry reads 16 books every 12 weeks. Carl has kept a table of his reading habits:

Sketch the graph of this proportional relationship.

Number of weeks Number of books read

12 30

12 16

48 32

60 80

Who reads the fastest?

Derek and Yuri are building robot kits. Derek can make 2 robots every 3 hours. Yuri can make 3 robots every 6 hours. a

Sketch the graphs of both proportional relationships on the same Cartesian plane.

Who is the faster builder?

The original of a printed image measures 8.5 cm in width and 51 cm in length. When a customer wants to print a copy of the original, they are offered prints in various sizes, but the width and length are in the same ratio as the original so that the photo does not appear distorted. a

Find the constant of proportionality in the equation length = k × width.

Find the length of a copy if the width of the copy is 14 cm.

The height, h, of a regular tetrahedron varies directly with its side length, s. A particular tetrahedron with a side length of 3 cm has a height of 2.45 cm. a

Find the constant of proportionality, k, to two decimal places.

Express h in terms of s.

Find the height of a different tetrahedron with a side length of 2 cm.

The amount A (in grams) of a substance dissolved in water is directly proportional to the capacity C (in litres) of water used. When 5 litres of water is used, 15 grams of the substance is dissolved. a

Find an equation relating the amount and capacity.

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4.09 Model linear equations After this lesson, you will be able to… • identify and differentiate between independent and dependent variables in linear equations. • model real-world situations using linear equations. • apply the concept of slope and y-intercept in the context of linear equations. • solve problems using linear equations.

Application of linear equations We can combine all the skills we have seen related to linear functions and apply them to real-world situations. These often involve the collection of data, which is commonly displayed in a table of values. The independent variable is the variable you can control or change, and is displayed in the ﬁrst row of the table, followed by the dependent variable, which is affected by this change, in the second row. Once we know that the relationship between the variables is linear, we can work out the gradient and y-intercept and express the relationship as a linear equation (or function).

Interactive exploration Discover this concept in action online

mathspace.co

We can identify relevant information through various features of the graph: • The slope of the line, expressed as a rate, will be the change in the dependent variable, for each unit change in the independent variable. • The y-intercept will be the initial value of the dependent variable.

Example 1 Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol in dollars: Number of litres (x) Cost of petrol ( y)

0 0

10 16.40

20 32.80

30 49.20

40 65.60

a How much does petrol cost per litre?

Create a strategy The cost per litre is a rate of change; this is the same as calculating the gradient. Divide the cost of the petrol ( y) by the number of litres (x).

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Apply the idea Divide 16.40 by 10

Evaluate

b By using your answer from part a, write an equation linking the number of litres of petrol pumped (x) and the cost of the petrol ( y).

Create a strategy

Apply the idea

The cost of petrol is equal to the cost per litre times the number of litres pumped.

We know the cost of petrol is $1.64 We can write the equation as y = 1.64x.

c How much would 47 litres of petrol cost at this unit price?

Create a strategy

Apply the idea

Using the equation from part (a), subtitute x = 47.

y = 1.64x

Write the equation

= 1.64 × 47

Substitute x = 47

= $77.08

Evaluate

d In the equation, y = 1.64x, what does 1.64 represent? A

The unit rate of cost of petrol per litre.

The number of litres of petrol pumped.

The total cost of petrol pumped.

Create a strategy

Apply the idea

By looking at part (a), 1.64 represents the rate at which y increases.

The correct answer is option A: The unit rate of cost of petrol per litre.

Example 2 Kerry currently pays $50 a month for her internet service. She is planning to switch to a ﬁbre optic cable service. a Write an equation for the total cost T of Kerry’s current internet service over a period of n months.

Create a strategy To ﬁnd the total cost, we need to multiply the cost of internet a month by the number of months.

Apply the idea Total cost = T, Cost of internet per month = $50, Number of months = n T = 50n

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b For the ﬁbre optic cable service, Kerry pays a one-off amount of $1200 for the installation costs and then a monthly fee of $25. Write an equation of the total cost T of Kerry’s new internet service over n months.

Create a strategy The $1200 represents an initial value, which can be represented with a y-intercept (c). To form the full equation, we multiply the monthly fee by the number of months, and then add it to the installation fee.

Apply the idea Total cost = T, Monthly fee = $25, Number of months = n, Installation fee = $1200 T = 25n + 1200

c Fill in the table of values for the total cost of the current internet service, given by T = 50n. n

T (dollars)

Create a strategy Substitute the value of n in T = 50n.

Apply the idea For n = 1, T = 50 × 1 = 50

Substitute n = 1 Evaluate

T = 50 × 6 = 300

Substitute n = 6 Evaluate

For n = 6,

For n = 12, T = 50 × 12 = 600

Substitute n = 12 Evaluate

For n = 18, T = 50 × 18 = 900

Substitute n = 18 Evaluate

For n = 24, T = 50 × 24 = 1200

Substitute n = 24 Evaluate

The complete table of values for the total cost of the current internet service: n T (dollars)

180

1 50

6 300

12 600

18 900

Mathspace Victorian Curriculum – Year 9 mathspace.co

24 1200

d Fill in the table of values for the total cost of the ﬁbre optic cable service, given by T = 25n + 1200. n

T (dollars)

Create a strategy Substitute the value of n in T = 25n + 1200.

Apply the idea For n = 1, T = 25 × 1 + 1200 = 1225

Substitute n = 1 Evaluate

T = 25 × 6 + 1200 = 1350

Substitute n = 6 Evaluate

For n = 6,

For n = 12, T = 25 × 12 + 1200 = 1500

Substitute n = 12 Evaluate

For n = 18, T = 25 × 18 + 1200 = 1650

Substitute n = 18 Evaluate

For n = 24, T = 25 × 24 + 1200 = 1800

Substitute n = 24 Evaluate

The complete table of values for the total cost of the ﬁbre optic internet service: n T (dollars)

1 1225

6 1350

12 1500

18 1650

24 1800

Idea summary • The independent variable is displayed in the ﬁrst row of the table, followed by the dependent variable, in the second row. • The slope is the constant rate of change. If the x-value is increasing by 1, then it will be the increase in the y-values. • The y-intercept is the value of y when x = 0. This may not actually be given in the table, so you may have to count backwards using the rate of change.

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4.09 Practice questions Understanding 1

When 4 is added to a certain number, the answer is 6.

Subtracting 10 from a certain number gives a result of 9.

Eleven times a certain number is 88.

A certain number divided by 5 gives a result of 25.

When 13 is subtracted from four times a certain number, the result is 21.

When 11 is added to eight times a certain number, the answer is −5.

When a certain number is subtracted from 20 and this result is then multiplied by 6, the result is −10.

The sum of −1 times a certain number and 4 is divided by 7, which gives a result of 9.

Which is shown ﬁrst in a table of values, the independent or the dependent variable? 3, 4, 7–10

3–5, 7–9, 11

1 7.6

2 14.4

3 21.2

Write a linear equation to represent the relationship between the cost of a call, y, and the length of the call, x.

Find the cost of a 6-minute call.

In a study, scientists found that the more someone sleeps, the quicker their reaction time. This table displays the ﬁndings: 0 6

1 5.8

2 5.6

3 5.4

4 5.2

5 5

How much does the reaction time decrease for each extra hour of sleep?

Write a linear equation relating the number of hours of sleep, x, and the reaction time, y.

Calculate the reaction time for someone who has slept 4.5 hours.

Calculate the number of hours someone sleeps if they have a reaction time of 5.5 seconds.

A racing car starts the race with 150 L of fuel. From there, it uses fuel at a rate of 5 L per minute. a

Complete this table of values: Number of minutes passed (x) Amount of fuel left in the tank ( y)

182

3, 5–8, 11, 12

The table shows the linear relationship between the length of a cellphone call and the cost of the call:

Number of hours of sleep (x) Reaction time in seconds ( y)

Trailblazer 1–2

Length of call in minutes (x) Cost in dollars ( y)

Adventurer 1–2

For each statement, write a linear equation using x to represent the unknown.

Fluency Ex 1

Explorer 1–2

Mathspace Victorian Curriculum – Year 9 mathspace.co

Write a linear equation linking the number of minutes passed, x, and the amount of fuel left in the tank, y.

How many minutes will it take for the car to run out of fuel?

A diver starts at the surface of the water and begins to descend below the surface at a constant rate. This table shows the depth of the diver, in metres, over 4 minutes: Number of minutes passed (x) Depth of diver in metres ( y)

7 Ex 2

0 0

1 1.4

2 2.8

3 4.2

4 5.6

Calculate the increase in depth each minute.

Write a linear equation for the relationship between the number of minutes passed, x, and the depth, y, of the diver.

Calculate the depth of the diver after 6 minutes.

Calculate how long the diver takes to reach 12.6 m beneath the surface.

Some friends spent a total of $5.00 on stickers and badges at a store. A sheet of stickers costs $0.50 and a badge costs $1.50. a

Write a linear equation that represents the relationship between the number of sheets of stickers, x, and the number of badges, y, the friends bought for $5.00.

If the friends bought 2 badges, how many sheets of stickers did they buy?

The amount of medication M (in milligrams) in a patient’s body gradually decreases over time t (in hours) according to the equation M = 1050 − 15t. a

After 61 hours, how many milligrams of medication are left in the body?

Calculate the number of hours it will take for the medication to be completely removed from the body.

A carpenter charges a callout fee of $150 plus $45 per hour. a

Write a linear equation to represent the total amount charged, y, by the carpenter as a function of the number of hours worked, x.

Find the total amount charged by the carpenter for 6 hours of work.

A particular restaurant has a ﬁxed weekly cost of $1300 and receives an average of $16 from each customer. a

Write an equation to represent the net proﬁt, y, of the restaurant for the week as a function of the number of customers, x.

Find the restaurant’s net proﬁt if it has 310 customers for the week.

Mario is running a 100 km ultramarathon at an average speed of 9 km/h. a

Write a linear equation to represent the distance Mario has left to run, y, as a function of the number of hours since the start, x.

Find the distance Mario will have left to run after 4.5 hours.

Mohamad is taking his new car out for a drive. He had only driven 50 kilometres in it before and is now driving it down the highway at 75 km/h. a

Write a linear equation to represent the total distance, y, that Mohamad had driven in his car as a function of the number of hours, x.

Find the total distance Mohamad will have driven in his car if his current drive begins at 5:10 p.m. and ﬁnishes at 7:25 p.m. 4.09 Model linear equations mathspace.co

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Reasoning 13

13, 14, 16

Paul has just purchased a prepaid phone, which he intends to use exclusively for sending text messages, and has purchased some credit along with it. After sending 11 text messages, he has $34.39 of credit remaining and after sending 19 text messages, he has $30.31 of credit remaining. The relationship between the amount of credit remaining, y, as a function of the number of text messages, x, is y = −0.51x + 40.

Explain what the slope of the line represents in this context.

Explain what the y-intercept represents in this context.

Find the amount of credit Paul will have left after sending 36 text messages.

A baseball is thrown vertically upward by a baseball player when they are standing on the ground, and the velocity of the baseball, V (in metres per second), after T seconds is given by V = 120 − 32T. When T = 4, there is a negative value of V. Explain what this means in context.

The table shows the water level of a well that is being emptied at a constant rate with a pump: Time in minutes (x) Water level in metres ( y)

2 26.8

5 25

8 23.2

The linear equation which represents the water level, y, as a function of the minutes passed, x, is y = −0.6x + 28

Explain what the slope of the function represents in this context.

State the y-intercept.

Explain what the y-intercept represents in this context.

After Anri starts running, her heartbeat increases at a constant rate. The equation which describes this relationship can be written as y = 3x + 49, where x represents the number of minutes passed, and y represents the heartbeat. a

Explain what the value of 49 represents in this context.

Explain what the slope represents in this context.

Problem-solving 17–19 17

184

18, 19, 21

17, 19–21

A mobile phone salesman earned $600 in a particular week during which he sold 26 phones, and $540 in another week during which he sold 20 phones. a

Write a linear equation to represent the relationship between the weekly earnings of the salesman, y, and the number of phones sold, x.

Find how much the salesman will earn in a week during which he sells 36 phones.

The cost of a taxi ride is given by C = 5.5t + 3, where t is the duration of the trip in minutes. a

Calculate the cost of an 11-minute trip.

For every extra minute the trip takes, how much more will the trip cost?

What could the constant value of 3 represent in this context?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Lucas is creating digital pixel art. On the ﬁrst day of his project, he created an artwork with a size of 8 megabytes. Inspired by minimalist art, he decided that each subsequent pixel art will be 0.5 megabytes smaller than the previous one.

What will be the size of the pixel art he creates on the 10th day?

Write a formula to represent the size, S, of the pixel art Lucas creates on the Dth day.

On which day will his digital pixel art’s size be just 1 megabyte?

A house is running a smart speaker in their household for eight hours every day of the year. The price of a smart speaker is $50 and they cost R dollars per hour to run.

Write an equation for the total cost (C dollars) of running this smart speaker for one year in terms of R.

Find the value of R if the total cost of running the smart speaker for one year is $650. Round your answer to two decimal places.

If the smart speaker is permanantly left on (i.e. 24 hours a day), show that the cost does not triple.

Claude is 32 years old and keeps active by running at their local park. a

Use this formula to ﬁnd their maximum heart rate in beats per minute (bpm). Maximum Heart Rate = 220 − Age in Years

Write a linear equation to represent the maximum heart rate, m, as a function of age, a.

Claude will get the most beneﬁt from this exercise if their heart rate is between 50% and 85% of their maximum heart rate. Between what two heart rates should Claude be aiming for to get the most beneﬁt from this exercise?

4.10 Linear inequalities After this lesson, you will be able to… • plot linear inequalities on a number line. • interpret worded problems into inequalities and solve them. • check whether a particular value satisfies a linear inequality. • solve real-world problems involving linear inequalities.

Solve linear inequalities Suppose we want to plot the solutions to the inequality 2 (3 + x) < 8 on a number line. 2 (3 + x) < 8 3+x<4 x<1

Divide both sides by 2 Subtract 3 from both sides

To test if x < 1 is correct, we can substitute some values into the original equation to check. Let’s check with 2 values: one less than 1 and one larger than 1. • When x = 0, we have 2 (3 + x) = 2 (3 + 0) = 6, which is less than 8. • When x = 2, we have 2 (3 + x) = 2 (3 + 2) = 10, which is not less than 8.

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To plot the solutions on a number line, we use a hollow circle for the endpoint (since the problem uses a “<” symbol, x = 1 is not included in the solution set): −8

−6

−4

−2

We solve inequalities just like we solve equations, except that the inequality sign flips direction if multiplying or dividing by a negative. When plotting inequalities: Symbol < and > ≤ and ≥

Endpoint Doesn’t include Does include

On a plot

Key words “less/greater than” “less/greater than or equal to”

●

“at least/at most” “no less/more than”

Example 1 Consider the inequality 3x + 1 > 4. a Solve the inequality.

Create a strategy Use inverse operations to ﬁnd the values of x that will satisfy the inequality.

Apply the idea Subtract 1 from both sides

Evaluate the subtraction

Divide both sides by 3

Evaluate

Reﬂect and check Test a value which satisﬁes our solution by substituting it into the original inequality. For example, choose x = 3 3×3+1>4 10 > 4 This is true, suggesting the solution is correct.

b Plot the solutions to the inequality 3x + 1 > 4 on a number line.

Create a strategy Plot the solution from part (a).

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Apply the idea To graph the inequality x > 1, plot a hollow circle on point 1 with a ray towards positive inﬁnity. −5

−4

−3

−2

−1

Example 2 Solve: 15 > 5 (2 − a)

Create a strategy Use inverse operations to isolate a on one side of the inequality.

Apply the idea 15 > 10 − 5a

Expand the brackets

5 > − 5a

Subtract 10 from both sides

−1 < a

Divide both sides by −5 (remembering to flip the inequality symbol)

a>−1

Make a the subject

Reﬂect and check Test a value which satisﬁes our solution by substituting it into the original inequality. For example, choose a = 0 15 > 5 (2 − 0) 15 > 10 This is true, suggesting the solution is correct.

Example 3 Consider this situation: “2 less than 4 groups of p is no more than 18”. a Construct and solve the inequality described above.

Create a strategy Translate the sentence into mathematical symbols, then solve the inequality by isolating x on one side.

Apply the idea The phrase “is no more than” has the same meaning as “is less than or equal to” (≤), “less than” means substracted from, and “groups of” means a multiplication operation.

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Write the inequality Add 2 to both sides Simplify Divide both sides by 4 Simplify

b What is the largest value of p that satisﬁes this condition? A

p=5

p = −5

There is no largest value.

p=4

Create a strategy Determine the largest value of the inequality from part (a) by recalling the deﬁnition of inequality symbols.

Apply the idea Using a number line, we can plot p ≤ 5 as: −5

−4

−3

−2

−1

The inequality p ≤ 5 means that p can take any value that is less than or equal to 5. So the largest value of p that can satisfy the inequality is 5. The correct answer is option A.

Idea summary We solve inequalities just like we solve equations, except that the inequality sign flips direction if multifying or dividing by a negative. When plotting inequalities: Symbol < and > ≤ and ≥

Endpoint Doesn’t include Does include

On a plot ●

Key words “less/greater than” “less/greater than or equal to” “at least/at most” “no less/more than”

Check solutions to a linear inequality In some cases, we don’t need to solve for all of the possible solutions. We may just want to know if a particular value will satisfy an inequality (if it is in the solution set). We can do this by substituting the value into the inequality and evaluating to see if the statement is true.

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Example 4 Does t = 6 satisfy the inequality 3t − 5 < 12?

Create a strategy Substitute t = 6 into the inequality, then evaluate.

Apply the idea Evaluate the left side of the inequality. 3t − 5 = 3 × 6 − 5

Substitute t = 6

= 18 − 5

Evaluate the multiplication

= 13

Evaluate

To test, we check if 3t − 5 < 12, when t = 6. This results to 13 < 12 which is false. The statement is false, so t = 6 does not satisfy the inequality.

Idea summary To decide if a particular value will satisfy an inequality (if it is in the solution set), substitute the value into the inequality and evaluate to see if the statement is true.

4.10 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

For each inequality: i

x ≥ 29

Plot the inequality on a number line.

Determine if each value satisfies them (makes them true). i

x = 27

x < −29

x = −28

iii

x = 29

x = −30

What happens to the inequality symbol when we multiply or divide both sides of an inequality by a negative number?

Is each statement true or false? a

The symbols ≤ and ≥ don’t include the endpoint, which we show with a hollow circle.

The symbols < and > don’t include the endpoint, which we show with a hollow circle.

The symbols < and > do include the endpoint, which we show with a ﬁlled-in circle.

The symbols ≤ and ≥ do include the endpoint, which we show with a ﬁlled-in circle.

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Fluency 4 Ex 1

5 Ex 2

5, 6a, 7d–l, 8–10, 12, 15, 17, 19, 20

6, 7g–p, 8, 9, 11, 14–16, 18–20

Solve: a

6x − 54 ≥ 30

21 + 7v ≥ 21

8.4 < 1.9c + 2.7

3x + 27 > 3

4 (b + 3) ≤ 12

3.3g + 4.9 ≥ 34.6

24.15 > 1.5s − 5.55

19.8 > 33 (0.4y + 1.208)

0.25 (x − 8.2) + 1.13 ≤ 1.13

Solve: a

8−y<6

4≤1−x

−9 − e > 4

3 − 2m < 5

−7b − 7 ≥ 7

−2h + 8 > − 14

−1.2m − 34.33 > 7.19

2.1 (0.37 − 4.63x) ≤ 10.5

−1.2(u + 3.7) ≥ −7.2

Solve: a

For each inequality: i

Solve.

Plot the solutions to the inequality on a number line.

3 < 4x − 1

7v + 4 > v + 16

0.75b − 1.3 ≤ 0.8

m Ex 3

4, 5a–c, 7a–e, 8–10, 12, 13, 17, 18, 20

−4y + 9 > 9

7.6 < 9.1 − 0.2s

3 (j − 1) < 9

−7a − 5 ≤ 2

2m + 6 > 30 − m

5.5a − 13.7 < −11.5

12.2y ≥ 15y − 14

Construct and solve an inequality for this situation: “The sum of 2 groups of x and 1 is at least 7.”

For “5 more than half of q is greater than 13”. a

Construct and solve the inequality.

Find the greatest value of q that satisﬁes this condition (if it exists).

Sophia has a budget for school stationery of $32, but has already spent $15.25 on books and folders. Write an inequality that shows how much Sophia can spend on other stationery, where p represents the amount she can spend, and solve for p.

Xavier struggles to save money and spent $33.40 last week. He needs $32.00 for a movie with friends on the weekend. Write an inequality representing the amount Xavier had at the beginning of the week, where A is the initial amount in his bank account, and solve for A.

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To get a grade of C, John needs a total score of at least 300 over four exams. He scored 63, 73, and 98 on the ﬁrst three exams. If x represents what he must score on the last exam to get a C or better, write an inequality and solve for x.

Skye was given $72 for a birthday present. This present, along with earnings from a Saturday job, is being set aside for a mountain bike. The job pays $5.00 per hour, and the bike costs $379. a

Write an inequality to ﬁnd the minimum number of hours that Skye needs to work to afford to buy the mountain bike, where h represents the number of hours worked.

Solve for h. Round your answer to one decimal place.

If Skye can only work her job for a whole number of hours, ﬁnd the minimum number of hours she must work to afford her bike.

Lachlan is planning on going on holiday. He has saved $2118.40, and spends $488.30 on his airplane ticket. a

Let x represent the amount of money Lachlan spends on the rest of his holiday. Write an inequality to represent the situation.

Solve for x.

Find the highest amount that Lachlan could spend on the rest of his holiday.

James wants to order some books from an online bookstore. Each book costs $13.30 and shipping for the entire order is a ﬂat rate of $29.05. James can spend no more than $74.00. a

If B represents the possible number of books that James can buy, write an inequality and solve for B. Round your answer to one decimal place, if necessary.

Find the maximum number of books that James can afford to buy.

A farmer wants to construct a rectangular enclosure alongside a barn. One side of the enclosure will be the length of the barn, which measures 36 metres. The farmer has a maximum of 80 metres of fencing material available for the project. If the width of the roped area is W, form an inequality and solve for the range of possible widths.

Ex 4

For 21 + 7x > 21: Do these values of x satisfy the inequality 21 + 7x > 21? a

x=1

x = 14

x=7

x = −6

x=6

x=0

x = 12

x = 6.8

Do these values of x satisfy the inequality 3.3 − 5.2x < − 3.2x − 7.5? a

Do these values of x satisfy the inequality 6 (x + 2) ≤ 48? a

x=0

x = 5.4

x = −5.5

When breeding certain types of fish it is recommended that the number of female fish is more than double the number of male fish. This can be expressed as the inequality f > 2m, where f is the number of female fish and m the number of male fish. Determine whether these values satisfy the recommendation. a

f = 10, m = 8

f = 23, m = 9

f = 13, m = 10

f = 7, m = 4

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Reasoning

21, 22

21–23

21–24

Explain the differences between the inequalities x < 4 and 0 ≤ x ≤ 4.

Isabelle tried to solve the inequality but made a mistake in her work.

25 − 2x < 31

Step 1:

2x < 6

Step 2:

x<3

Determine which step is incorrect and explain the error.

Write the correct working to solve the inequality.

Percy tried to plot the solution to the inequality 4x + 28 ≥ − 8 on a number line, however his answer is incorrect. −10 −9 −8 −7 −6 −5 −4 −3 −2 −1

Identify the error(s) and explain how to rectify them. 24

Ryan wants to save up enough money to buy a new sports equipment set, which costs $40.00. Ryan has $22.10 that he saved from his birthday. To make more money, he plans to wash neighbours’ windows for $2 per window. The expression which represents the number of windows, w, required to be washed is w ≥ 8.95. Determine whether each statement is correct. Explain your reasoning. a

Ryan must wash more than 9 windows to be able to afford the equipment.

Ryan must wash at least 8 windows to be able to afford the equipment.

If Ryan washed 8 windows and 95% of another window, he could afford the equipment.

The number of windows Ryan must wash to be able to afford the equipment must be greater than or equal to 9.

Problem-solving 25, 26 25

25–27

Jimmy spends $120 to produce some rockmelons. He is able to sell each one for $9. a

If n represents the number of rockmelons he must sell to make a proﬁt, write an inequality and solve for n.

Find the least number of rockmelons that Jimmy must sell to make a proﬁt.

The table shows the times of six swimmers in a 1500 m race. Only those who achieve a time of 22 minutes or below qualify for the final. Use the inequality t ≤ 22, where t is the time of each swimmer, to ﬁnd all of the swimmers who qualify for the ﬁnal.

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Swimmer James Harry Lachlan Adam Glen Kenneth

Time (minutes) 21 24 22 27 31 23

Neville is saving up to buy an LED TV that is selling for $950. He has $650 in his bank account and expects a nice sum of money for his birthday next month. a

If the amount he is to receive for his birthday is represented by x, write an inequality that models the situation where he is able to afford the LED TV.

How much money would he have in total if his parents were to give him $310 for his birthday?

Would he have enough to buy the plasma TV if his parents were to give him $310 for his birthday?

A particular bus can carry a maximum weight of 4300 kg on board. If the average adult passenger weighs 75 kg and the driver weighs 81 kg, then the number of passengers on the bus at any one time must satisfy 75a + 81 ≤ 4300, where a represents the number of adult passengers. a

There are 58 adults already on board and 8 more at the bus stop waiting to get on. Find the combined weight of everyone on board if the driver lets all of the new passengers on.

Can the driver safely let all the passengers waiting at the bus stop to get on board?

A particular beef mince brand must contain between 10% and 15% fat before it is shipped. To make 1500 grams of mince which is acceptable to be shipped, the processors mix w grams of 18% fat mince with mince containing 3% fat. Write an inequality for the number of grams of fat in the mix, and solve this to ﬁnd the least and greatest acceptable values of w.

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4 Chapter review 1

Find the coordinates of the midpoint M of segment AB. a

A (−5, −1) and B (3, 5)

A (−5, 3) and B (1, −7)

y 7 6 5 4 3 2 1

x 1 2 3 4

1 2

A (−2, 2) and B (−2, −2)

A (−4, 9) and B (5, 9)

B (4, 8) and M (−1, 7)

B (−4, −9) and M (2, −3)

Describe the gradient of the lines as negative, zero, positive, or undefined. a

y 2

−2 −1

−1

−7 −6 −5 −4 −3 −2 −1

x 1

−1

−3

−4

−5

−6

−7

4 3

1 −1

x 1

4 3

−2 −3

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−4

x 1

−2

−4 −3 −2 −1

194

−8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8

If the midpoint of points A and B is M, what are the coordinates of A? a

Find the midpoint of segment AB. a

3 2 1

−6 −5 −4 −3 −2 −1 −1 A −2 −3

x 1

Find the gradient of the line that passes through the given points: a

(5, 1) and (2, 7)

(−9, −5) and (−9, −3)

and

(−1.6, 8) and (0.2, 8)

Complete the coordinates using the points and gradient provided.

(3, −3) and (5, ⬚), gradient = −3

Use Pythagoras’ theorem to find the length of side AC, rounded to two decimal places.

2 1 1

1 1

−2

3 −3

−5 5 − −4 4 − −3 3 − −2 2 − −11 −1

−2

−3 −2 −1 −1

−3

−4 4

−4

Use the distance formula to find the length of the interval AB. Write your answer in exact form. a 6

−4 −3 −2 −1 −1

2 1 −4 −3 −2 −1 −1

x 1

−3

−4

−2

x 1

−5

Find the perimeter of the parallelogram whose vertices are A(−5, 1), B(−14, 3), C(−17, −1) and D(−8, −3). Round your answer to one decimal place.

For each equation: i

What is the gradient, m?

What is the value of the y-intercept, c?

y = −4x + 12

5x − 2y = 17

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hat are the values of the gradient, m, W and the y-intercept, c?

y 5

Write the equation of the line in gradient-intercept form.

Find the value of y when x = 20

4 3 2 1 −2

−1

x 1

−2

For the equation y = −2x + 5: a

Complete the table of values.

Sketch the graph of the line on a Cartesian plane. x y

−3

−1

Sketch the line that has a y-intercept of 3 and whose gradient is .

For each equation:

Find the x-intercept.

iii

Sketch the line.

y = 4x − 6

3x − 5y = 15

y = −3x + 5

y = 3x + 2

3x + y = −5

y = 3x

4y = x

Are these lines perpendicular to y = −4x + 1? a

Find the y-intercept.

Are these lines parallel to y = 3x + 5? a

x+5

y = −4x + 5

x−2

Find the equation of the line described. a

Parallel to y = −5x + 4 and passes through the point (−1, 3).

Perpendicular to y = 2x + 7 and passes through the point (6,−3).

This pair of lines is parallel. Find the value of g. 0 = 6x − 5y − 2 0 = gx + 2y − 1

196

Find the equation of each line. a

A line that passes through the point (−7, −13) and has a gradient of 4.

A line that passes through the points (−9, 5) and (−12, 14).

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Find the coordinates of M, the midpoint of AB.

Find the equation of the line.

Could the table represent a directly proportional relationship between x and y? If yes, write the equation relating x and y. a

and passes through the midpoint of A (5, 4) and B (−1, −10).

A line has a gradient of

Find the constant of proportionality, k.

Write the equation for y in terms of x.

0 0

25 $1

50 $2

75 $3

3 27

4 48

100 $4

How much does frozen yogurt cost per gram?

How much would a cup of frozen yogurt that weighs 315 g cost?

If a person paid $11.24 for their cup of frozen yogurt, how much frozen yogurt did they have in their cup?

A driver starts driving 150 km from home. They are driving at a rate of 80 km per hour. Complete this table of values: 0

0.5

1.5

Write a linear equation linking the number of hours passed, x, and the number of kilometres left to drive, y.

How many minutes will it take for the car to reach home?

Solve: a

7x − 28 ≥ 35

8.4 > 1.9c + 2.7

−8x + 64 < 24

Do these values of x satisfy the inequality 1 + 11x > 15? a

2 12

Number of hours passed (x) Number of kilometres left to drive ( y)

1 3

A self-serve, frozen yogurt company charges customers by the amount of frozen yogurt in grams. The table of values shows the costs of various weights.

x y

when x = 4:

If y varies directly with x, and

Amount in grams (x) Total cost ( y)

x 1 3 5 7 y 2 6 10 14

x=0

x = −2

x=4

Jacob spends $100 to buy some pineapples. He sells each one for $9. a

If p represents the number of pineapples he must sell to make a profit, write an inequality and solve for p.

What is the whole number of pineapples that Jacob must sell to make a profit?

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Big ideas • Understanding quadratic relationships requires identifying their key features and exploring different representations of functions that highlight various characteristics of the function. • Quadratic equations can have zero, one, or two solutions, which correspond to the x-intercepts that can be solved graphically or algebraically. • Changes in the equation of a function result in specific graphical transformations.

Curriculum Outcomes VC2M9A05 identify and graph quadratic functions, solve quadratic equations graphically and numerically, and use null factor law to solve monic quadratic equations with integer roots algebraically, using graphing software and digital tools as appropriate VC2M9A06 use mathematical modelling to solve applied problems involving change, including financial contexts involving simple interest; formulate problems, choosing to use either linear or quadratic functions or other simple variations; interpret solutions in terms of the context; evaluate the model and report methods and findings VC2M9A07 experiment with the effects of the variation of parameters on graphs of related functions, using digital tools, making connections between graphical and algebraic representations, and generalising emerging patterns

Quadratics Chapter outline 5.01 5.02 5.03 5.04 5.05 5.06 5.07 5.08 5.09 5.10 5.11

Quadratic relationships Graphs and features of quadratics Investigation: Transformations of quadratics Basic quadratics Vertical translations Horizontal translations Turning point form Factorised form Standard form Solutions to quadratics Solve algebraically Solve by factorisation Investigation: Other non-linear graphs Chapter 5 review

200 209 221 223 230 236 242 254 263 271 280 286 293 295

Quadratics can model real-life situations, like the projectile motion of a basketball flying through the air!

5.01 Quadratic relationships After this lesson, you will be able to… • define and recognise a quadratic equation and its standard form. • distinguish quadratic equations from non-examples. • identify quadratic relationships from their equations, graphs, and tables of values. • define what is a parabola and its relationship to quadratic equations. • determine whether a relationship is linear, quadratic, or neither, based on a graph or a table of values.

Quadratic equations Equations that have terms where the highest power of x is 2, and all powers are positive integers, are called quadratic equations. A quadratic equation can appear in many different forms but the standard form is y = ax2 + bx + c where a ≠ 0 Examples of quadratic equations y = 5x2 y= 2x + 7 − 18x2 = 0 −2(x − 3)2 + 7 = 0 Non-examples y=

Reason Not a quadratic as

Not a quadratic as there is a square root

Has a degree of 3, not 2

y=x +x

Example 1 Which two of these equations represent a quadratic relationship between x and y? A y = 1 − x2

y = (x2 − 1)3

E y = 2x2

Create a strategy

Apply the idea 2

Choose options that have x as the highest power. The correct answers are option A: y = 1 − x2 and 2 Remember that quadratic equations do not have option E: y = 2x . terms in x with negative indices or under square root signs.

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Idea summary A quadratic equation contains a term of x with a power of 2, but no higher. A quadratic equation can appear in many different forms but the standard form is y = ax2 + bx + c, where a ≠ 0.

Quadratic relationships We can identify a quadratic relationship from its equation, its table of values, and by the shape of its graph. The most basic quadratic equation is y = x2. We can create a table of values for this equation and then plot the points to see the general shape. The curve that is formed is known as a parabola. x y

−3 (−3)2 = 9

−2 (−2)2 = 4

(−3, 9)

−1 (−1)2 = 1

y (3, 9)

8 7 6 5 (−2, 4)

(2, 4)

3 2 (−1, 1)

−5 −4 −3 −2 −1 −1

(1, 1) x (0, 0) 1 2 3 4 5

0 0 =0 2

1 1 =1 2

2 2 =4 2

3 3 =9 2

This graph shows a quadratic relationship. • The graph is U-shaped. In this case, it is open upwards. We call this concave up. • The graph is symmetrical about a vertical line. In this case, the graph is symmetrical about the line x = 0, the y-axis. We call this line the axis of symmetry. • It has a turning point (a vertex) which could be a minimum or maximum value for y. In this example, the turning point is (0, 0) and we can see that there are no values of y below this. This is a minimum turning point.

We can also identify a quadratic relationship by looking at the table of values. A reminder that with linear relationships there is a constant ﬁrst difference for evenly spaced x-values. The ﬁrst difference is the difference between consecutive y-values.

Linear relationship x

−3

−2

−1

−2

−5

First Difference 7 − 10 = −3 4 − 7 = −3 1 − 4 = −3 −2 − 1 = −3 −5 − (−2) = −3 −8 − (−5) = −3

−8 In linear relationships, the first difference is constant. 3

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The second difference is the difference between consecutive ﬁrst differences. If the x second difference is a non-zero −3 constant, we have a quadratic −2 relationship. −1 Let’s look at a table of values for 0 y = x2. 1

y 9 4 1 0 1

2 3

4 9

Quadratic relationship First Difference Second Difference 4 − 9 = −5 1 − 4 = −3 0 − 1 = −1 1−0=1

−3 − (−5) = 2 −1 − (−3) = 2 1 − (−1) = 2

4−1=3 9−4=5

5−3=2

3−1=2

In quadratic relationships, the second difference is constant. We can see the ﬁrst differences are: −5, −3, −1, +1, +3, +5. The ﬁrst differences have a difference of 2 each time, this is the second difference.

Example 2 Are the graphs of the relationships shown linear, quadratic or neither? a 4

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

Create a strategy Linear forms a line, quadratic forms a parabola (U-shape) and other shapes are neither.

Apply the idea

Reﬂect and check

The graph is an upside down U-shape, has a maximum value of y = 4, and is symmetrical, so is quadratic.

We could make a table of values, and examine how the y-values are changing for a consistent change in x.

x y

−3 4

−2 1.5

−1 −1

0 −3.5

1 −6

2 −8.5

3 −11

Create a strategy

Apply the idea

Determine if the table of values has constant ﬁrst differences (linear) or non-zero constant second differences (quadratic) or neither.

The y-values are decreasing by a constant value of 2.5 for each consecutive increase in x by 1. Therefore, the table shows a linear relationship.

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Example 3 Complete this table that represents a quadratic relationship: x y

−3

−2 0

−1 −3

0 −4

1 −3

3 5

Create a strategy Determine if the minimum or maximum value is shown in the table. If it is, we can use the property of symmetry to help complete any missing values. To find any remaining values, we want to find the first and second differences and use them to complete the table.

Apply the idea In this case since the y-values of x = 1 and x = −1 are equal, the minimum value occurs at the x-value directly between −1 and 1. The minimum value occurs at (0, −4). Since the quadratic is symmetrical about this point, we can use this fact to ﬁnd other points. We are given the point (−2, 0). Using symmetry, we know (2, 0) must also be a point. For the same reason, we know when x = −3, the output is 5. x y

−3 5

−2 0

−1 −3

0 −4

1 −3

2 0

3 5

To find the last two values, we want to find the first and second differences. x

−3 −2

−1

−3 −4 −3

−5 −3

−1

+3 +5

+2 +2 +2 +2 +2 This means the second difference is +2, so we can now ﬁnd the remaining values. When x = 4, the output will be 5 + 2 = 7 more than the output for when x = 3: 5 + 7 = 12 The output for x = 5 will be 7 + 2 = 9 more than the previous output. 12 + 9 = 21 x y

−3 5

−2 0

−1 −3

0 −4

1 −3

2 0

3 5

4 12

5 21

Reﬂect and check In this case the quadratic equation describing this relationship is y = x2 − 4. We will learn techniques to derive this a little later, but for now you can test against the table of values.

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Idea summary • Quadratic relationships have non-zero, constant second differences. • The graph of a quadratic relationship is a parabola. It is U-shaped, symmetrical, and has a minimum or maximum turning point (vertex).

−4 −3 −2 −1 −1

−2

−3

−4

Minimum turning point

Maximum turning point

5.01 Practice questions Understanding 1

Explorer 1–3

1–3

204

Trailblazer 1–3

Copy and complete this table representing a quadratic relationship. x 0 1 2 3

Adventurer

y First Difference Second Difference −1 −1 − (−1) = 0 −1 0 2 3−2=1 5−2=3 5

Do these statements describe a linear or quadratic relationship? a

Has a constant rate of change.

The graph is symmetrical about a vertical line.

First differences between the y-values is a constant.

Second differences between the y-values is a non-zero constant.

The graph is a straight line.

Second differences between the y-values is zero.

The graph is U-shaped.

Has a changing rate of change.

Has a turning point on the graph.

The equation has a squared term.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Create a table of values for step number versus number of squares.

Describe the pattern in words.

Find the number of squares for Step 5.

Step 1

Fluency Ex 1

4, 5, 6ac, 7ab, 8ab

Step 4

4, 5, 6ab, 7ab, 8bc

4, 5, 6cd, 7cd, 8bc

2x − 3y − 2 = 0

y = 2x2 − 3x − 2

y = 5x3 − x2

y = 2x − 3

y = (x2 − 3)2

y = 1 − 3x − 5x2

y = 6x2 − 2x + 7

y = x2 + 4x

m 2x + 5x2 = y

y = (x2 − 5)2

y = 30x2

o 5

Step 3

Which of these equations represent a quadratic relationship between x and y?

Ex 2a

Step 2

4 y = 8 x2 − 9 x + x

Do these graphs show relationships that are linear, quadratic or neither? a

8 6

4 2 x −5

x −5

−2

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y 5

x −5

5 −5

−5

−10 −15

5 x

x −5

−5

Ex 2b

Do these tables of values show relationships that are linear, quadratic, or neither? a

Ex 3

206

x −3 −2 −1 0 1

y 27 12 3 0 3

x −3 −2 −1 0 1

y −9 −6 −3 0 3

x −2 −1 0 1 2

y 8 2 0 2 8

x 0 1 3 6 10

Complete these tables that represent quadratic relationships. a

x y

−3 ⬚

−2 5

−1 2

0 1

1 2

1 ⬚

5 1

−1 2

3 ⬚

4 4

−2 ⬚

2 ⬚

0 3

1 2

2 −1

−3 ⬚

−2 ⬚

−1 1

0 −1

1 1

Mathspace Victorian Curriculum – Year 9 mathspace.co

2 ⬚

3 10

6 0

7 1

3 ⬚

4 ⬚

2 7

3 ⬚

4 ⬚

5 ⬚

8 ⬚

9 ⬚

4 ⬚

5 ⬚

6 ⬚

10 ⬚

y 9 4 1 4 9

For these patterns: i

Describe the pattern in words.

How many squares or hexagons will be in the next ﬁgure?

iii

Is the relationship between ﬁgure number and number of shapes linear or quadratic?

Reasoning

9, 10

9–11

10–12

Explain why this table of values does not show a quadratic relationship. x y

−2 2

−1 0

0 −2

1 0

2 2

Compare these equations, tables and graphs. Do they have a quadratic or linear relationship? Justify your answer. a

y1 = 3x + 1 and y2 = x2 − 5

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x x y1 −3 −8 −3 −2 −5 −2 −1 −2 −1 0 1 0 1 4 1 2 7 2 3 10 3

y2 4 −1 −4 −5 −4 −1 4 y2

x −5

−5

5 −5

−5

Create a quadratic relationship using an equation, table of values, or a graph. Justify that it is quadratic in two different ways.

Identify if these statements about this incomplete table of values are always, sometimes or never true and explain why. x y

−2

−1

If the y-values are all positive, then it is not quadratic.

If the y-values go from positive, to negative, and back to positive, then it cannot be a linear relationship.

If the y-values are all the same number, then it is a quadratic relationship.

Problem-solving 13, 14

13–15

13–16

A quadratic pattern has an equation y = x2 + 3. It starts with x = 1 and y = 4. Draw ﬁgures to show the ﬁrst four steps geometrically using shapes.

The perimeter of a rectangular community garden is 60 m. a

Copy and complete the table for all missing measurements. Width (m) 6 12 15 18 22 24

208

Length (m) 24 ⬚ ⬚ ⬚ ⬚ ⬚

Mathspace Victorian Curriculum – Year 9 mathspace.co

Area (sq m) ⬚ ⬚ ⬚ 216 ⬚ ⬚

Determine the length and the width that would produce the largest possible area.

For any width w, determine an expression to ﬁnd the area in terms of w.

Consider the table shown: x y

−2 16

−1 4

0 0

1 4

2 16

Describe the pattern of the output in words.

Determine if the rule would be a linear or quadratic relationship.

Write the rule that relates x and y.

A quadratic relationship has a second difference of 3. Create a table of values that could represent the relationship.

5.02 Graphs and features of quadratics After this lesson, you will be able to… • identify the key features of a parabola: intercepts, turning points, and the axis of symmetry. • graph quadratic relationships using a table of values.

Identify key features Intercepts are where the parabola crosses the horizontal (x) and vertical ( y) axes. • The x-intercepts are where the parabola crosses the x-axis. This occurs when y = 0. • The y-intercept is where the parabola crosses the y-axis. This occurs when x = 0. Note, that there can be 0, 1, or 2 x-intercepts.

x-intercepts 4 2 −6 −4 −2 −2 −4

x 2

y-intercepts

For now, we will just identify the intercepts from a graph, later we will look at doing it algebraically. Parabolas can have a maximum or minimum value. We call this a turning point, or vertex. If a parabola opens upwards, we call this concave up and it will have a minimum value. If a parabola opens downwards, we call this concave down, and it will have a maximum value.

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1 −3 −2 −1

x 1

−1

Max value (Maximum turning point)

−2

−3 −2 −1

−3 −4

Min value (Minimum turning point)

−5

Concave up, with minimum value

−1

x 1

−2 −3

Concave down, with maximum value

Parabolas are symmetrical. The axis of symmetry is the dotted line pictured here. It has an equation of the form x = ⬚, and you can see that this x-value will be the same as the x-value for the maximum or minimum turning point.

3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

Example 1 For each graph, identify: i)

the turning point

ii) whether the turning point is a maximum or minimum iii) the equation of the axis of symmetry iv) any intercepts that can be identiﬁed a

8 6 4 2 −6

210

−4

−2

−2 −4 −6 −8 −10 −12 −14 −16

x 2

Mathspace Victorian Curriculum – Year 9 mathspace.co

x 1

Apply the idea i)

The graph has a turning point at (−2, −16).

ii) This graph has a minimum turning point because it is concave up. iii) The equation of the axis of symmetry is the vertical line through the turning point. The equation is x = −2 iv) The graph has x-intercepts at (−6, 0) and (2, 0) and a y-intercept at (0, −12).

y 50 25 x −10−8 −6 −4 −2 −25

2 4 6 8 10

−50 −75 −100

Apply the idea i)

The graph has a turning point at (−5, 0).

ii) This graph has a maximum turning point because it is concave down. iii) The equation of the axis of symmetry is the vertical line through the turning point. The equation is x = −5 iv) The graph has an x-intercept at (−5, 0) and a y-intercept at (0, −25).

Idea summary The key features of a quadratic graph are: • Turning point: point where the graph has a maximum or minimum and it changes between increasing and decreasing. • The minimum or maximum value of the graph is the y-value of the turning point. • y-intercept: point where the graph crosses the y-axis. • x-intercept(s): point(s) where the graph touches or crosses the x-axis, there may not be any. • Axis of symmetry: vertical line that cuts the graph in half; it passes through the turning point.

Plot from table of values If we have an equation for a quadratic function we can create a table of values to create a series of points to plot. 5.02 Graphs and features of quadratics mathspace.co

211

Example 2 For y = x2 − 2x + 1, complete the table of values and graph the parabola. x y

−2

−1

Create a strategy To complete the table, substitute each x-value into the equation. Once the table of values is completed, we can plot the ordered pairs and connect them with a smooth curve.

Apply the idea Here is how we can obtain y, when x = −2: y = (−2)2 − 2(−2) + 1 y=4+4+1 y=9 Repeat this process for each x-value in the table. x y

−2 9

−1 4

0 1

1 0

2 1

3 4

4 9

Now we can use these points to graph the quadratic function y 9

8 7 6 5 y

4 3 2 1 −2

−1

x 1

Reﬂect and check Having the turning point in your table is useful, since it tells you where the parabola will change direction. Sometimes the table of values you select will not include the turning point of the parabola, depending on the quadratic function being graphed. If you plot your initial table of values and ﬁnd you are unsure where the parabola changes direction, you can add additional values to your table until you can identify where the curve changes direction.

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Idea summary The graph of a quadratic relationship is called a parabola. To graph it, we can complete a table of values and plot the points on the coordinate plane.

5.02 Practice questions Understanding 1

Explorer

Adventurer

1, 2

1–3

Describe these curves as concave up or concave down. a 4

1 −1

−2

−3

−4

10 5

x 1

−4 −3 −2 −1 −5

−2

−10

−3

−15

−4

−1

−4 −3 −2 −1 −1

−2

−1

Trailblazer

Determine if the statements are true or false. a

The axis of symmetry is a vertical line passing through the turning point.

A minimum turning point is the highest point on the graph.

The x-intercepts are the points where the graph cuts the y-axis.

The y-intercept occurs where x = 0.

Describe the relationship between the turning point and the axis of symmetry.

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213

Fluency 4

4–6, 8ab, 9ab, 10, 11ac

4–6, 8bc, 9ac, 10, 11bc

4, 6, 7, 8bd, 9cd, 10, 11ad

For each graph: i

Describe the graph as concave up or concave down.

Is the turning point a maximum or minimum point?

y 5

x −5

−5

5 x

−5

x −5

5 −5

−5

For each graph, identify the x and y-intercepts. a

9 y 8 7 6 5 4 3 2 1

−5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

214

x 1 2 3 4 5

Mathspace Victorian Curriculum – Year 9 mathspace.co

9 y 8 7 6 5 4 3 2 1 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

x 1

11 y

10 9 8 7 6 5 4 3 2 1

−3 −2

Ex 1

−1

−1 −2 −3 −4 −5 −6 −7 −8 −9 −10

−1−1 −2 −3

2 1

2 3 4 5 6 7 8

For each graph, identify: i

If the graph is concave up or down.

The turning point.

iii

The equation of the axis of symmetry.

The x and y-intercepts.

y −4 −3 −2 −1

x 4

−5

−20 −25

−2 −1−1

x 1 2 3 4 5 6 7 8

x 1

−2 −3 −4 −5 −6 −7 −8 −9

−15

13 12 11 10 9 8 7 6 5 4 3 2 1

−3 −2 −1−1

−10

4 3 2 1

2 −8 −6 −4 −2 −2 −4 −6 −8 −10 −12 −14 −16 −18

y x

5.02 Graphs and features of quadratics mathspace.co

215

For each graph, identify: i

The turning point.

The equation of the axis of symmetry.

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1

14 12 10

1 2 3 4 5

6 4 2

55 50 45 40 35 30 25 20 15 10 5

216

16 14 12 10 8 6 4 2

−9 −8 −7 −6 −5 −4 −3 −2 −1

Complete the table of values. x y

1 2

For each equation: i

−10−9 −8 −7 −6 −5 −4 −3 −2 −1

2 4 6 8 10 12 14 16 18 Ex 2

y 16

−3

−2

−1

Plot the points and join with a parabolic shaped line.

iii

Identify the axis of symmetry.

Mark any easily identiﬁable y and x-intercepts.

y = 3x2

y = −x2 + 6

y = x2 + 2x + 1

For each equation: i

Create a table of values with at least 5 points.

Plot the points and join with a parabolic shaped line.

iii

Identify the equation for the axis of symmetry.

Mark any easily identiﬁable y and x-intercepts.

y = −x2 − 5

y = 2x2 + 3

Mathspace Victorian Curriculum – Year 9 mathspace.co

y = −x2 + 2x + 8

x 1

This diagram shows only part of the graph of a quadratic relationship. a

Complete the table of values for the graph. −3

x y

−2

−1

Determine the x-values of the x-intercepts.

State the coordinates of the vertex in the form (a, b).

Does the vertex represent a minimum or a maximum point?

If we knew the equation of the curve, what would be the sign of the coeﬃcient of x2?

16 14 12 10 8 6 4 2 −1 −2 −4 −6 −8

x 1 2 3 4 5 6 7 8 9 10

For each function: i

Graph the function.

Find the equation of the axis of symmetry.

iii

Find the turning point.

What are the coordinates of the x and y-intercepts?

y = 2x2

Reasoning

y = x2 – 1

12, 13, 15, 16

y = −x2

y = 4 − x2

12, 13, 15, 16

12–16

Explain why the graph of y = x2 has a minimum value of 0.

Determine whether this statement is always true, sometimes true, or never true. Justify your answer with examples or counter-examples. “A parabola has either a maximum or minimum value, but not both.”

Justify the signiﬁcance of the y-value of the turning point in determining the range of values of the y-values of a quadratic function.

Match each equation to its graph. A

y = x2

y = x2 + 4

y = 2x2

1 −1

−2

y = −x2 + 4

x 1

−2

−1

x 1

5.02 Graphs and features of quadratics mathspace.co

217

iii

1 −2

b 16

−1

−2

−1

Explain what feature(s) of the graph helped you deduce the equations for each.

Match the given quadratic equations with their corresponding graphs and features. Equations: A

y = x2

y = −x2

y = x2 + 1

y = x2 − 1

Parabola concave down, turning point at (0, 0), symmetric about the y-axis.

Graphs and features: a

Parabola concave up, turning point at (0, 1), symmetric about the y-axis.

−2

−1

−2

−3 x

−2

−1

−4

Parabola concave up, turning point at (0, −1), symmetric about the y-axis.

Parabola concave up, turning point at (0, 0), symmetric about the y-axis.

3 x

−2

−1

−1 −2

218

Mathspace Victorian Curriculum – Year 9 mathspace.co

1 x −2

−1

Problem-solving 17, 20, 21, 23 17

17, 18, 20, 21, 23

The height of a tennis ball a number of seconds after it is thrown is shown in the graph: a

Determine the coordinates of the y-intercept.

Interpret the y-intercept in this context.

Determine the coordinates of the x-intercept.

Interpret the x-intercept in this context.

17–23

height

16 14 12 10 8 6 4 2 1

seconds 6

A parabola represents the path of a ball thrown into the air. The equation of the parabola is given by y = −2x2 + 8x. a

Is the parabola concave up or down? Does it have a minimum or maximum value?

Find the coordinates of the turning point (vertex) of the parabola.

What is the minimum or maximum value of the parabola?

The proﬁt of a company is modelled by the quadratic equation P = −3x2 + 12x − 5, where P is the proﬁt in thousands of dollars and x is the number of items produced. a

Is the parabola concave up or down? Does it have a minimum or maximum value?

Find the coordinates of the turning point (vertex) of the parabola.

What is the minimum or maximum value of the parabola?

If the axis of symmetry of a quadratic graph is x = −2, and one of the x-intercepts is (6, 0) what is the other?

If the x-intercepts for a quadratic graph are x = −12 and x = 7, what is the axis of symmetry?

5.02 Graphs and features of quadratics mathspace.co

219

Painting a multi-storey building requires extra effort to set up each additional floor. So a painter decides to charge proportionally to the square of the number of floors. He charges $1000 for a 1-floor building, $4000 for a 2-floor building and so on. a

Complete the table for the amount charged for 2, 3, … 6 ﬂoors. Number of ﬂoors Amount charged

1 $1000

2 $4000

Plot the graph for the amount charged against the number of ﬂoors.

Determine the equation for the amount charged ( y) as a function of the number of ﬂoors (x).

What would be the cost to paint an 8-ﬂoor building?

The painter charged $400 000 for a building. How tall was the building?

In a science experiment a substance is heated over a ﬂame and then cooled. The graph shows the temperature in degrees Celsius (c) over time in hours (h). Use the graph to approximate answers to the following:

220

What was the initial temperature?

When did the temperature reach 0 °C?

What was the maximum temperature reached?

Did the temperature rise and fall at similar rates?

When did the temperature reach room temperature (18 °C)?

Mathspace Victorian Curriculum – Year 9 mathspace.co

50 40 30 20 10 −10 −20 −30 −40 −50

h 2 4 6 8 10 12 14 16 18 20

INVESTIGATION

Transformations of quadratics Investigate transformations of the parabola y = x2 to understand the link between its graph and algebraic form using digital tools.

Objectives • To analyse coeﬃcients’ effects on a quadratic graph • To explore how parameters shift a quadratic graph • To correlate values in a quadratic equation with intercepts

Coefficients and transformations y = ax2 + k Using this form of a quadratic function, we will explore compression, enlargement, reﬂections, and vertical translations of the graph.

Interactive exploration Discover this concept in action online

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Parameters and transformations y = a (x − h)2 + k Using this form of a quadratic function, we will continue to explore transformations by investigating horizontal shifts.

Interactive exploration Discover this concept in action online

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Intercepts and transformations y = a (x – b) (x – c) Using this form of a quadratic function, we will explore the relationship between key features and the equation of the function.

Interactive exploration Discover this concept in action online

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Investigation: Transformations of quadratics mathspace.co

221

Discussion 1.

Compare the transformations of quadratic functions to those of linear functions.

How would you explain the difference in graph transformation between y = ax2 + k and y = a (x − h)2 + k to someone who has never studied quadratics before?

In the context of y = a (x – b) (x – c), which gives the x-intercepts, how might knowing the vertex of a quadratic be beneﬁcial, even if the equation form doesn’t explicitly provide it?

Did you know?

Headlight reflectors in cars are designed with precision optics to direct light exactly where it’s needed. These reflectors often have a parabolic shape, which focuses and amplifies the light from the bulb, making the headlights brighter and more efficient while minimising glare for other drivers. In fact, some modern cars use complex reflector designs to ensure that every part of the road is well-lit without blinding oncoming traffic!

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5.03 Basic quadratics After this lesson, you will be able to… • understand and explain the concept of vertical reflections in quadratic graphs. • interpret and analyse the effect of a negative coefficient on the graph of a quadratic equation. • understand and explain the concept of dilation in quadratic graphs. • interpret and analyse the effect of changing the coefficient of the quadratic term on the graph of a quadratic equation. • apply understanding of these concepts in solving problems and interpreting real-world scenarios.

Basic form Interactive exploration Discover this concept in action online

The basic parabola is represented by the equation y = x2. It forms a U-shaped curve that opens upwards.

y 4 3 2 1 −4 −3 −2 −1 −1

mathspace.co

x 1

−2 −3 −4

If the coefficient of x2 is negative (meaning the equation looks like y = −x2 ), the parabola flips upside down. This is called a vertical reflection over the x-axis. In the basic parabola, y = x2, all the y-values are positive or zero. When the parabola is reflected (flipped), like in y = −x2, the y-values become negative or zero.

The features of y = x2 and y = −x2 are: Turning point Axis of symmetry x-intercept(s), also called the roots or solutions to the quadratic Description

y = x2 Minimum turning point at (0, 0) x=0

y = −x2 Maximum turning point at (0, 0) x=0

(0, 0)

The graph of y = x2 represents a basic parabola.

The graph of y = x2 that is vertically reflected across the x-axis (flipped upside down)

5.03 Basic quadratics mathspace.co

223

Dilations Interactive exploration Discover this concept in action online

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A dilation stretches or compresses a graph based on a scale factor. The terms vertical enlargement and vertical compression describe how a graph changes in relation to the x-axis: • Vertical enlargement occurs when the graph moves away from the x-axis, making it appear narrower. This happens for a > 1. • Vertical compression happens when the graph moves towards the x-axis, making it appear wider. This happens when 0 < a < 1. y 8

1 −3 −2

−1

x 1

A vertical enlargement makes the graph narrower

−3 −2

−1

x 1

A vertical compression makes the graph wider

The features of y = ax2 and y = −ax2 are: Turning point Axis of symmetry x-intercept(s), also called the roots or solutions to the quadratic

Description

y = ax2 Minimum turning point at (0, 0) x=0

y = −ax2 Maximum turning point at (0, 0) x=0

(0, 0)

The graph of y = x2 that has been dilated by a factor of a.

The graph of y = −x2 that has been dilated by a factor of a.

Vertically compressed for 0 < a < 1

Vertically compressed for 0 < a < 1 Vertically enlarged for a > 1

Vertically enlarged for a > 1

Example 1 Graph y = −x2 and y = −4x2 on the same Cartesian plane. Mark the turning point and intercepts on the graph. Which graph is vertically enlarged? 224

Mathspace Victorian Curriculum – Year 9 mathspace.co

Create a strategy To graph the parabolas, create a table of values and plot the points on the same Cartesian plane. To decide which is vertically enlarged, look for the graph that is narrower.

Apply the idea Create a table of values by substituting x-values into each equation. x 2

y = −x y = −4x2

−2 −4 −16

−1 −1 −4

0 0 0

1 −1 −4

2 −4 −16

Both graphs have a turning point at (0, 0). This is also the x and y-intercept for both quadratics. y −4 −3 −2 −1 −2

x 1

−4 −6 −8 −10 −12 −14 −16

The graph of y = −4x2 is narrower so it is vertically enlarged.

Reﬂect and check When looking at the table of values, notice that the y-values for y = −4x2 are multiplied by 4, which moves them further away from the x-axis. This is what causes the parabola to become steeper, or narrower.

Example 2 Order the quadratic equations from most narrow (vertically enlarged), to least steep (vertically compressed).

Create a strategy Compare the values of a in each equation. When a = 1, the graph is the basic parabola. When a > 1, the graph is narrower (vertically enlarged) than the basic parabola. When 0 < a < 1, the graph is less steep (vertically compressed) than the basic parabola.

5.03 Basic quadratics mathspace.co

225

Apply the idea For the equation y = x2, a = 1. This is the graph of the basic parabola. 1 2 1 , the graph will be less steep (vertically For the equation y = x , a = . Since 2 2 compressed) than the basic parabola. For the equation y = 5x2, a = 5. Since 5 > 1, the graph will be narrower (vertically enlarged) than the basic parabola. In order from most narrow to least steep:

Reﬂect and check We can check our answer by graphing all three quadratics on the same Cartesian plane. y 6

y = 5x2

y = x2

1 2 x 2

4 3 2 1 −5

−4

−3

−1

−2

x 1

Idea summary The equation y = ax2 represents a vertical dilation of y = x2 by a factor of a. • a > 1 represents a vertical enlargement • 0 < a < 1 represents a vertical compression 8

1 −3 −2

−1

A vertical enlargement makes the graph narrower

226

x 1

Mathspace Victorian Curriculum – Year 9 mathspace.co

−3 −2

−1

x 1

A vertical compression makes the graph wider

5.03 Practice questions Understanding

Explorer

Adventurer

1–4

Trailblazer

1–4

List two equations for quadratics that are narrower (vertically enlarged) than y = x2.

List two equations for quadratics that are wider (vertically compressed) than y = x2.

What is different and what is the same when comparing the graphs y = x2 and y = −x2?

The graph of y = x2 is shown with a dashed line. Does the parabola with a solid line show a vertical enlargement or a vertical compression? a

1 −4 −3 −2 −1

x 1

−4 −3 −2 −1

Fluency Ex 1

5, 6a, 7a, 8–10

−4 −3 −2 −1

5, 6a, 7a, 8–10

5, 6b, 7b, 8–10

Graph both quadratics on the same Cartesian plane. Mark the turning point and intercepts on the graph. Which graph is vertically enlarged? a

x 1

−4 −3 −2 −1

y = x2 and y = 3x2

y = −2x2 and y = −x2

Graph both quadratics on the same Cartesian plane. Mark the turning point and intercepts on the graph. Which graph is vertically compressed? a

y = x2 and

and y = −x2

5.03 Basic quadratics mathspace.co

227

Ex 2

Order the quadratic equations from most narrow (vertically enlarged), to least steep (vertically compressed). a

y = 0.2x2, y = 2x2, y = x2

Match each graph to the equation: i

y = 2x2

y = −x2

iii b

228

−2

−3

−4

y = x2

2 1

x 1

−4 −3 −2 −1 −1

−2

−4 −3 −2 −1 −1

y = −3x2

−4 −3 −2 −1 −1

−2

−3

−4

Match each description to the equation: i

y = −2x2

Narrower than y = x2, a reﬂection of y = x2 on the x-axis.

Wider than y = x2, shares the same turning point as y = x2.

Narrower than

Same vertical enlargement as part (a) and is a reﬂection of part (a) on the x-axis.

iii

y = 2x2

but wider than y = 2x2. Has the same x-intercept as y = 2x2.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Each of the graphs shown have an equation of the form y = ax2. By using a point on the graph, ﬁnd the value of a, and the equation of each graph. a

y 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

4 3 2 1

Reasoning

4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

5 4 3 2 1

11, 12

1 2 3 4 5

x 1 2 3 4 5

12, 13

Using a table of values, explain why y = x2.

Compare the similarities and differences in the graphs of y = 2x2 and y = (2x)2 and y = −2x2.

Without plotting the graph, explain why y = −3x2 is non-positive for all values of x.

Problem-solving 14, 16a–c 14

is wider (vertically compressed) compared to

14, 16c–e, 17

14, 15, 16d–f, 17

A square paving tile has side length s cm and a height of 3 cm. a

Write an expression for the volume of 10 square paving tiles stacked on top of each other.

If the volume of the 10 square paving tiles is 1920 cm3, ﬁnd the side length, s, of each paving tile.

Graph y = x2 and the line y = 2x + 2 on the same Cartesian plane. Shade the area between the two graphs. Find an estimate for the area of the shaded region.

5.03 Basic quadratics mathspace.co

229

Find the equation of the quadratic function of the form y = ax2 for quadratics that have the given coordinates on the curve. Leave a as a fraction if needed.

(−9, 9)

(4, 8)

(10, −20)

(1, −4)

(−2, 40)

A social media company’s total advertisement sales (S, in thousands of dollars per week) is proportional to the square of the number of countries (C) they have a presence in.

If S = 16 thousand when C = 3, then: a

What is the quadratic relationship between S and C?

How many countries does the company need to be in to triple the sales from 16 thousand?

5.04 Vertical translations After this lesson, you will be able to… • interpret and describe the effects of vertical translations on quadratic graphs. • identify the equation of a quadratic function given certain points or intercepts. • understand and apply combinations of transformations including vertical translations, compressions, enlargements, and reflections. • find the equation of a specific quadratic function using given transformations.

Vertical translations Interactive exploration Discover this concept in action online

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The features of y = x2 + k and y = −x2 + k are: Turning point Axis of symmetry x-intercept(s), also called the roots or solutions to the quadratic Description

230

y = x2 + k Minimum turning point at (0, k) x=0 For k > 0, no x-intercepts For k ≤ 0, the x-intercepts are ( , 0) and (− , 0) This is the graph of y = x2, that has been vertically translated by k units: up (for positive k) and down (for negative k)

Mathspace Victorian Curriculum – Year 9 mathspace.co

y = −x2 + k Maximum turning point at (0, k) x=0 For k ≥ 0, the x-intercepts are ( , 0) and (− , 0) For k < 0, no x-intercepts This is the graph of y = −x2, that has been vertically translated by k units: up (for positive k) and down (for negative k)

Example 1 Identify the vertical translation amount and direction from y = x2, for each of these: a y = x2 + 2

Create a strategy The vertical translation is the amount up or down that the graph has shifted from the base y = x2. 4

3 2 1

−4 −3 −2 −1 −1

−2 −3 −4

Apply the idea The graph has been vertically translated up by 2 units.

Reﬂect and check Show the graph of y = x2 also on it. y 4 y = x2 + 2

3 2

y = x2

x −4 −3 −2 −1 −1

−2

We can see the turning point has moved up 2 units from (0, 0) to (0, 2).

5.04 Vertical translations mathspace.co

231

b y = x2 − 3

Create a strategy The vertical translation is the amount up or down that the graph has shifted from the base y = x2.

Apply the idea In equations of the form y = ax2 + k, the k value is the vertical translation. This equation has a vertical translation of 3 units downwards.

Reﬂect and check Show the graph with the graph of y = x2 on it. y 4 3 2 1

y = x2 1 2 3

−4 −3 −2 −1 −1 −2 −3

x 4

y = x2 − 3

−4

We can see the turning point has moved down by 3 units from (0, 0) to (0, −3).

Example 2 Find the equation of the quadratic in the form y = x2 + k that passes through the point (1, 7).

Create a strategy Using the point (1, 7) we can substitute the values of x and y into the equation y = x2 + k and ﬁnd the value of k.

Apply the idea y = x2 + k

Start with the equation

7 = (1) + k

Substitute in the values (1, 7) for (x, y)

7=1+k

Evaluate the square

6 = k

Subtract 1 from both sides 2

The equation is y = x + 6.

232

Mathspace Victorian Curriculum – Year 9 mathspace.co

Idea summary In this lesson, we explored vertical translations of quadratic functions in the form of y = x2 + k and y = −x2 + k. The value of k determines the vertical translation: a positive k shifts the graph upwards, while a negative k shifts the graph downwards. The turning point of the quadratic is affected by the vertical translation, but the axis of symmetry remains the same. To ﬁnd the equation of a translated quadratic function, we can substitute the given point into the equation and solve for the unknown constant.

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Example 3 Find the equation of the quadratic in the form y = ax2 + k that has a y-intercept of −4 and passes through the point (2, −16).

Create a strategy Using the y-intercept provided, we know that the k value in the equation will be −4. We can then use the other point to ﬁnd the value of a.

Apply the idea y = ax2 – 4

Start with the equation

−16 = a(2) – 4

Substitute in the values (2, −16) for (x, y)

−12 = 4a

Add 4 to both sides, and evaluate the square

−3 = a

Divide both sides by 4 2

The equation is y = −3x − 4.

Idea summary In this lesson, we explored the effects of various transformations on quadratic functions of the form y = ax2 + k. We learned that translating a graph both vertically and horizontally, vertically reflecting it, and applying vertical compressions or enlargements can create new quadratic functions with different characteristics. By using given points and intercepts, we can determine the values of a and k to find the equation of a specific quadratic function.

5.04 Vertical translations mathspace.co

233

5.04 Practice questions Explorer

Understanding

Trailblazer

1–5

List two equations for a quadratic that has been vertically translated upwards from y = x2.

List two equations for a quadratic that has been vertically translated downwards from y = x2.

What changes and what stays the same, when comparing the graphs of y = x2 and y = x2 + 7?

What happens to the x-intercepts when the graph of y = −x2 + 1 is vertically translated downwards by 2 units?

What is the connection between the y-intercept and the value of k in y = x2 + k? 6, 7, 8ab, 9ab, 10ab, 12

Fluency Ex 1a

Adventurer

1–5

6, 7, 8bc, 9bc, 10bc, 11, 12

Identify the vertical translation amount and direction from y = x2, for these graphs: a

−6

−3

x 3

−3

−6

−3

−6

3 2

234

x 1

−4 −3 −2 −1 −1

−2

−3

−4

x 1

Identify the vertical translation amount and direction from y = −x2, for these equations: a

−3

−4 −3 −2 −1 −1

x 3

−6

Ex 1b

6, 7, 8cd, 9cd, 10cd, 11, 12

y = −x2 + 4

y = −x2 + 2

y = −x2 − 3

For each of these quadratic equations: i

Sketch the graph on a Cartesian plane alongside the graph of y = x2.

Find the value of the y-intercept.

Mathspace Victorian Curriculum – Year 9 mathspace.co

y = −x2 − 1

Ex 2

iii

Find the turning point.

y = x2 − 5

Ex 3

y = x2 + 3

y = x2 − 12

y = 2x2 + 10

Find the equation of the quadratic in the form y = x2 + k that passes through the point: a

(1, 1)

(−4, 5)

(7, 0)

(2, −4)

Find the equation of the quadratic in the form y = ax2 + k that has the y-intercept provided and passes through the point given. Then describe the transformations that have occurred from the base graph of y = x2. a

y-intercept at y = 3, passes through (1, 5)

y-intercept at y = −1, passes through (3, 8)

y-intercept at y = −4, passes through (−6, −184)

y-intercept at

, passes through

A stone is thrown vertically into the air. Its height above the ground is given by the equation h = −5t2 + 20, where h is the height in metres, and t is the time in seconds. Determine the maximum height the stone reaches and the time it takes to reach the ground.

The height of a rocket above the ground is described by the equation h = 4t2 + 10, where h is the height in metres, and t is the time in seconds since the rocket was launched. a

Calculate the rocket’s height 3 seconds after it was launched.

Sketch the rocket’s trajectory for the ﬁrst 3 seconds of ﬂight.

What was the initial height of the rocket above the ground? What feature of the graph does this relate to?

Reasoning 13

13, 14

13–15

With each pair of quadratic equations, identify the quadratic that has the largest maximum value: a

y = −x2 + 3, y = −x2 − 3

y = −2x2 + 8, y = −2x2 − 10

y = −x2 − 1, y = −5x2 − 2

y = −x2, y = −x2 − 1

With each pair of quadratic equations, identify the quadratic that has the smallest minimum value: a

y = x2, y = x2 + 1

y = 3x2, y = 3x2 − 4

y = x2 − 3, y = 5x2 + 1

y = x2 + 12, y = x2 − 12

Do vertical translations always result in a new turning point?

Problem-solving 16, 17 16

13–15

16, 17, 19

16–19

A toy rocket is launched vertically from an elevated platform. The height of the rocket above the ground can be modelled by the equation h = −4t2 + 30, where h is the height in metres and t is the time in seconds. The rocket launches from a platform that is 10 metres above the ground. a

How does the vertical translation of 10 metres affect the equation modelling the toy rocket’s height? What is the new equation?

Using the new equation, ﬁnd the maximum height the toy rocket reaches above the ground.

Calculate the time it takes for the toy rocket to return to the ground. Round your answer to two decimal places.

Sketch a graph showing the movement of the toy rocket. 5.04 Vertical translations mathspace.co

235

Water is sprayed into the air from a fountain, forming a parabolic arc. The height of the water above the fountain can be described by the equation h = −2t2 + 8, where h is the height in metres, and t is the time in seconds.

What do the values of the coeﬃcient −2, and the constant term +8 reﬂect in the context of the problem?

Determine the maximum height the water reaches.

What is the time it takes for the water to hit the ground?

Another fountain nearby has a similar equation, h = −2t2 + 10. Describe the similarities and differences between the two fountain arcs.

The elevation of a hill can be modelled by the equation h = Ax2 + B, where h is the elevation in metres and x is the horizontal distance from the base of the hill in metres.

5 metres from the base of the hill the elevation is 65 metres, and the elevation is 10 metres at the base, what is the equation?

As we move further and further away from the hill, what happens to the elevation? Is there a problem with the model?

A water tank is being drained, and the water level y in centimetres after x minutes is described by the quadratic equation y = −3x2 + 1500.

What is the initial water level in the tank?

Find the time it takes for the tank to be completely drained.

What parameter in the equation (the dilation or the vertical translation) affects: i

The amount of water in the tank

The rate the water drains

5.05 Horizontal translations After this lesson, you will be able to… • define horizontal translation and explain its effect on a graph. • understand and interpret the equation y = (x − h)2 to denote a quadratic function that has been horizontally translated by h units. • identify the direction of horizontal translation based on the sign of h. • explain how the turning point and axis of symmetry are affected by horizontal translation. • apply the concept of horizontal translation to find the equation of a translated quadratic function.

Horizontal translation Interactive exploration Discover this concept in action online

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mathspace.co

The features of y = (x − h)2 are: Turning point Axis of symmetry x-intercept(s), also called the roots or solutions to the quadratic

Description

y = (x − h)2 Minimum turning point at (h, 0) x=h

y = −(x − h)2 Maximum turning point at (h, 0) x=h

(h, 0)

This is the graph of y = x2 which has This is the graph of y = x2 which has been horizontally translated by h been vertically reflected across the x-axis units: and horizontally translated by h units: – right for h > 0, (x − h)2

– right for h > 0, (x − h)2

– left for h < 0, (x − h)2

– for h = 0, no horizontal translation

Example 1 For this graph:

7 6 5 4 3 2 1

−7 −6 −5 −4 −3 −2 −1 −1

1 2 3

−2

a Identify the horizontal translation amount and direction from y = x2.

Create a strategy The horizontal translation is the amount left or right that the graph has shifted from the base y = x2. Count from the vertex.

y 4 3 2 1 −6 −5 −4 −3 −2 −1 −1

x 1

2 3 4

−2 −3 −4

Apply the idea The graph has been horizontally translated left by 4 units. 5.05 Horizontal translations mathspace.co

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b Identify the horizontal translation amount and direction from y = x2 to y = (x − 7)2.

Create a strategy In equation y = (x − h)2, the h value is the horizontal translation.

Apply the idea Notice that in the translated equation, the h term inside the brackets has been replaced by 7. This equation has a horizontal translation of 7 units to the right.

Example 2 From the base y = x2, write the equation of the quadratic function in the form y = (x − h)2 using the description: a Horizontally translated by 9 units to the left

Create a strategy The h value in y = (x − h)2 is the horizontal translation. y 4 3 2 1 −11−10−9 −8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4

Apply the idea The function has been translated left by 9 units so h = −9. So the equation is y = (x − (−(9))2 which becomes y = (x + 9)2.

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x 1 2

b Horizontally translated 11 units to the right and has a vertical reflection across the x-axis

Create a strategy The h value in y = (x − h)2 is the horizontal translation. A vertical reflection means multiplying the entire function by −1.

Apply the idea The function has been translated right by 11 units, so h = 11. The equation is y = (x − 11)2 which is then multiplied by −1, giving us the final equation y = − (x − 11)2.

Idea summary Horizontal translation is the shifting of a graph left or right from its base function. The equation y = (x − h)2 represents a quadratic function that has been horizontally translated by h units. When h is positive, the graph is translated to the right, while when h is negative, the graph is translated to the left. The turning point and axis of symmetry are also shifted accordingly. To ﬁnd the equation of a quadratic function with a horizontal translation, substitute the given point into the equation and solve for h. This will give you the equation in the form y = (x − h)2.

5.05 Practice questions Understanding

Explorer 1–5

Adventurer

Trailblazer

1–5

List two equations for a quadratic that has been horizontally translated left from y = x2.

List two equations for a quadratic that has been horizontally translated right from y = x2.

What changes and what stays the same when comparing the graph of y = x2 and the graph of y = (x − 2)2?

What happens to the y-intercept when the graph of y = −x2 is horizontally translated left?

What is the connection between the x-value of the turning point and the value of h in the equation y = (x − h)2?

Fluency 6

6, 7, 8a–d, 9ab, 10ab, 11, 12a–c

6, 7, 8c–f, 9bc, 10bc, 11, 12b–d, 13, 14

6, 7, 8e–h, 9cd, 10bd, 11, 12cef, 13, 14

Has the given equation been translated up, down, left or right from y = x2? a

y = (x − 8)2

y = (x + 9)2

y = −(12 − x)2

y = (0.75 + x)2

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239

Ex 1a

Identify the horizontal translation amount and direction from y = x2 for these graphs: a

7 6

1 −3 −2 −1 −1

Ex 2

240

2 3 4

−2

7 6 5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3

Ex 1b

−5 −4 −3 −2 −1 −1

1 2 3 4 5 6 7

−2

7 6 5 4 3 2 1

x 1

−7 −6 −5 −4 −3 −2 −1 −1 −2 −3

x 1 2 3

Identify the horizontal translation amount and direction from y = x2 for these equations: a

y = (x − 1)2

y = (3 + x)2

y = −(x + 3)2

y = (x − 10)2

y = (x + 5)2

y = (−6.5 + x)2

y = −(5 − x)2

For each of these quadratic equations: i

Sketch the graph on a Cartesian plane alongside the graph of y = x2.

Find the value of the y-intercept.

iii

Find the turning point.

y = (x + 1)2

y = (x − 1)2

y = (x + 3)2

y = −(x + 2)2

For each of these quadratic equations: i

Sketch the graph on a Cartesian plane.

Identify the axis of symmetry.

iii

Find the turning point.

y = −(x + 1)2

y = −(x − 5)2

y = −(6 + x)2

Write the equation of the quadratic function that has the form y = (x − h)2 using the description of the horizontal translation from the base y = x2. a

Horizontally translated left 7 units

Horizontally translated right 13 units

Mathspace Victorian Curriculum – Year 9 mathspace.co

Horizontally translated left

Horizontally translated right 4.6 units

Horizontally translated 3 units left and has a vertical reﬂection across the x-axis

Horizontally translated 2 units right and has a vertical reﬂection across the x-axis

For these equations, describe the transformations that have taken place from the base y = x2. a

y = (x + 6)2

e 13

units

y = (x − 9)2

y = (x − 0.15)2

y = −(x − 3)2

y = (13 + x)2

Find the equation of the quadratic that: • Passes through the point (5, −1) • Is concave down

A photographer is taking a picture of a tall building, and the camera lens is positioned at H = (D − 1)2 metres above the ground. If the height of the building (H) is 40 metres, ﬁnd the horizontal distance from the photographer to the building.

Reasoning

15, 16

16, 17

Explain why with a quadratic of the form y = (x − h)2, the y-values can never be negative.

What is the maximum y-value the quadratic y = −(x − h)2 + 5 can have? Explain why this is the maximum y-value for any value of h.

When subtracting 3 on a number line we move our starting point 3 units to the left. But when we go from y = x2 to y = (x − 3)2 we translate the graph 3 units to the right. Explain why these are different.

Problem-solving 19 18

18, 19

18–20

A marksman is practicing at a shooting range with a target positioned at y = (x − 3)2 metres, where y is the height above the ground, and x is the horizontal distance from the shooter. a

If the bullseye is at a height of 16 metres, ﬁnd the horizontal distance from the shooter to the centre of the bullseye.

If the marksman wants to hit a target 25 metres away, at what height should the target be placed?

A hiker travels along a straight path in the forest for 6 hours at a speed modelled by s = (t − 2)2, where s represents the speed in km/hour and t is the time in hours. a

What is the hiker’s initial speed?

At what time is the hiker travelling at a speed of 9 km/h?

When does the hiker stop momentarily?

Draw a graph to determine the hiker’s fastest speed.

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A company’s monthly sales data is modelled by the equation R = (m − 4)2, where R represents the sales revenue in thousands of dollars, and m is the month. The model is only accurate for months 1 through 12. a

Determine the month in which the company had its highest sales revenue

Calculate that maximum revenue.

Find the month when sales ﬁrst reached $6000.

What happens to the revenue at month 4?

Graph the model and describe what happens to revenue over the 12-month period.

5.06 Turning point form After this lesson, you will be able to… • understand the turning point form of a quadratic equation. • identify the coordinates of the turning point of a parabola from the turning point form of a quadratic equation. • determine the direction of opening and the scale factor of a quadratic graph from the turning point form of its equation. • graph quadratic equations in turning point form. • solve problems involving quadratic equations in turning point form.

Turning point form A quadratic equation can be written in turning point form (also called vertex form): y = a (x − h)2 + k This form helps us easily find the turning point (or vertex) of the parabola, which is the point where the graph changes direction. It also shows us whether the parabola opens upwards or downwards and how “wide” or “narrow” it is.

Interactive exploration Discover this concept in action online

mathspace.co

• (h, k) are the coordinates of the turning point (where the graph reaches its highest or lowest point). • The value of a shows whether the graph is stretched or compressed (this is called a dilation). If a > 1, the graph is narrower (enlarged), and if 0 < a < 1, the graph is wider (compressed). • The value of k moves the graph up or down (vertical translation). • The value of h moves the graph left or right (horizontal translation).

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y = a (x − h)2 + k (h, k) h k a

coordinates of the turning point is the horizontal translation is the vertical translation scale factor

Example 1 Which equation matches the graph?

A y = 3(x − 2) − 4

B y = 3(x + 2)2 − 4

C y = −3(x − 2)2 − 4

D y = (x − 2)2 − 4

1 −1

−1

x 1

−2 −3 −4

Create a strategy We know that turning point form is y = a(x − h)2 + k The graph has a turning point at (2, −4), so we are looking for an equation with h = 2 and k = −4. If we cannot distinguish using the turning point, we will need to look at the scale factor.

Apply the idea B: y = 3(x + 2)2 − 4, has h = −2 and k = −4. The value of h is incorrect, so we can eliminate B. A, C and D: These three graphs have h = 2 and k = −4, so they are all possible solutions at this stage. The graph is concave up, so we know that a > 0, so we can eliminate C. To ﬁnd the value of a, we need to use another known point on the graph. Let’s use (3, −1): y = a (x − 2)2 – 4

Start with our known equation

−1 = a (3 − 2) – 4 Substitute (3, −1) into the equation −1 = a(1)2 – 4

Evaluate the brackets

−1 = a – 4

Evaluate the square

3=a

Add 4 to both sides

So the answer is A: y = 3(x − 2)2 − 4.

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Example 2 Find the equation of each of the graphs below in turning point form. a

y 7 6 5 4 3 2 1 −1

−1

x 1

Create a strategy We know that turning point form is y = a(x − h)2 + k We can identify h and k from the coordinates of the turning point, and substitute a point on the parabola into the equation to ﬁnd a.

Apply the idea The turning point of the given curve is at (3, 2). So h = 3 and k = 2. This gives the equation y = a (x − 3)2 + 2. To ﬁnd the value of a, we need to use another known point on the graph. Let’s use (2, 3): y = a (x − 3)2 + 2

Start with our known equation

3 = a (2 − 3) + 2

Substitute (2, 3) into the equation

3 = a (−1)2 + 2

Evaluate the brackets

3=a+2

Evaluate the square

a=1

Subtract 2 from both sides

So, the equation is y = (x − 3)2 + 2. b

y 8 6 4 2 x −7 −6 −5 −4 −3 −2 −1 −2

−4

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Create a strategy We know that turning point form is

y = a(x − h)2 + k

We can identify h and k from the coordinates of the turning point, and substitute a point on the parabola into the equation to ﬁnd a.

Apply the idea The turning point of the given curve is at (−3, 7). So h = −3 and k = 7, which gives us: y = a(x + 3)2 + 7 To ﬁnd the value of a, we need to use another known point on the graph. Let’s use (−1, 3): y = a (x + 3)2 + 7 2

3 = a (−1 + 3) + 7 2

Start with our known equation Substitute (−1, 3) into the equation

3 = a (2) + 7

Evaluate the brackets

3 = 4a + 7

Evaluate the square

−4 = 4a

Subtract 7 from both sides

−1 = a

Divide both sides by 4

So, the equation is y = −(x + 3)2 + 7.

Example 3 For y = (x + 2)2 − 3. a Find the turning point.

Create a strategy For y = a (x − h)2 + k, the turning point is (h, k).

Apply the idea We can rewrite y = (x + 2)2 − 3 as y = (x − (−2))2 − 3. From here, we can ﬁnd that the turning point is (−2, −3). b Find the y-intercept.

Create a strategy To ﬁnd the y-intercept, we need to set x = 0 and solve for y.

Apply the idea y = (x + 2)2 – 3 2

Write the equation

= (0 + 2) – 3

Substitute x = 0

Evaluate

So, the y-intercept is at (0, 1). 5.06 Turning point form mathspace.co

245

c Will the graph be concave up or down?

Create a strategy Remember that if a > 0, the parabola will be concave up and if a < 0, the parabola will be concave down.

Apply the idea a = 1, so a > 0 The graph will be concave up.

d Find the axis of symmetry.

Create a strategy The axis of symmetry is the x-coodinate of the turning point.

Apply the idea The turning point is at (−2, −3). The axis of symmetry is x = −2.

Reflect and check The axis of symmetry is helpful because for any point on the parabola, we can find another point by reflecting it across the axis of symmetry. For example, for the y-intercept, (0, 1), we can: 1. Find the distance from the y-intercept to the 2. Reflect this distance across the axis of axis of symmetry: 2 units symmetry: y

1 −4

−3

−2

−1

−4

−3

−2

−1

−2

−3

−4

So (−4, 1) is also on the parabola.

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x 1

e Sketch the graph.

Create a strategy Plot the points we have found and draw a curve through the points.

Apply the idea

Reﬂect and check y

1 −4

−3

−2

−1

x 1

−4

−3

−2

−1

x 1

−2

−3

−4

We can check our graph by looking at the concavity and the axis of symmetry to make sure all of our information matches. Our graph is concave up and the axis of symmetry cuts the graph in half as it should.

Idea summary Turning point form, also known as vertex form, is:

y = a (x − h)2 + k (h, k)

coordinates of the turning point, indicating the horizontal translation (h) and the vertical translation (k) from the base y = x2 a scale factor If a is negative, the quadratic is concave down. If a is positive, the quadratic is concave up. If the magnitude of a is between 0 and 1 then it is a vertical compression. The graph moves towards the x-axis and appears wider. If the magnitude of a is greater than 1, then it is a vertical enlargement. The graph moves away from the x-axis and appears narrower. The value of a can be found by substituting in any other known point into the equation.

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5.06 Practice questions Explorer

Understanding 1

Adventurer 1ab, 2–4

1–3

Trailblazer 1–4

Identify the coordinates of the turning point for each quadratic function.

−2 −1 −2

12 10

x 1 2 3 4 5 6 7 8

−4

−6

−8

4 2

−10

−8 −7 −6 −5 −4 −3 −2 −1 −2

−12

1 2

−14

y 12

x −1

−2

−4

−6

2 x

−8

Has the given equation been translated up, down, left or right from y = x2? y = (x − 8)2

a 3

−9 −8 −7 −6 −5 −4 −3 −2 −1

y = x2 − 5

y = (x + 9)2

y = x2 + 0.75

Consider the table of values of a function. x y

1 −6

2 0

3 2

4 0

5 −6

Select the graph that could represent the relationship: A

3 2 1

2 1 −1

−1

x 1

−2 −3

−1

−2 −3

−5

−4 −5

−6

−4

248

−6 −5 −4 −3 −2 −1

Mathspace Victorian Curriculum – Year 9 mathspace.co

x 1

3 2 1

2 1 −6 −5 −4 −3 −2 −1

−1

x −1

−3 −4 −5

−4 −5

−6

A pumpkin is launched at a medieval fair. The graph shows its ﬂight path. Match points A, B and C to: a

Feature of the graph i

The y-intercept

The x-intercept

iii

The turning point Maximum height is reached

Pumpkin is launched

iii

Pumpkin hits the ground 5–8, 9ace, 10acd, 11a–c, 12a–c

Fluency Ex 1

y = −(x + 5)2 + 25 2

y = (x − 5) + 25

y = (x + 5)2 − 25

y = −(x + 5)2 − 25

Height (m) B 20

10 5 A 1

5–8, 9ace, 10ace, 11b–d, 12a–d, 13

For the given graph, which equation represents the relationship? A

Moment in the pumpkin’s journey i

x 1

−1

−2 −3

−2

C Time (s) 3

5–8, 9bdf, 10bdf, 11d–f, 12acef, 13, 14 y 5 x

−12 −10 −8 −6 −4 −2 −5

−10 −15 −20 −25

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249

For the given graph, which equation represents the relationship? A

p = −10 (x − 1)2 − 4

p = −10 (x + 1)2 − 4

p = −10 (x − 1)2 + 4

p = −10 (x + 1)2 + 4

2 −1

p x 1

−2 −4 −6

−8 −10 −12 −14 −16

For the given graph, which equation represents the relationship? A B

y 8

y = 2 (x + 3)2 + 8

y = 2 (x − 3) + 8

y = 2 (x − 3) − 8

y = −2 (x − 8)2 − 3

−3−2 −1 −2

x 1 2 3 4 5 6 7 8 9

−4 −6 −8

Ex 2

Write the equation of the graph in the form y = (x − h)2 + k. a

1 −1 −1

x 1

−2

−3

−4 −5

−1

y 5 4 3 2 1

−7 −6 −5 −4 −3 −2 −1 −1

x 1

−2 −3

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x 1

10 9 8 7 6 5 4 3 2 1 −10−9 −8 −7 −6 −5 −4 −3 −2 −1

Write the equation of the graph in the form y = a (x − h)2 + k. a

−4 −3 −2 −1 −1 −2 −3 −4 −5 −6

x 1 2 3 4 5 6 7

8 7 6 5 4 3 2 1

1 2 3 4 5 6 7 8

−7 −6 −5 −4 −3 −2 −1 −1 −2

4 3 2 1

−9 −8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

Ex 3

y −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10

1 2 3 4 5 6

y −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10

4 3 2 1

5 4 3 2 1

x 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x 1 2 3

x 1 2 3 4 5

For each quadratic equation: i

What is the turning point?

What is the y-intercept?

iii

What is the axis of symmetry?

Will the graph be concave up or down?

Sketch the graph.

y = −(x − 1)2 + 5

y = −(x + 2)2 + 6

y = (x − 4) + 2

y = (x + 8)2 − 4

y = (x + 0.5)2 − 0.6

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251

For each quadratic equation: i

What is the turning point?

What is the y-intercept?

iii

What is the axis of symmetry?

Will the graph be concave up or down?

Sketch the graph.

y = 2 (x − 3)2 + 7

y = −3(x + 1)2 + 4

y = −2(x − 1)2 − 1

y = 0.5(x + 4)2 + 8

y = 5(x − 0.2)2 − 0.7

For the following equations, describe the transformations that have taken place from the base equation y = x2. a

y = (x + 4)2 − 3

y = −(x − 2)2 + 5

y = 2 (x − 3) + 7

y = −4(x + 1)2 + 6

y = 5(x − 0.3)2 + 0.8

A diver jumps off a platform into a pool, creating a splash. The height of the splash above the water can be modelled by the equation h = −2(t − 2)2 + 10, where h is the height in metres, and t is the time in seconds since the jump. Determine the time it takes for the splash to reach its highest point, the maximum height of the splash, and the initial height of the platform.

A company’s profit from selling a product is given by the equation P = −3(x − 5)2 + 300, where x is the number of units sold, and P is the profit in dollars. Determine the number of units the company needs to sell to maximise its profit, calculate the maximum profit, and find the initial cost incurred by the company.

Reasoning

16, 18

15, 16, 18

15–18

How many quadratic equations could share a common turning point? Explain.

Using the diagram: Describe a situation that could be modelled by the following three quadratic relationships. For each part, make sure to identify the turning point and another possible point on the curve. For example, this could be the path of a dolphin, jumping out of the water and diving back down.

252

Consider the family of quadratic equations of the form y = a(x − b)2 − c where a, b and c are chosen from the integers 2, 3, and 5, each number can only be chosen once. a

Using these integers, write two quadratic equations with the lowest possible minimum value.

Describe the similarities and differences between your answers to part (a).

Mathspace Victorian Curriculum – Year 9 mathspace.co

Houssam says the equation of this graph is y = −(x − 4)2 − 2.

0.5 −1 −0.5 −1 −1.5 −2 −2.5 −3 −3.5 −4 −4.5 −5 −5.5

Explain and correct his error.

Problem-solving 19, 20

x 1 2 3 4 5 6 7 8 9

19, 20, 22

19–22

Ashtyn throws a rock into Crescent Lake. The height of the rock above ground is a quadratic function of time. The rock is thrown from 2 m above ground. After 2 seconds, the rock reaches its maximum height of 8 m. The graph of the motion is going to be drawn with time, t seconds on the horizontal axis, and height, h, on the vertical axis. Write the quadratic equation in turning point form.

A roller coaster has a part that is modelled by a quadratic function. Hau loves roller coasters and wants to be able to build a small model with his 3D printer. Assume that the roller coaster passes through points (0, 0) and (20, 0) and reaches a maximum height of 30 m. Help build the model of the roller coaster by writing an equation for the parabola in turning point form.

Karima and Riley are playing soccer. Karima just kicked a soccer ball and Riley is playing goalie 2 m in front of the goal post. The diagram shows the situation on a Cartesian plane. The soccer ball is kicked at (1, 0.36) and the goal post is 13 m away from Karima. The maximum height of the soccer ball is 4 m and this occurs when it is 8 m away from Karima.

0 10

5 Height (m)

(8, 4)

4 3 2 1

(1, 0.36)

15 5 10 Write a quadratic function that models Distance (m) the height, h, of the soccer ball in metres in terms of the horizontal distance from Karima’s starting position. Give your answer in turning point form.

Riley can block soccer balls that are between 1.3 m and 2.8 m high. Will Riley be able to block Karima’s kick? Explain your answer.

A soccer goal is 2.44 m tall. If the ball gets past Riley, will it land in the goal or go over?

A gardener is planting ﬂowers in a rectangular garden. The area of the garden, A, is given by the equation A = −4(x − 3)2 + 36, where x is the width of the garden in metres. Find the width of the garden that maximises the area, calculate the maximum area, and determine the length of the garden corresponding to this width. 5.06 Turning point form mathspace.co

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5.07 Factorised form After this lesson, you will be able to… • understand the factorised form of a quadratic equation. • graph quadratic equations in factorised form. • write quadratic equations in factorised form from sketched graphs. • solve problems involving quadratic equations in factorised form.

Quadratic equations in factorised form The x-intercepts of a parabola are where it crosses or touches the x-axis.

The x-intercepts occur where the y-value is equal to 0. x1

For quadratic equations, there can be: • Two x-intercepts, • One x-intercept (if the parabola just touches the x-axis), or • No x-intercepts (if the parabola does not touch the x-axis at all). The x-intercepts are also called the solutions to the quadratic. To find these, we set y = 0 and solve for x.

Exploration Consider the graph, which shows the vertical position, y in metres, of a water balloon being tossed from a diving board over time, x in seconds.

The equation representing the projectile motion of the water balloon is: y = −3 (x + 1) (x − 2)

254

9 8 7 6 5 4 3 2 1 −1

−1

x 1

From the graph, determine the x-intercepts.

How might the equation y = −3 (x + 1) (x − 2) relate to the x-intercepts?

From the graph, determine the y-intercept.

How might the equation y = −3 (x + 1) (x − 2) relate to the y-intercept?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Quadratic equations can be represented in factorised form. This form allows us to identify: • x-intercepts • whether the parabola is concave up or concave down • the scale factor

y = a (x − x1) (x − x2)

2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8

x1, x2

x-values of the x-intercepts

scale factor

y x 1 2 3 4 5

The equation is y = 2 (x + 1) (x − 3); a > 0 and it is concave up

7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3

x 1 2 3 4 5

The equation is y = −(x + 2) (x − 3); a < 0 and it is concave down

Example 1 For this parabola:

y 4 3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

a Identify the x and y-intercepts.

Create a strategy The x-intercepts occur when y = 0 and the y-intercept occurs when x = 0.

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Apply the idea 4

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

The x-intercepts are (−2, 0) and (3, 0). The y-intercept is (0, 2).

b Find the equation of the quadratic relationship in factorised form.

Create a strategy Substitute the x-intercepts for x1 and x2 in the equation y = a(x − x1) (x − x2), then use any other point on the graph to substitute for x and y and solve for a.

Apply the idea Since the x-values of the x-intercepts are −2 and 3, we know that the factorised form will be: y = a(x + 2) (x − 3) for some value of a. We can ﬁnd a by substituting in the coordinates of the y-intercept into the equation. To ﬁnd a: Factorised form

Substitute (0, 2)

Evaluate the addition and subtraction

Evaluate the multiplication

Divide both sides by −6

The equation of the quadratic relationship in factorised form:

Reﬂect and check Notice that the signs of the x-intercepts are opposite in the equation compared to the graph.

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Example 2 For the parabola y = (2 − x) (x + 4): a Find the y-intercept.

Create a strategy Substitute x = 0 into the equation.

Apply the idea y = (2 − 0) (0 + 4)

Substitute 0 to x

= 2(4)

Evaluate inside the brackets

Evaluate

The y-intercept is (0, 8).

b Find the x-intercepts.

Create a strategy The x-intercepts occur when y = 0. We can use the factorised form of the equation and make each factor equal 0.

Apply the idea The equation is y = (2 – x) (x + 4). When y = 0, we have (2 – x) (x + 4) = 0, which means that either (2 – x) = 0, or (x + 4) = 0. This means that the x-intercepts occur at (2, 0) and (–4, 0).

Reﬂect and check If we substitute each of the x-values of the x-intercepts into the original equation, we should get y = 0. y = (2 − x) (x + 4)

Original equation

y = (2 − 2) (2 + 4)

Substitute x = 2

y = (0) (6)

Simplify the brackets

y=0

Multiply

y = (2 − x) (x + 4)

Original equation

y = (2 − (−4)) (−4 + 4)

Substitute x = −4

y = (6) (0)

Simplify the brackets

y=0

Multiply

This conﬁrms that (2, 0) and (−4, 0) are the x-intercepts.

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c Determine the equation of the axis of symmetry.

Create a strategy Use the fact that the axis of symmetry is exactly halfway between the two x-intercepts.

Apply the idea

Reﬂect and check Get the average of x-values

This is also the x-coordinate of the turning point.

Evaluate

The equation of the axis of symmetry is x = −1.

d Find the coordinates of the turning point.

Create a strategy Use the fact that the turning point lies on the axis of symmetry.

Apply the idea We substitute x = −1 into the equation to ﬁnd the y-coordinate of the turning point. y = (2 − (−1)) (−1 + 4)

Substitute x = 3

= (3) (3)

Evaluate inside the brackets

Evaluate the multiplication

This means that the coordinates of the turning point are (−1, 9).

e Graph the parabola.

Create a strategy Use the key features found from parts (a) to (d) to graph the parabola.

Apply the idea 11 y 10 (−1, 9) 9 8 (0, 8) 7 6 5 4 3 2 1 x (−4, 0) (2, 0) −4 −3 −2 −1−1 1 2 3 4 −2 −3 −4

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Reﬂect and check The parabola is concave down, which matches the negative coeﬃcient of the rearranged equation y = −(x − 2) (x + 4).

Idea summary To write the equation of the graph of a quadratic relationship in factorised form: 1. Identify the x-intercepts from the graph. 2. Substitute the x-values of the x-intercepts for x1 and x2 into the equation y = a (x − x1) (x − x2). 3. Use any other point on the graph to substitute for x and y and solve for a, the scale factor. To graph a parabola from an equation in factorised form: 1. Identify the x-intercepts from the equation. 2. Find the y-intercept. 3. Find the axis of symmetry which is halfway between the x-intercepts. 4. Find the y-value of the turning point. 5. Plot the x-intercepts, y-intercept, and turning point, then join with a smooth curve. 6. Check that the concavity matches the value of a.

5.07 Practice questions Understanding 1

Explorer 1–3

Trailblazer 1–3

Which quadratic equations are in factorised form? a

y = 2 (x + 3) (x + 5)

c 2

Adventurer 1–3

y = −2 (x − 4) + (x + 3)

y = −3 (x − 7) (x + 11)

Identify each key feature from the graph. a

Turning point

y-intercept

x-intercepts

Axis of symmetry

Concavity

Sign of scale factor

9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1

x 1 2 3 4 5

Is each statement true or false? a

To ﬁnd the y-intercept, substitute x = 0 into the quadratic equation.

To ﬁnd the x-value of the turning point, add the x-intercepts.

The turning point only requires the x-value.

To draw a parabola, we generally need three different points on the graph, such as an x-intercept, y-intercept and the coordinates of the turning point. 5.07 Factorised form mathspace.co

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Fluency 4

4–6, 7c–f, 8d–h, 9

y = (x − 1) (x − 4)

y = (x − 3) (x + 7)

y = x (x − 4)

y = 2 (x − 1) (x + 5)

y = −4 (5 − x) (x + 6)

y = (x + 3)2

y = x (x + 9)

What is the y-intercept for each equation? y = 3 (x − 1) (x − 8)

Match each equation with one of the graphs. i

y = −4 (x + 1) (x − 3)

y = 0.5 (x + 3) (x − 1)

iii

y = 2 (x − 1) (x − 3)

y = −(x + 1) (x + 3)

3 2 1 −6 −5 −4 −3 −2 −1 −1

x 1

−2 −4

5 4

3 2 1 −6 −5 −4 −3 −2 −1

−1

7 6 5 4 3 2 1 −3 −2 −1 −1 −2 −3

−3

x 1

−2 −3

260

4acef, 5, 6, 7c–f, 8e–j, 9

What are the x-intercepts for each equation?

a 6

4–6, 7a–d, 8a–e

Mathspace Victorian Curriculum – Year 9 mathspace.co

x 1 2 3 4 5 6 7

18 16 14 12 10 8 6 4 2 −5 −4 −3 −2 −1 −2

x 1 2 3 4 5

Ex 1

For each graph: i

What are the x-intercepts?

What is the y-intercept?

iii

Write the equation of the parabola in factorised form.

b 6

1 −2 −1 −1

x 1

−2

−3

−4 −3 −2 −1 −1

−4

−2

d 16

x 2

−4

4 2

−8

−2 −1

4 x 6

y 8

x 1

−6

−8 −6 −4 −2 −2

2 −8 −6 −4 −2 −2

−4

−6

−8

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Ex 2

For each quadratic equation: i

What is the y-intercept?

What are the x-intercepts?

iii

What is the axis of symmetry?

What are the coordinates of the turning point?

Graph the parabola.

y = (x − 2) (x + 4)

y = (x + 3) (x − 4)

y = (2x − 4) (x + 1)

y = 4x (x − 2)

y = (2 − x) (x − 4)

y = −(x + 1) (x − 1)

y = 3 (x + 2) (x − 2)

y = −(x − 3) (x − 1)

y = −2 (x − 1) (x − 5)

y = (1 − x) (4 + x)

The x-values of the x-intercepts of a parabola are x = −4 and x = −7. The parabola also passes through the point (−3, 12). Write the quadratic equation in factorised form.

Reasoning 10

10–11, 13

10–14

10–15

Explain how to determine the equation of the quadratic relationship in factorised form based on the given graph.

1 −2 −1 −1

y x 1

−2 −3 −4 −5 −6 −7

Explain how to graph the quadratic equation y = −(x − 6) (x + 2).

What key features do y = 4 (x − 2) (x + 3) and y = (x − 2) (x + 3) share? Explain.

Identify the error in ﬁnding the x-intercepts of y = (x + 2)(x − 3) and explain the correct approach.

(x + 2) (x − 3) = y (x + 2) (x − 3) = 0

x = 2, −3

Substitute y = 0 Solve for x

Identify the x-intercept(s) of y = (x − 2) (2 − x) and describe what you notice.

What are the advantages of having a quadratic equation in factorised form?

Problem-solving 16–19

16–19

16, 18, 20, 21

A parabola passes through the points (−2, 0), (9, 0) and (0, −6). Write the quadratic equation in factorised form.

Sketch the graph of y = (2 − 2x) (x + 3).

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Burnell jumps up and off a 4-metre high springboard into a diving pool below.

Burnell’s jump can be represented by the equation y = −2 (x − 2) (x + 1) where y is Burnell’s height above the water and x is the horizontal distance from the springboard towards the pool (both in metres). a

Graph the equation modelling Burnell’s jump.

Use the model to predict where Burnell will enter the water. Explain your answer.

Xia notices that the Sunshine State Arch is in the shape of a quadratic function. She knows that the arch has a height of 110 m and the feet of the arch are 100 m apart. Xia chooses to let the x-intercepts of the arch be the origin and (100, 0).

What are the coordinates of the turning point of the arch?

Write a quadratic equation for Xia’s representation of the Sunshine State Arch in factorised form.

Wilson throws an eraser into the air and counts how long it takes for the eraser to return to his hand. He throws it 6 metres into the air and it takes 2 seconds to return to his hand. Wilson assumes a quadratic relationship where height (h) in metres is graphed against the time (t) in seconds. t = 0 represents the time when he initially throws the eraser, and h = 0 represents the height of his hand in metres.

Find the t-intercepts and the h-intercept of the graph.

Assuming that the path of the eraser is symmetric going up and coming down, ﬁnd the coordinates of the turning point of the relationship.

Graph the relationship.

Write the equation of the parabola in factorised form.

A quadratic relationship has the equation y = k (x − x1) (x − 10). If the parabola has a turning point at (12, 20), determine the values of x1 and k.

5.08 Standard form After this lesson, you will be able to… • understand the standard form of a quadratic equation. • identify the y-intercept of a parabola from its equation in standard form. • find the axis of symmetry for a given quadratic equation in standard form. • find the turning point of a parabola given in standard form. • graph quadratic equations in standard form. • solve problems involving quadratic equations in standard form.

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263

Quadratics in standard form Interactive exploration Discover this concept in action online

mathspace.co

The standard form of a quadratic equation, where a, b, and c are real numbers is: y = ax2 + bx + c The standard form of a quadratic equation allows us to quickly identify the y-intercept and whether the parabola is concave up or down. If a > 0, the parabola will be concave up (open upwards) and have a minimum turning point. If a < 0, the parabola will be concave down (open downwards) and have a maximum turning point. The y-intercept is (0, c). We can also find the axis of symmetry of the parabola using this formula:

This gives the x-coordinate of the turning point (the highest or lowest point of the graph). To find the full turning point, we can substitute this value into the original equation to get the y-coordinate.

Example 1 For y = 3x2 − 6x + 8: a What is the equation of the axis of symmetry?

Create a strategy Use the formula for the axis of symmetry.

Apply the idea Write the formula

Substitute b = −6 and a = 3

Simplify the denominator

Divide

Simplify

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b What are the coordinates of the turning point?

Create a strategy Use the axis of symmetry from part (a) for the x-coordinate of the turning point, then substitute this into the original equation to get the y-coordinate.

Apply the idea From part (a), we know that the axis of symmetry is x = 1, so the x-coordinate of the turning point is x = 1. y = 3x2 − 6x + 8

Write the quadratic equation

y = 3 (1)2 − 6 (1) + 8 Substitute x = 1 y = 3 (1) − 6 (1) + 8

Evaluate the index

y=3−6+8

Multiply

y=5

Simplify

The turning point is (1, 5).

c What are the coordinates of the y-intercept?

Create a strategy The x-value of the y-intercept is 0 and when x = 0 is substituted into the equation, we are left with c. So c is the y-intercept of an equation written in standard form.

Apply the idea In this case, the value of c in the equation is 8. So the coordinates of the y-intercept are (0, 8).

Reﬂect and check We could also substitute x = 0 in the original equation to check. y = 3x2 − 6x + 8

Original equation

y = 3 (0) − 6 (0) + 8

Substitute x = 0

y=0−0+8

Evaluate index and multiply

y=8

Simplify

This conﬁrms that (0, 8) is the y-intercept.

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d Draw a graph of the parabola.

Create a strategy Use the features found in parts (a), (b) and (c), and the axis of symmetry as a line of reﬂection.

Apply the idea 10 9 8 7 6 5 4 3 2 1

Axis of symmetry: x = 1

Turning point: (1, 5) y-intercept: (0, 8) If we reﬂect the y-intercept across the axis of symmetry, we get another point: (2, 8)

x 1

Reﬂect and check We know that the parabola must open upwards because a > 0. From the graph, we can identify that the turning point form of the equation would be: y = 3(x − 1)2 + 5 If we expand this, we get y = 3x2 − 6x + 8, the original equation.

Idea summary The standard form of a quadratic shows the y-intercept.

y = ax2 + bx + c a scale factor b linear coeﬃcient c y-value of the y-intercept If a > 0, the parabola will be concave up and have a minimum turning point. If a < 0, the parabola will be concave down and have a maximum turning point. If a > 1 or a < −1, the parabola will be narrower than y = x2. If −1 < a < 1, the parabola will be wider than y = x2. The axis of symmetry is the line:

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5.08 Practice questions Understanding 1

Explorer

Adventurer

1–3

Trailblazer

1–3

The standard form of a quadratic is y = ax2 + bx + c. a

What feature of the graph is determined by the value of c?

If a = 2, will the graph be wider or narrower than y = x2?

If a = −1, will the graph have a minimum or maximum?

Which of a, b and c are used to ﬁnd the axis of symmetry?

What is the equation for the axis of symmetry of y = ax2 + bx + c?

For y = x2 + 10x + 16: a

Complete the table of values. x y

−3

−2

−1

Select the correct graph. A

−6

−4

−2

−4

−6

−4

−2

D 6

4 x

x 2

2 −6

−8

y 8

2 −8

−2

−4

−6

−8

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Fluency 4

4ab, 5, 6ab, 7ace, 8ace

For each quadratic equation: i

Complete the table of values

Sketch the parabola.

y = 2x2 + x − 3

−3

−2

4cd, 5, 6bd, 7cef, 8cdf

−1

y = −3x2 + x

y = 4x2 + 2x − 15

iii

y = x2 + 4x + 6

y = x2 + 6x + 4

Match each equation to the correct graph. i

y = x2 + 6x − 4

y = −x2 + 6x + 4 5 4 3 2 1

9 8 7 6 5 4 3 2 1

−7 −6 −5 −4 −3 −2 −1

268

x y

y = −x2 + 2x − 5

−1

y 2 −7 −6 −5 −4 −3 −2 −1 −2

x 1

−4

−7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

4ac, 5, 6bc, 7bce, 8b–d

−6

−8 −10 −12 −14

y 14 12 10 8 6 4 2

−1 −2

x 1

y = x2 − 12x + 46

For each quadratic: i

Would the parabola have a minimum or a maximum?

What is the y-intercept?

iii

What is the equation of the axis of symmetry?

What are the coordinates of the turning point?

y = x2 − 4x + 8

y = −x2 − 4x − 9

Mathspace Victorian Curriculum – Year 9 mathspace.co

y = x2 + x + 1

Ex 1

For each quadratic: i

What is the y-intercept?

What is the equation of axis of symmetry?

iii

What are the coordinates of the turning point?

Sketch the parabola.

y = x2 − 2x + 5

y = 2x2 + 24x + 75

y = −x2 + 6x − 8

y = 2x2 + 2x + 9

y = 4x2 − 64

Using technology: i

Find the x-intercept(s).

Find the coordinates of the turning point.

iii

Draw a graph of the parabola.

y = x2 − 3x − 10

y = x2 − 9

y = 4x2 + 8x − 5

y = −6x2 + 10x + 4

Reasoning 9

9a, 10ac

y = 4 − 3x − x2

9a, 10ac, 12

y = 2x2 + 12x + 18

9b, 10bc, 11

Using technology, graph each pair of equations on the same set of axes. Then: i

Compare the x-intercepts of the two graphs.

Compare the concavity.

iii

Find the y-value of the minimum/maximum of each parabola.

Compare the axis of symmetry of the parabolas. Are they the same or different?

y = x2 + 3x – 4

y = x2 + x – 2

y = −x2 − 3x + 4 10

Graph using technology and answer these questions: i

Does the parabola have a maximum or a minimum value?

Find the y-value of the minimum/maximum.

iii

Describe where the axis of symmetry lies.

Use technology to graph y = x2 on the same set of axes. Describe the similarities and differences between this graph and that of the given equation in terms of the axis of symmetry and minimum/maximum values.

y = 2x2

y = x2 − 7

Kelly wants to understand how the turning point form and standard form of a quadratic equation are related. Turning point form is: y = a (x − h)2 + k a

Expand a (x − h)2 + k and write the equivalent equation in the form y = ax2 + bx + c.

Explain how the equations are related.

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269

Johnny and Derek both ﬁnd the turning point of y = −0.5x2 + x + 7.5. Identify and explain their error(s). Johnny’s work: 1

Finding axis of symmetry

2 3

Finding y-value of turning point

4 5 Turning point is (−1, 6). Derek’s work: 1

Finding axis of symmetry

2 3

Finding y-value of turning point

4 5 Turning point is (2, 7.5).

Problem-solving 13–16 13

13–17

rite an equation in standard form to represent W this parabola.

Write an equation in factorised form to represent this parabola.

Expand the factorised form to conﬁrm your equation in part (a).

13, 16–19

16 14 12 10 8 6 4 2 −1−2 −4 −6 −8 −10

x 1

2 3 4 5 6 7 8

Jackie is starting a summer car wash business. The given equation relates the total proﬁt, P, to the rate she charges for each car wash, x. The rate and proﬁt are in dollars. P = −x2 + 50x − 95

270

Jackie wants to make at least $525 this summer. Determine if she could make enough money based on her quadratic proﬁt model.

Draw a graph of P.

For y = −2x2 + 10x − 3, where x represents time in minutes and y represents the height of a ball from the ground in metres: a

Draw a graph of the corresponding parabola.

Using your graph, predict when the ball will be 3 m above the ground.

Mathspace Victorian Curriculum – Year 9 mathspace.co

A frisbee is thrown upward and away from the top of y a cliff that is 240 metres above the ground. The height 350 y (in metres) of the frisbee at time x (in seconds) is 300 given by: y = −20x2 + 80x + 240 The graph of this relationship is shown:

Identify and interpret the y-intercept of the graph.

Identify and interpret the x-intercept shown.

The frisbee reaches its maximum height after 2 seconds. Determine the maximum height reached by the frisbee.

250

200 150 100 50

x 1

Bella is competing at her high school swim meet. She dives off a springboard that is 3 m above the pool surface. She reaches a maximum height of 8 metres after 1 second. a

Determine the equation in standard form that represents Bella’s diving path.

Determine after how many seconds Bella will be at the same height as the springboard again.

If the pool is 12 metres deep, determine how many seconds after jumping it will take Bella to reach the bottom of the pool.

Winnie throws a bag of cookies to his friend, Peach, from a height of 3.25 m. The cookies reach a maximum height of 19.25 m, 1 second after being thrown. a

Determine the equation in standard form that represents the situation.

Graph the corresponding parabola.

If Peach caught the cookies after 2 seconds, determine the height of the catch.

Fill in the boxes using the digits 1 to 5 at most once, to create a quadratic equation with the lowest possible minimum value. Use only single digit numbers. y = ⬚x2 + ⬚x + ⬚

5.09 Solutions to quadratics After this lesson, you will be able to… • identify the solutions of a quadratic equation from its graph. • relate the solutions of a quadratic equation to its x-intercepts. • understand the possibilities of having one, two, or no real solutions for a quadratic equation. • apply the knowledge of finding solutions to quadratic equations in solving problems.

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Solutions to quadratics Exploration This is the graph of y = x2 + 2x − 8. 1.

What are the x-intercepts of the graph?

Substitute the x-values of the x-intercepts into the equation x2 + 2x − 8 = 0. What do you notice?

How are the solutions to x2 + 2x − 8 = 0 related to the graph?

y x

−6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

1 2 3 4

Recall that the standard form of a quadratic equation is written ax2 + bx + c = 0 where a, b, and c are real numbers. Finding the solutions to a quadratic equation is the same as ﬁnding the x-intercepts of a quadratic function. We ﬁnd the x-intercepts by setting y = 0 and solving. A quadratic equation can have one, two, or no real solutions, as shown in these graphs. y

One real solution

Two real solutions

No real solutions

Example 1 Consider the function y = (x − 2)2 − 9. a Draw a graph of the function.

Create a strategy The function is given in turning point form. Use the function and a table of values to draw a graph that shows the key features (intercepts and turning point).

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Apply the idea From the turning point form of the function we know the turning point is at (2, −9). The axis of symmetry is x = 2. Substituting x = 0 gives the y-intercept at (0, −5). Reﬂecting this point across the axis of symmetry gives us another point (4, −5). All of these points are below the x-axis, so we need more x-values to ﬁnd the x-intercepts. Try −2 ≤ x ≤ 6. y 6 4 2 −3 −2 −1 −2

x 1 2 3 4 5 6 7

−4 −6 −8 −10

Reflect and check Graphing using technology is often the most accurate way to find solutions from a graph. If zeros do not appear in the table of values, then we may need to use technology to find them.

b Determine the solution(s) to the equation (x − 2)2 = 9.

Create a strategy The equation (x − 2)2 = 9 can be rearranged to (x − 2)2 − 9 = 0, so its solutions can be found using the x-intercepts of the graph from (a).

Apply the idea The solutions are x = −1 and x = 5.

Reﬂect and check We can check each solution by substituting them back into the original equation, for example: Left-hand side = (x − 2)2 = (−1 − 2)2 Substitute x = −1 = (−3)2

Evaluate the brackets

Evaluate the square

This is the same as the right-hand side, so the value of x = −1 is correct.

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Example 2 Consider the function y = (x + 3) (x − 1). a Draw a graph of the function.

Create a strategy The function is given in factorised form where we can easily ﬁnd the x-intercepts and then the axis of symmetry to draw a graph that shows the key features (intercepts and turning point from axis of symmetry).

Apply the idea From the factorised form of the function we know the x-intercepts are (−3, 0) and (1, 0). The axis of symmetry is exactly halfway between the two x-intercepts,

. So the axis of symmetry has an

x-coordinate of x = −1. Substitute x = −1, into the equation to identify the y-value of the axis of symmetry. (x + 3)(x − 1) = (−1 + 3)(−1 − 1) = 2 × −2

y 6 4 2 −6 −5 −4 −3 −2 −1 −2

x 1 2 3 4

−4 −6 −8 −10

= −4 Hence the turning point is (−1, −4). Substituting x = 0 gives the y-intercept at (0, −3). Connecting all these points using a parabolic shaped curve will give the graph of the factorised form y = (x + 3) (x − 1)

b Determine the solution(s) to the equation 0 = (x + 3) (x − 1).

Create a strategy The solution to the equation 0 = (x + 3)(x − 1) can be found using the x-intercepts of the graph from (a).

Apply the idea The solutions are x = −3 and x = 1.

Reﬂect and check Recall that the factorised form is useful because it immediately identiﬁes the x-intercepts. Unless required to draw a graph, we can jump straight to the solutions from this form.

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Idea summary A quadratic equation ax2 + bx + c = 0 can be solved by graphing y = ax2 + bx + c and finding its x-intercepts (roots or zeros). We can graph using technology if we cannot accurately find the solutions from a hand-drawn graph. A quadratic equation can have one, two, or no real solutions.

5.09 Practice questions Explorer

Understanding

Adventurer

1–4

Trailblazer

1–4

Describe the similarities and differences for the terms roots, solutions, x-intercepts.

How many solutions does each quadratic equation have? a

y = x2 + 3x − 10

y = −x2 + 4x + 4 y

y 20

5 x

−4 −2

5 4

−20

y = x2 − 6x + 15

y = −2x2 + 12x − 18

5 x

−4 −2

2 −5

−10

5 x 2 −5

−15

−2

−10

−5

−8 −6 −4 −2 −5

−15 −20

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Is each statement always, sometimes, or never true? a

If a table of values has all positive y-values, then there is no solution to the related equation.

If the related parabola is concave up and has a turning point at (2, 4), then there are two solutions.

If the related parabola is concave down and has an axis of symmetry at x = −2, then there are no solutions.

A quadratic equation can be rearranged to the form ax2 + bx + c = 0, with a ≠ 0.

Which of these are solution(s) to 2x2 + 2x − 12 = 0? A

x=2

Fluency 5

Ex 1

Ex 2

x=0

5a–c, 6acd, 7, 8ace, 9

x=3

5c–e, 6b–d, 7, 8bdf, 9

x = −3

5d–f, 6bdef, 7, 8bdf, 9

For each function, i

Sketch the graph.

Use the graph to determine the solution(s) to the quadratic equation when y = 0.

y = x2

y = −4x2

3x2 + 3 = y

3x2 − 3 = y

y = −x2 + 9

y = 8 − 2x2

For each function, i

Sketch the graph.

Use the graph to determine the solution(s) to the quadratic equation when y = 0.

y = (x + 1)2.

y = −(x − 3)2

y = 5(x − 2)2

y = 2(x + 4)2 + 32

y = (x + 1)2 − 9

y = −(x − 2)2 + 4

Consider the function y = (x + 5) (x − 3) a

Sketch the graph of the function.

Determine the solutions to the equation when y = 0.

Solve each equation by graphing using technology. Round your answer to two decimal places. a e

276

x2 − 4x + 4 = 0 2

3x − 4x + 2 = 0

b f

x2 + 5x + 3 = 0 2

2x + 1.5x − 1 = 0

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2x2 − 2x + 1 = 0

4x2 + 12x + 1 = 0

Select the correct graph of the function given the key features. a

• Solutions to the equation of the parabola are −2 and 6. • Minimum: y = −16 A

y 8 6 4 2

−8 −6 −4 −2 −2 −4 −6 −8 −10 −12 −14 −16

8 6 4 2

x 2

8 6 4 2

x 2

−8 −6 −4 −2 −2 −4 −6 −8 −10 −12 −14 −16

• No real solutions • Axis of symmetry: x = 10 • Maximum: y = −6 A

y 2

−2 −4 −6 −8 −10 −12 −14 −16 −18

B x

10 12 14 16

2 −2 −4 −6 −8 −10 −12 −14 −16 −18

y x 2

10 12 14 16

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y −2 −4 −6 −8 −10 −12 −14 −16 −18

10 12 14 16

11, 14

−2 −4 −6 −8 −10 −12 −14 −16 −18

x 4

10, 11, 13, 14

10 12 14 16

10–13

Look at the table of values for y = −2x2 + 5: x y

Reasoning 10

−3 −13

−2 −3

−1 3

0 5

1 3

2 −3

3 13

How many real solutions does 2x2 = 5 have? Explain.

Draw the graph of the function.

Use the graph to estimate the solutions to 2x2 = 5.

A quadratic equation has one solution. Is each statement true or false? Explain your answer. a

For the related parabola, the turning point will be on the x-axis.

If we add 2 to both sides of the equation, then it will have two solutions.

What values of k will ensure the equation x2 + k = 0 has zero, one and two real solutions? Justify your answer using the graph of y = x2 + k.

Imran is told that one of the solutions to (x − 3)2 = 4 is x = 1. Explain how Imran can use the axis of symmetry of the function y = (x − 3)2 − 4 to ﬁnd the other solution.

The graph represents the path a magpie swoops from y a tree to the ground with y representing the height of the magpie in metres, and x representing the horizontal 6 distance from the tree in metres. a

What is the height of the tree?

What is the horizontal distance the magpie lands away from the tree?

Explain why the graph only displays one solution.

2 x 2

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Problem-solving 15–18 15

15, 16, 18, 19

15, 16, 19–21

Find the solution(s) using the table of values. a

x2 + 7x + 12 = 0 x y

−6 6

−5 2

−4 0

−3 0

−2 2

−1 6

−1 −24

0 −27

3x2 − 27 = 0 x y

−4 21

−3 0

−2 −15

1 −24

2 −15

3 0

4 21

A rainbow sign is to be placed on a cafe entrance. The rainbow sign is in the shape of a parabola and can be modelled by the quadratic equation y = −x2 + 8x − 7, where both x and y can be measured in metres. A beam will be placed underneath to support it. What is the minimum length required of the beam?

A bowl follows the shape of a parabola represented by the equation y = (x − 5)2 where y cm is the height of the bowl and x cm is the horizontal distance away from the edge of the table. Find out how far the base of the bowl is from the edge of the table.

A weather tracker device is thrown upward and out of an air balloon at a height of 100 metres as part of a controlled experiment. The height, h, of the device at time t in seconds is given by the equation: h = −20t2 + 80t + 100 a

Complete the table of values for h = −20t2 + 80t + 100: t h

Graph the relationship y = −20t2 + 80t + 100.

Find the time it takes for the compass to hit the ground.

Jessie uses large sprinklers to water crops. The stream of water takes the path of a parabola that reaches a maximum height of 16.2 m above the ground, 18 m from the sprinkler. The path of water sprayed from the sprinkler is shown in the graph, where (0, 0) is the position of the sprinkler, x is the horizontal distance from the sprinkler and y is the height above the ground. a

In which form would it be easiest to write the equation of the quadratic function, standard form or turning point form? Explain your answer. Write the quadratic function for the path of the water.

State the solutions of the quadratic equation when y = 0.

Jessie wants to adjust the sprinkler so that it reaches a maximum height of 10 m above the ground, 40 m from the sprinkler. What is the quadratic function that Jessie should use?

y 16 14 12 10 8 6 4 2

x 5

15 20 25 30 35

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When an object is dropped, its height in metres above the ground (h) can be modelled with , where g is the acceleration due to gravity, t is time in seconds, and

the function

h0 is the height it was dropped from. The values for g are shown for three different planets. Planet Gravity value

Mercury

Earth

Neptune

3.7 m/s2

9.8 m/s2

11.15 m/s2

10 9 8 7 6 5 4 3

An object is dropped from a height of 10 m on all three planets and the graphs which represent this are given: a

Determine which graph corresponds to each planet. Justify your answer.

Identify and interpret the x-intercepts of the functions. Give your answer to two decimal places.

y3 y2

2 1

y1 0.5

t 1.5

2.5

A gardener is designing two rectangular gardens. The area of the ﬁrst garden is represented by the equation A = (x − 3) (x + 2) square metres, and the area of the second garden is represented by the equation B = (x + 1) (x − 4) square metres. If the gardener wants the second garden to have 10 m2 more area than the ﬁrst garden, what should be the dimensions (width and length) of each garden?

5.10 Solve algebraically After this lesson, you will be able to… • understand that a quadratic equation can have one, two, or no real solutions. • apply algebraic manipulation to solve quadratic equations. • solve a variety of quadratic equations using algebraic methods.

Solve algebraically Recall that a quadratic can have one, two, or no real solutions. y

One real solution

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Two real solutions

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No real solutions

Some forms of quadratics can be solved using a few steps of algebraic manipulation. The key thing to remember for solving algebraically is that when you take the square root of a number

, the answers are

and

Example 1 Solve

for x.

Create a strategy Use inverse operations to isolate x on one side of the equation.

Apply the idea Write the equation

Add 3 to both sides

Multiply both sides by 16

Square root both sides

Evaluate the square root

Reﬂect and check Our two solutions are x = 12 and x = −12. To check these, substitute each into evaluate to see if it is a true statement. For example, checking x = −12:

and

Example 2 Solve (2x + 5)2 = 36 for x.

Create a strategy Use inverse operations to isolate x on one side of the equation.

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Apply the idea Write the equation

Square root both sides

Subtract 5 from both sides

We now have 2 linear equations to solve

Evaluate

Divide both sides by 2

Reﬂect and check Our two solutions are

and

. To check these, substitute each into (2x + 5)2 = 36 and

evaluate to see if it is a true statement. For example, checking

Example 3 Solve −3(x − 4)2 + 12 = 0 for x.

Create a strategy Use inverse operations to isolate x on one side of the equation.

Apply the idea Write the equation

Subtract 12 from both sides

Divide −3 on both sides

Square root both sides

Evaluate the square root

Add 4 to both sides

We now have 2 linear equations to solve

Evaluate

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Reﬂect and check Our two solutions are x = 2 and x = 6. To check these, substitute each into −3(x − 4)2 + 12 = 0 and evaluate to see if it is a true statement. For example, checking x = −3((2) − 4)2 + 12 = 0: −3((2) − 4)2 + 12 ≟ 0 −3(−2)2 + 12 ≟ 0 −3(4) + 12 ≟ 0 −12 + 12 ≟ 0 0 =0

Idea summary Quadratic equations can have up to two solutions. Quadratic equations in the form x2 = k (where k > 0) have two solutions.

5.10 Practice questions Understanding

Explorer 1–3, 4ab

Adventurer 1–3, 4cd

Trailblazer 1–3, 4d

Identify the similarities and differences between a linear equation and a quadratic equation.

Are the following statements true or false? a

Linear equations have only one solution.

Quadratic equations can potentially have three solutions.

The symbol ± means “plus or minus.”

To solve for x we must do the inverse operation.

The equation x2 − 144 = 0 has two solutions. One of those solutions is x = 12. What is the other solution?

Evaluate and round to three decimal places. a

Fluency 5

5, 6a–f, 7a–h, 8a–d, 9a–d, 5, 6c–i, 7c–j, 8b–e, 9c–f, 10a–c, 11ab 10ace, 11bc, 13

5, 6d–l, 7e–l, 8e–h, 9e–h, 10bdf, 11cd, 12, 13

Solve: a

x2 = 2

x2 = 25

x2 = 121

x2 = 294

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283

Ex 1

Ex 2

Solve: a

4y2 = 100

x2 − 10 = 15

81k2 + 8 = 24

−25v2 + 64 = 0

81x2 − 16 = 0

9x2 − 4 = 0

x2 − 121 = 0

(x − 3)2 = 64

(x − 6)2 = 2

(2 − x)2 = 81

(x − 7)2 = 81

(7 − x)2 = 81

(8x + 9)2 = 256

(x + 3)2 = 49

(10x − 9)2 = 0

(4x − 9)2 = 0

(−3 + 7x)2 = 0

(3y + 9)2 = 100

(5y − 9)2 = 25 y = −3x(2x + 6)

Find the solution(s) to these quadratics when y = 0. y = x(x + 7)

y = (2x − 9)2

y = 4x(x + 5)

y = 0.4x(1.8x − 7.3) g

y = −x(x − 5)

Find the solution(s) to these quadratics when y = 0. a

y = (x − 5) (x + 7)

y = (x + 1)2

y = (2x + 4) (x − 5)

y = (−3x − 12) (x − 1)

y = (8x − 5) (3x − 7)

y = (3x + 8) (5x − 7)

y = (3x − 17) (2x − d)

Solve for y = 0: a

y = −2 (x − 3)2 + 8

y = 4 (x + 1)2 − 16

y = 5 (x − 2)2 − 100

y = 4 (x − 6)2 − 36

y = −7 (x − 2)2 + 700

e 11

−3k2 = −12

Ex 3

Solve:

25y2 = 36

y = 6 (x − 4) − 24

For each equation: i

What is the turning point?

Identify any x-intercepts.

iii

Sketch the graph of the parabola.

y = (x − 5)2 − 25

y = −(x + 4)2 + 16

y = −3 (x − 6)2 + 27

y = 3 (x + 1) + 12

A dolphin jump can be described as a quadratic relationship where h represents height in metres above sea level and t is time in seconds: h = −2 (t − 6)2 + 18 When will the dolphin be above sea level?

The Widget and Trinket Emporium has released the forecast of its revenue over the next year. The revenue R (in dollars) at any point in time t (in months) is described by the equation: R = −(t − 12)2 + 4 When will the revenue be zero?

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Reasoning

14–16

14–17

Explain why the equation x2 − 9 = 0 has two solutions.

Explain why the equation x2 + 4 = 0 has no real solutions.

Identify the step where Ted made an error and explain what he should have done to correct the error: 1

Divide both sides by 25

Take the square root of both sides

Identify the step where Samantha made an error and explain what she should have done to correct the error: 1 2

Add 8 to both sides

Divide both sides by 36

Take the square root of both sides

Problem-solving 19, 20 18

19, 20

18–21

Solve the equation for x, in terms of a, b and c. a(x2 + b) − c = 0

Rehana thinks of a number. If she multiplies her number by four more than the number, she gets zero. a

Write an equation representing this situation, where Rehana’s number is x.

What could Rehana’s number be?

A square has sides of length (x + 3) cm and an area of 25 cm2. a

Write an equation for the area of the square, A, in terms of x.

Substitute A = 25 and solve for x, stating both possible solutions.

One of the solutions from (b) does not give a realistic length. What is the side length of the square?

A stone is thrown up and into a river, and its height (in metres) above the water can be modelled by the equation h = −3.2(t − 0.44)2 + 1.61, where t represents the time (in seconds) after the stone is thrown. a

What was the height at the time it was thrown?

At what times will the stone be at the same height above the water as it was when it was thrown?

If the ratio of the horizontal distance to the time it takes to reach the water is 5 : 3, what is the horizontal distance the stone travels before hitting the water? 5.10 Solve algebraically mathspace.co

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5.11 Solve by factorisation After this lesson, you will be able to… • understand and apply the null factor law in the context of factorising and solving quadratic equations. • solve quadratic equations using factorisation methods.

The null factor law The null factor law states that if the product of two or more factors is equal to zero (null), then at least one of those factors must be equal to zero. This means that if xy = 0, then either x = 0 or y = 0. We can use the null factor law to solve quadratic equations when they are written in factorised form. If a quadratic equation looks like this: 0 = a(x − x1) (x − x2) Then either: x − x1 = 0 or x − x2 = 0 By solving these equations, we get the solutions x = x1 and x = x2.

Example 1 Find the solutions to y = x (4x + 12)

Create a strategy To ﬁnd solutions, set y = 0 and solve by applying the null factor law.

Apply the idea x=0

First factor Second factor

4x + 12 = 0

Equate the ﬁrst factor to 0 Equate the second factor to 0

4x = 0 – 12

Subtract 12 from both sides

4x = −12

Evaluate

The solutions are x = 0 and x = −3

x = −3

Divide both sides by 4

Reﬂect and check To check the solutions, we substitute x = 0 and x = −3 into the equation y = x(4x + 12). Check x = 0 y = 0 (4 × 0 + 12)

y=0

Confirmed, x = 0 is a solution. Check x = −3

y = −3(4(−3) + 12)

y = −3(−12 + 12)

y = −3(0)

y=0

Confirmed x = −3 is a solution. 286

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Example 2 Find the solutions y = (x − 6) (x + 7)

Create a strategy To ﬁnd the solutions, set y = 0 and solve by applying the null factor law.

Apply the idea First factor

Second Factor

x−6=0

Equate the ﬁrst factor to 0

x=0+6

Add 6 to both sides

x=6

Evaluate

x+7=0

Equate the second factor to 0

x=0–7

Subtract 7 from both sides

x = −7

Evaluate

The solutions are x = 6 and x = −7

Reﬂect and check To check the solutions, we substitute x = 6 and x = −7 into the equation y = (x − 6)(x + 7). Check x = 6

y = (6 − 6)(6 + 7) y = 0 × 13 y=0

Confirmed, x = 6 is a solution. Check x = −7

y = (−7 − 6)(−7 + 7) y = −13 × 0 y=0

Confirmed x= −7 is a solution.

Idea summary The null factor law states that if the product of two or more factors is equal to 0, then at least one of those factors must be equal to 0. If (x − a) (x − b) = 0 then we know that either x − a = 0 or x − b = 0. We can solve the quadratic equation by solving each of these linear equations.

Solve by factorisation When solving a factorised quadratic such as y = (x − x1) (x − x2), we know two key points: • If we set y = 0, the solutions are the same as the x-values of the x-intercepts. • If (x − x1) (x − x1) = 0, we know, using the null factor law, that x = x1 or x = x2 are the solutions. Factorising a quadratic equation is one method we can use to ﬁnd its solutions.

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Example 3 The equation 3y = 15y2 has two solutions. Solve for both values of y.

Create a strategy Move all terms to one side, then factorise, and use the null factor law to solve.

Apply the idea 3y = 15y2

Write the equation

3y − 15y = 0

Subtract 15y2 from both sides

3y(1 − 5y) = 0

Factorise the quadratic expression

First Factor

Second Factor

3y = 0

Equate the ﬁrst factor to 0

y=0

Divide both sides by 3

1 − 5y = 0

Equate the second factor to 0

5y = 1

Add 1 to both sides

Divide both sides by 5

So the solutions are y = 0 and y =

Idea summary Factorise equations such as ax2 + bx = 0 into x (ax + b) = 0 and use the null factor law to solve.

Solve quadratic trinomials A quadratic equation with three terms (ax2 + bx + c) is called a quadratic trinomial. If a = 1, we can call this a monic quadratic trinomial. To factorise a monic quadratic trionomial x2 + bx + c, we need to ﬁnd numbers p and q so that b = p + q and c = pq Then, the quadratic can be factorised by grouping: x2 + bx + c = x2 + ( p + q)x + pq 2

Since b = p + q and c = pq

= x + px + qx + pq

By expanding the middle term we have four terms

= x(x + p) + q(x + p)

Factorise the ﬁrst two and the last two terms

= (x + q) (x + p)

Finish the factorisation

If you need to practice factorising, return to Chapter 3 on Algebra. We can see that, if we ﬁnd p and q so that b = p + q and c = pq, then the monic quadratic x2 + bx + c is factorised to (x + p) (x + q). We can use this as a shortcut, so that we don’t need to do the working for factorisation by grouping each time. 288

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Example 4 Solve x2 + 9x + 20 = 0 for x.

Create a strategy Factorise, then use the null factor law to solve.

Apply the idea x2 + 9x + 20 = 0

Write the equation

To factorise the quadratic expression, we must ﬁnd two integers p and q which add to 9 and multiply to 20. Start by considering the factors of 20. We try pairs of factors of 20 to see which gives the correct sum: p

1 2 4

20 10 5

p+q 21 12 9

The required values are p = 4 and q = 5.

(x + 4) (x + 5) = 0

First factor Second factor

Factorise the quadratic expression

x+4=0

Equate the ﬁrst factor to 0

x = −4

Subtract 4 from both sides

x+5=0

Equate the second factor to 0

x = −5

Subtract 5 to both sides

So the solutions are x = −4 and x = −5.

Reﬂect and check We can check the factorisation by expanding: (x + 4) (x + 5) = x2 + 4x + 5x + 20 = x2 + 9x + 20 We can check the solutions by substitution. For example, for x = −4: x2 + 9x + 20 = (−4)2 + 9(−4) + 20 = 16 − 36 + 20 =0

Example 5 Solve x2 = 17x + 60 for x.

Create a strategy Move all terms to the left, factorise, then use the null factor law to solve.

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Apply the idea x2 = 17x + 60 2

x − 17x − 60 = 0

Write the equation Move all the terms to the left hand side

To factorise the quadratic expression, we must ﬁnd two integers p and q which add to −17 and multiply to −60. Start by considering the factors of 60. However, since the product is negative, one of the factors must be negative. The sum is also negative, which tells us that the greater of the factors must be negative. We try pairs of factors of 60 (with the greater factor negative), to see which gives the correct sum: p

1 2 3

−60 −30 −20

p+q −59 −28 −17

The required values are p = 3 and q = −20. (x + 3) (x − 20) = 0 First factor Second factor

Factorise the quadratic expression

x+3=0

Equate the ﬁrst factor to 0

x = −3

Subtract 3 from both sides

x − 20 = 0

Equate the second factor to 0

x = 20

Add 20 to both sides

So the solutions are x = −3 and x = 20.

Reﬂect and check We can check the factorisation by expanding: (x + 3) (x − 20) = x2 − 20x + 3x − 60 = x2 − 17x − 60 We can check the solutions by substitution. For example, for x = −3: x2 − 17x − 60 = (−3)2 − 17(−3) − 60 = 9 + 51 − 60 =0

Idea summary An expression of the form x2 + bx + c is a monic quadratic trinomial. To factorise expressions like this, we ﬁnd a pair of numbers p and q such that p + q = b and pq = c. Then the factorisation is x2 + bx + c = (x + p) (x + q).

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5.11 Practice questions Understanding

Ex 3

Ex 4

Ex 5

Trailblazer 1–2

Can we use factorisation as a method to solve all quadratics? Which of these represent quadratic equations in factorised form? A

0 = x2 − 2x + 1

0 = x (x − 2)

0 = (x + 2)2 + 3

0 = (x − 2) (x + 4)

Why is the factorised form of a quadratic useful when trying to solve it? 3a–d, 4ace, 5ac, 6a–c, 7a–d, 8a–f, 9a–d, 10, 11ab

Fluency

Ex 2

Adventurer 1–2

Ex 1

Explorer 1–2

3c–f, 4b–d, 5bc, 6b–e, 7b–f, 8c–h, 9c–f, 10, 11b–d

3c–g, 4c–e, 5cd, 6c–f, 7c–h, 8e–l, 9e–h, 10, 11c–f

Find the solution(s) to: a

y = x (x − 1)

y = 2x (x + 3)

y = (x − 1) (x + 1)

y = (x + 4) (x + 2)

y = (2x + 1) (x + 3)

4m (m + 5) = 0

y = (x + 1) (x − 2)

Solve using the null factor law: a

x (x + 7) = 0

b (10b − 90) = 0

x (2x − 9) = 0

Solve using the null factor law: a

(x − 5) (x + 7) = 0

(8x − 5) (3x − 7) = 0

(3x + 8) (5x − 7) = 0

(3x − 17) (2x − d) = 0

Solve: a

m2 − 14m = 0

5m2 + 2m = 0

2y − 6y2 = 0

2y − 14y2 = 0

−3m (m + 2) = 0

3 (n2 + 8) + 18n = 0

Solve: a

m2 + 8m + 12 = 0

m2 + 11m + 24 = 0

y2 + 5y + 6 = 0

y2 + 9y + 14 = 0

x + 8x + 15 = 0

x2 + 7x + 12 = 0

x2 + 3x + 2 = 0

x2 + 9x + 18 = 0

Solve: a

m2 − m − 20 = 0

m2 − 7m + 6 = 0

m2 − 11m + 30 = 0

m2 − 27m + 182 = 0

x2 + 13x + 42 = 0

x2 + 4x − 21 = 0

x2 − 5x − 50 = 0

x2 − 22x + 121 = 0

x2 + 8x − 20 = 0

x (x + 18) + 80 = 0

x (x + 2) − 48 = 0

x − 13x − 114 = 0

Solve: a

m2 = 3m + 10

−6y = y + 8

b f

x2 = 4 − 3x 2

10y = y + 24

c g

x2 − 12x = −20 2

−m − 7m = −18

m2 + 5m = 14

−n2 − 5n = −84

Find the coordinates of the x-intercepts of a parabola with equation y = x2 + x − 6. 5.11 Solve by factorisation mathspace.co

291

Find the solution(s) when y = 0 for: a e

y = x2 + 3x y = −4x (x + 4)

Reasoning

y = 2x2 + 12x

y = 4x2 + 12x

y = −x2 + 18x

y = −x − 15x

12–14

12–15

Explain why this equation (x − 2) (x − 2) = 0 has only one solution, and how this will affect the graph of y = (x − 2) (x − 2).

Explain why the quadratic equation cannot be factorised using integers. x2 − 4x + 5 = 0

Identify the error in the factoring attempt and explain the correct approach.

x2 + 5x + 6 = 0 (x + 1) (x + 5) = 0 15

x = −5, −1

Factorise Solve for x

Explain why the quadratic equation will always have two solutions and state the two solutions. ax2 + bx = 0

Problem-solving 16, 18ab 16

17, 18cd, 19

A kingﬁsher dives in search of food in a lake and follows a quadratic pathway h = t2 − 9t + 8, where h is the height in metres above sea level and the t is time in seconds. a

Graph the equation including the t-intercepts and turning point.

Find the times at which the kingﬁsher goes under the surface of the lake and then reappears on the surface of the lake.

How long was the kingﬁsher under the water?

Find the value of a if: a

The non-zero solution of the equation x (x + a) = 0 is x = 10.

The non-zero solution of the equation x (x + a) = 0 is x = −1.

Solve: a

16, 18bc

Software engineers are designing a self-serve checkout system for a supermarket. They notice that the traﬃc through the store during the day is described by the function: C = −t (t − 12) where C is the number of customers and t is the number of hours after the store opens. To meet the peak demand, the engineers allow for an extra checkout machine to automatically turn on when the number of customers ﬁrst reaches 27 people, and to automatically turn off when it next falls below 27 people.

292

Find the times t when the number of customers is equal to 27 people.

After how many hours will the extra checkout machine turn on after opening?

How many hours will the extra machine be on for?

Mathspace Victorian Curriculum – Year 9 mathspace.co

INVESTIGATION

Other non-linear graphs Investigate how different functions change and move graphically when their equations change.

Objectives To identify and describe patterns found in the transformed functions

Other non-linear graphs We have explored four types of function transformations: reﬂections in the x-axis, vertical dilations, horizontal shifts, and vertical shifts. For example, a transformed quadratic function generally takes the form y = a(x − h)2 + k

Interactive exploration Discover this concept in action online

For the function

mathspace.co

, describe what values of a, b, or c must change in order to:

• Reﬂect in the x-axis • Stretch it vertically • Slide it to the left or right • Move it up or down The function

represents a square root.

How does the graph look different when you apply each transformation compared to the function? Consider the function y = x3 • When you explore y = x3, does it have any similarities with the quadratic function y = x2? • How does it behave with the transformations? Examine the functions y = 2x and • How are the graphs of each different? • What patterns do you notice when they undergo transformations?

Investigation: Other non-linear graphs mathspace.co

293

Investigate 1.

Were there any transformations that had a similar effect across all functions?

Were there any transformations that had very different outcomes depending on the function?

How do the transformations of these non-linear functions compare to the transformations you explored with quadratic functions? Are there any notable differences or similarities?

After adjusting the sliders, did you ever notice a function behaving in a way you didn’t expect? Describe what surprised you and why.

Discussion 1.

Predict how another function, not in this investigation, might transform. For example, how might y = ∣x∣ (an absolute value function) change with the sliders?

Some functions have points or lines where they don’t exist, like

, which doesn’t

exist at x = 0. How do the transformations affect these special points or lines?

Did you know?

Quadratic equations are the key to perfect basketball shots! When a player shoots, the ball’s path forms a parabolic arc, modeled by a quadratic equation. The ball’s height, angle, and distance all work together to create that perfect swish. Scientists and engineers use quadratics to analyse and improve shooting techniques. So, next time you see a three-pointer, remember—quadratics are behind it! 294

Mathspace Victorian Curriculum – Year 9 mathspace.co

5 Chapter review 1

Does the equation represent a quadratic relationship between x and y? a

4x − 6y − 3 = 0

y = 2x3 − 3x2 + x

x y −2 16 −1 9 0 4 1 1 2 0

−1 0 1 2 3

1 2 4 8

For each graph, identify: i

If the graph is concave up or down.

The turning point.

iii

The equation for the axis of symmetry.

The x and y-intercepts.

y −8 −4

−8

−12

−16

−20

2 −8 −6 −4 −2

−4

y = (3x2 − 1)2

Does the table of values show a relationship that is linear, quadratic, or neither? a

y = 2x2 − x + 3

x 2

−24

In an environmental study, the temperature of the water in a lake is monitored throughout the day. The graph shows the temperature (c) in Celsius change over time (d) in days. Use the graph to approximate answers to each question. a

What was the initial temperature?

When did the temperature reach 0 °C?

What was the maximum temperature reached?

When is the temperature decreasing?

30 20 10 −10 −20

d 2 4 6 8 10 12 14 16 18 20

−30 −40 −50

Chapter 5 review mathspace.co

295

Match each description to the equation. i iii

y = 3x2 y = −x2

y = −0.5x2 y = 0.5x2

Narrower than y = x2 and opens upwards.

Wider than y = x2 and has a minimum point.

Wider than y = −x2 and opens downwards.

Narrower than y = −0.5x2 and has a maximum point.

ii iv

The graph shown has an equation of the form y = ax2. By using a point on the graph, ﬁnd the value of a, and hence the equation of the graph.

4 3

2 1 −5 −4 −3 −2 −1 −1 −2

x 1 2 3 4 5

−3 −4 −5

Each equation represents a translation of the graph of y = x2. Describe the direction and the number of units it has been translated. a

y = x2 + 5

y = (x − 4)2

y = x2 − 2

Write the equation of the graph in the form y = (x − h)2 + k. a

y 5 4 3 2 1

−7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

296

y = (x + 1)2

x 1 2 3 4

Mathspace Victorian Curriculum – Year 9 mathspace.co

11 10 9 8 7 6 5 4 3 2 1 −3−2 −1−1

x 1 2 3 4 5 6 7 8 9

For each graph: i

What are the x-intercepts?

What is the y-intercept?

iii

Write the equation of the parabola in factorised form.

−8 −6 −4 −2 −2

4 2

−6

−8

For each quadratic equation: i

What is the turning point?

What is the y-intercept?

iii

Sketch the graph.

Use the graph to determine the solution(s) to the quadratic equation when y = 0.

y = (x − 1)2 − 4

y = 2 (x + 3)2 − 8

y = 3x (x − 1)

y = (2 − x) (3 + x)

y = −x2 + 5x + 6

Solve: a

16y2 = 64

(x − 8)2 = 49

(7x + 10)2 = 121

0 = 4 (x − 3)2 − 96

5 (x − 5)2 − 25 = 0

3m (m + 6) = 0

(8x − 6) (3x − 8) = 0

Solve: a

b2 − 22b = 0

t2 − 14t + 45 = 0

y2 − 10y = −24

−2x2 − 9x + 5 = 0

3 (h2 + 8) + 22h = 0

e 13

−4

−8 −6 −4 −2 −2

Identify the error in the factoring attempt and explain the correct approach.

x2 + 6x + 7 = 0 (x + 2) (x + 5) = 0

x = −5, −2

Factorise Solve for x

Chapter 5 review mathspace.co

297

The equation 4x2 + kx − 15 = 0 has two solutions. One solution is x = −3. a

Find the value of the coefficient k.

Find the second solution.

A gardener is planting ﬂowers in a rectangular garden. The area of the garden, A, is given by the equation A = −4 (x − 3)2 + 36, where x is the width of the garden in metres. a

Find the width of the garden that maximises the area.

Calculate the maximum area.

Determine the length of the garden corresponding to this width.

A ball is thrown upward and away from the top of a tower that is 260 metres above the ground. The height y (in metres) of the ball at time x (in seconds) is given by: y = −25x2 + 85x + 260 The graph of this relationship is shown.

Identify and interpret the y-intercept of the graph.

350

Identify and interpret the x-intercept shown.

300

The ball reaches its maximum height after 1.7 seconds. Determine the maximum height reached by the ball.

250 200 150 100 50

x 1

298

Mathspace Victorian Curriculum – Year 9 mathspace.co

Alex has a fountain in his garden. The stream of water takes the path of a parabola that reaches a maximum height of 15 m above the ground, 15 m from the fountain. The path of water sprayed from the fountain is shown in the graph, where (0, 0) is the position of the fountain, x is the horizontal distance from the fountain and y is the height above the ground.

y 16 14 12

Write the quadratic function for the path of the water.

State the solutions of the quadratic equation. when y = 0.

Alex wants to adjust the fountain so that it reaches a maximum height of 8 m above the ground, 20 m from the fountain. What is the quadratic function that Alex should use?

8 4 2

x 5

Chapter 5 review mathspace.co

299

Big ideas Accurate calculation and interpretation of interest, earnings, and profit are crucial for effective money management in earnings, investments, and loans. Curriculum Outcomes VC2M9A06 use mathematical modelling to solve applied problems involving change, including financial contexts involving simple interest; formulate problems, choosing to use either linear or quadratic functions or other simple variations; interpret solutions in terms of the context; evaluate the model and report methods and findings

Finance Chapter outline 6.01 6.02 6.03 6.04

Investigation: Finances Cost, profit, revenue Income Simple interest Re-investing with simple interest Chapter 6 review

302 304 314 325 334 347

Long-term saving is a great way to set yourself up in the future. If you saved & invested $1 each week from your 5th birthday until you turned 65 , how much would you have?

INVESTIGATION

Finances Investigate how to create a budget for initiating a sustainable garden, taking into account various costs, and potential proﬁts.

Objectives To solve problems involving financial contexts

Finances Budgeting for a sustainable garden Urban gardening and sustainable farming are growing trends in Australia, driven by a desire for fresh, local produce and a commitment to environmental sustainability. For this investigation, you will engage in research and a cost analysis for your own garden. Research the types of plants for your garden and the potential costs to create your garden: 1.

Suitable plants: Research which plants are suitable for your local climate and are beneﬁcial for the environment. List at least three plants, noting their price per seed/seedling.

Tools and essentials: Make a list of essential gardening tools (e.g., shovel, rake, watering can) and their respective prices.

Soil and fertilisers: Find out the cost of organic soil and compost, considering how much each bag can cover in terms of area.

Watering solutions: Look into the cost of different watering systems like drip irrigation or soaker hoses, and estimate their running costs.

Seed and plant budgeting • With a plant budget of $100, how many of each plant could you buy? • Which combination gives you the most diversity of plants? Soil coverage estimation: • If you wish to have a garden that is 5 metres by 3 metres and 25 millimetres in depth, how many bags of soil would you need? • How would this number change if you decided to expand your garden? Watering system cost analysis • Calculate the cost of installing and running your chosen watering system for a month. • How would the cost differ if you watered by hand?

302

Mathspace Victorian Curriculum – Year 9 mathspace.co

Investigate 1.

Calculate your expected total spending for the initial setup of your garden, taking into account the seeds, tools, soil, and watering system.

Assume you’ll harvest your plants in three months. Estimate how much produce (e.g., kilograms of tomatoes) you might obtain.

Calculate the price per kilogram you’d need to set to recoup your initial costs in three months if you decide to sell 75% of your harvest.

Discussion 1.

Imagine you’re hired to set up a similar garden for a local school. Create a quote for your services, including a call-out fee, labor costs, and materials.

Consider the water sources for your garden. How could you incorporate methods to reduce water usage or employ sustainable water sources (like harvested rainwater)? Estimate potential costs or savings associated with these methods.

Did you know?

In finance, markets are often described as either bull or bear depending on the overall trend. A bull market signals rising stock prices and investor confidence, while a bear market reflects falling prices and pessimism. The terms come from how the animals attack: bulls charge upward, symbolizing rising markets, and bears swipe downward, representing falling markets. Bulls thrive during economic growth, while bears dominate in downturns. Bull markets tend to last longer, but both are part of the financial cycle. Investigation: Finances mathspace.co

303

6.01

Cost, profit, revenue After this lesson, you will be able to… • model revenue and cost using linear equations. • define what is a break-even point. • apply the formulas to calculate profit and loss. • solve problems involving profit, revenue and costs.

Revenue and cost Businesses keep track of two important things: • Revenue: This is the money the business earns from selling products or services. • Cost: This is the money the business spends to operate, such as buying materials or paying workers. There are two types of costs: • Fixed costs: Costs that do not change, no matter how many items are sold (like rent). • Variable costs: Costs that change depending on how many items are sold (like the cost of materials). We can model both revenue and cost using linear equations. In these equations: • The number of items sold (or services provided) is the independent variable (x). • The revenue or cost in dollars is the dependent variable ( y). Key points: • The slope of each line represents how much the cost or revenue increases for each item sold (cost per item or revenue per item). • The y-intercept (where the line crosses the y-axis) represents the fixed costs (costs that exist even when no items are sold). y 10 9 8 7 6 5 Break-even point 4 Cost 3 2 Revenue 1 x 1 2 3 4 5 6 7 8 9 10

304

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A business breaks even when the revenue and cost are the same - meaning the business isn’t losing or making money yet, but it’s covering its costs. In a graph, the break-even point is where the revenue line and cost line meet. To find this point, we set the revenue equal to the cost: Revenue = cost By solving this equation, we can determine when a business will break even.

Example 1 Luana sells handmade scarves from a stall at her local market for $25. Each scarf costs Luana $15 to make, and the cost to have a stall at the market for a day is $100. a Determine the revenue from selling 20 scarves.

Create a strategy Multiply the revenue of $25 per scarf by 20 scarves.

Apply the idea R = Selling price per scarf × Number of scarves

Find revenue

= 25 × 20 Substitute known values = $500 Evaluate

b Write an equation for the daily revenue, R, as a function of x, the number of scarves sold in a day.

Create a strategy Multiply the revenue of $25 per scarf by x.

Apply the idea R = 25x

Write the equation

c Write an equation for the daily cost, C, as a function of x.

Create a strategy The cost is $15 per scarf plus a ﬁxed $100 for the stall, so the equation needs to multiply 15 by x and add 100.

Apply the idea C = 15x + 100

Write the equation

Example 2 Sven decides to start teaching a ﬁtness class. The predicted cost (C) and revenue (R) from running the class are shown on the graph.

72 66 60 54 48 42 36 30 24 18 12 6

Dollars

C = 6x + 24

R = 12x

Participants 1

6.01 Cost, profit, revenue mathspace.co

305

a Find the revenue and cost if Sven has 2 participants.

Create a strategy Use the graph to ﬁnd the revenue and cost when x = 2.

Apply the idea 72 66 60 54 48 42 36 30 24 18 12 6

On the revenue line, when x = 2, y = 24, so the revenue when there are 2 participants is $24.

Dollars

C = 6x + 24

On the cost line, when x = 2, y = 36, so the cost when there are 2 participants is $36.

R = 12x

Participants 1

Reﬂect and check Alternatively, we could have used the equations: R = 12x

Write the equation for revenue

= 12 × 2

Substitute x = 2

= $24

Evaluate

C = 6x + 24

Write the equation for the cost

= 6 × 2 + 24 Substitute x = 2 = $36

Evaluate

b How many participants does Sven need to break-even?

Create a strategy Use the graph to ﬁnd the point of intersection of the revenue and cost lines.

Apply the idea The point of intersection is (4, 48). So if Sven has 4 students, he will break-even.

Reﬂect and check Alternatively, we could have solved this algebraically:

306

R=C

Equate revenue and cost

12x = 6x + 24

Substitute the expressions

6x = 24

Subtract 6x from both sides

x=4

Divide both sides by 6 to solve for x

Mathspace Victorian Curriculum – Year 9 mathspace.co

Idea summary Revenue and cost can be modelled using linear equations. The break-even point, where the two lines intersect, is when revenue and cost are the same.

Profit and loss If revenue is greater than cost, a business is making money (making a profit (P)). If revenue (R) is less than cost (C), a business is losing money (making a loss). Profit can be calculated using the formula: Profit = Revenue − Cost If the profit is zero, then the business is breaking even. If the profit is negative, then the business is making a loss. A profit of −$1000 is a loss of $1000. If the profit is positive, then the business is making a profit.

Interactive exploration

Discover this concept in action online

mathspace.co

Example 3 Thao umpires hockey games on the weekend and is paid $50 per game. His costs work out to around $10 per game in travel, plus a ﬁxed $210 for yearly registration to the hockey association. Thao umpires 52 games in a year. a Calculate the annual revenue.

Create a strategy Multiply the revenue per game by the number of games.

Apply the idea Revenue = Pay per game × Number of games = 50 × 52

Substitute the values

= $2600

Evaluate

b Calculate the annual cost.

Create a strategy Multiply the cost per game by the number of games and add the ﬁxed cost.

6.01 Cost, profit, revenue mathspace.co

307

Apply the idea Cost = Cost per game × Number of games + Flat registration fee = 10 × 52 + 210

Substitute the values

= $730

Evaluate

c Calculate the proﬁt that Thao will make in a year.

Create a strategy Subtract the cost from the revenue.

Apply the idea Profit = Revenue – Cost

Proﬁt formula

= 2600 – 730

Substitute the values

= $1870

Evaluate

In a year of umpiring, Thao will earn a proﬁt of $1870.

Example 4 Tayna has a small business selling bunches of cut ﬂowers. The weekly cost (C) and revenue (R) for the business, based on the number of bunches sold (x), are shown in the graph.

Dollars 600 500 400 C = 20x + 200

300 200

a Determine the proﬁt (or loss) that Tayna makes if she sells 10 bunches in a week.

100

R = 60x Bunches 1 2 3 4 5 6 7 8 9 10

Create a strategy Find the difference between the revenue and the cost when x = 10.

Apply the idea On the cost line, when x = 10, y = 400. So the cost when selling 10 bunches is $400.

Dollars 600

On the revenue line, when x = 10, y = 600. So the revenue when selling 10 bunches is $600.

500 400 300

Cost = 20x + 200

200 100

Revenue = 60x Bunches 1 2 3 4 5 6 7 8 9 10

308

Mathspace Victorian Curriculum – Year 9 mathspace.co

The revenue is greater than the cost, so Tayna will make a proﬁt. The difference between 600 and 400 is 200, so the proﬁt is $200.

Reﬂect and check Alternatively, we could have solved this algebraically: Profit = revenue – cost

Write the equation for proﬁt

= 60x − (20x + 200)

Substitute the expressions

= 40x – 200

Simplify

= 40 × 10 – 200

Substitute x = 10

= $200

Evaluate

b Determine the proﬁt (or loss) that Tayna makes if she sells 2 bunches in a week.

Create a strategy Find the difference between the revenue and the cost when x = 2.

Apply the idea On the cost line, when x = 2, y = 240. So the cost when selling 2 bunches is $240.

Dollars 600

On the revenue line, when x = 2, y = 120. So the revenue when selling 2 bunches is $120.

500 400 300

Cost = 20x + 200

200 100

Revenue = 60x

The revenue is less than the cost, so Tayna will make a loss. The difference between 240 and 120 is 120, so the loss is $120.

Bunches 1 2 3 4 5 6 7 8 9 10

Reﬂect and check Alternatively, we could have solved this algebraically: Profit = 4x – 200

Write the simpliﬁed equation for proﬁt from (a)

= 40 × 2 – 200

Substitute x = 2

= −$120

Evaluate

A proﬁt of −$120 is a loss of $120.

Idea summary Profit can be calculated using the formula: Profit = Revenue − Cost A profit of zero dollars is breaking even. A negative profit is a loss.

6.01 Cost, profit, revenue mathspace.co

309

6.01 Practice questions Understanding 1

5 Ex 1a

Ex 1b

Ex 1c

310

Explorer

1–3

Adventurer

1–3

Trailblazer

Identify the formula for calculating proﬁt: A

Profit = Cost – Revenue

Profit = Revenue − Cost

Profit = Revenue – Loss

Profit = Loss − Revenue

Money spent is called loss.

Is each statement true or false? a

Money earned is called revenue.

A proﬁt of −$500 is a loss of $500.

The break-even point is when revenue and cost are equal.

When given the graph for revenue and cost, how can we ﬁnd the break-even point?

Fluency 4

1–3

4–11, 13, 15

4–11, 13, 15, 17

4–10, 12, 14, 16, 17

Determine the proﬁt (or loss) made if: a

The revenue is $6590 and the cost is $1250.

The cost is $250 and the revenue is $980.

The revenue is $12 500 and the cost is $13 000.

The cost is $720 and the revenue is $240.

Calculate the revenue if: a

68 drinks are sold for $4.80 each.

A personal trainer runs a group session with 12 people, earning $9 per person.

15 pairs of jeans are sold for $84 per pair.

Write an equation for revenue (R) if: a

A restaurant sells x meals for $26 each.

A computer store sells x computers for $1890 each.

Calculate the cost if: a

A pop-up clothing store spends $12 per shirt for 150 shirts, plus a ﬁxed cost of $450 to hire space in a shopping centre.

A bakery spends $1800 on rent for the month and purchases 540 kilograms of ﬂour at $1.50 per kilogram.

Write an equation for the cost (C) in each scenario: a

An artist creates x paintings. Supplies like canvas and paint cost $170 for each painting, and they also spend a ﬁxed $220 on advertising.

A furniture maker rents a workshop for $440 and purchases materials to make x items. The cost of materials is $520 per item.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Ex 2

Fatu makes cakes and sells them at a market stall. The graph shows her cost (C) and revenue (R) where x is the number of cakes sold. a

Find the revenue if Fatu sells 20 cakes.

Find the cost if Fatu sells 20 cakes.

What is the proﬁt or loss from selling 20 cakes?

How many cakes does Fatu need to sell to break-even?

What are the ﬁxed costs for Fatu?

Dollars 400 300 C = 10x + 100

200 100

R = 20x

2 4 6 8 10 12 14 16 18 20

Each week Vince sells x motors. The weekly revenue and cost for his business are shown in the graph. a b c

Find the cost and revenue if Vince doesn’t sell any motors. Find the cost and revenue if Vince sells 8 motors. How many motors must Vince sell to cover his costs?

Dollars 60 000 50 000 40 000 30 000

C = 2500x + 10 000

20 000 10 000

R = 5000x 2

x 8 10 12 14 16

The revenue and cost for a business are given by R = 120x and C = 40x + 380 respectively, where x is the number of items sold. a

Graph both the revenue (R) and cost (C) lines on the same Cartesian plane.

Find the revenue per item.

Find the fixed part of the cost.

Find the revenue if 5 items are sold.

Find the cost if 12 items are sold.

For each of the businesses with revenue (R) and cost (C) descriptions, ﬁnd: i

The revenue per item

The ﬁxed costs

iii

The proﬁt or loss value for 3 items

The proﬁt or loss for 10 items

Business A selling ﬂoor lamps: R = 150x, C = 40x + 400

Business B selling USB drives: R = 25.5x, C = 1250 + 10x

Business C selling digital keyboards: R = 3500x, C = 10 000 + 1250x

6.01 Cost, profit, revenue mathspace.co

311

13 Ex 3

Ex 4

Aser has a website design business with a revenue of $450 per website. His costs are $80 per website, plus a ﬁxed $240 for the week. Next week, Aser will design 4 websites. a

Determine next week’s revenue.

Determine the cost for the week.

Calculate the proﬁt that Aser will make next week.

Wasim sells gourmet mushrooms to local restaurants for $40 per box. Each box of mushrooms costs them $20 to produce, and the cost to market and distribute the mushrooms to restaurants each week is $120. In a slow week, Wasim sells just 5 boxes. a

Determine the revenue for the week.

Determine the cost for the week.

Does Wasim make a proﬁt or a loss?

Calculate the proﬁt (or loss) that Wasim makes that week.

Sasha collects washing machines which are no longer Dollars working. After ﬁxing the machines she sells them on. 2500 Her monthly cost (C) and revenue (R) are shown in the graph, where x is the number of washing 2000 machines sold. a

Determine the proﬁt (or loss) that Sasha makes if she sells 4 washing machines in a month.

Determine the proﬁt (or loss) that Sasha makes if she sells 1 washing machine in a month.

C = 250x + 500

1500 1000 500 R = 500x 1

Each week, Kathie makes x reusable bags and sells them online. Her weekly revenue (R) and cost (C) are shown in the graph. a

Determine the proﬁt (or loss) that Kathie makes if she sells 20 bags in a week.

Determine the proﬁt (or loss) she makes if she sells 0 bags in a week.

Solve the equation 35x = 10x + 100 to ﬁnd the number of bags Kathie must sell in a week to break-even.

312

700 Dollars 650 600 550 500 450 400 350 300 250 200 150 100 50 R = 35x 5

x 5

C = 10x + 100

x 10

Ismene runs a catering business that supplies platters of food for $50 each. Ingredients cost her $25 for each platter and other ﬁxed costs like advertising come to $200 for a month. a

Write equations for the monthly cost C and revenue R, if Ismene sells x platters in a month.

How many platters does Ismene need to sell each month to break-even?

How much proﬁt (or loss) will Ismene make each month if she sells 4 platters?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Reasoning 18

19, 20

For this graph: a

Using this example of cost and revenue, explain why 800 it can be diﬃcult for some businesses when the proﬁt 720 640 margin per item is so small and costs start to rise.

Consider what might happen to the break-even point for example if the cost per item went up from 10 to 11 and they were unable to raise the revenue per item. b

18–20

Is it harder for a business to break even if the ﬁxed price rises from $120 to $200, or if the cost per item rises from $10 to $10.50?

y = 10x + 120

560 480 400 320 240 160 80

y = 12x 10

20 30 40 50 60

The cost of most businesses has a variable (per item) part and a ﬁxed part. a

Consider the following example, where the cost is purely variable. R = 20x C = 5x

Will the business ever make a loss? Justify your answer.

Give an equation for the proﬁt (P) in terms of x.

Consider the following example, where the cost is purely ﬁxed. R = 20x C = 500

Describe the graph of the cost.

Give an equation for the proﬁt (P) in terms of x.

Naia is thinking about opening a smoothie stand. Her predicted weekly revenue (R) and cost (C), where x is the number of smoothies sold, are: R = 8x C = 9x + 600 a

Graph the revenue (R) and cost (C) lines on the same Cartesian plane.

Explain why Naia will never break-even with this model.

What does Naia need to change in order for the smoothie stand to make a proﬁt?

Problem-solving 21, 22 21

21, 22

21–23

Salvio is considering opening a restaurant. He has calculated his ﬁxed costs at $8 000 each month. He expects the cost to produce each meal will be $15. a

Write an equation for Salvio’s monthly cost, C, if he sells x meals.

Determine the number of meals that will result in a monthly cost of $20 000.

What revenue per meal will cause Salvio’s restaurant to break-even when the monthly cost is $20 000?

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Helene makes wooden bowls. Each bowl costs her $20 to make and she sells them for $70 each. She also spends money on advertising and rent for the workshop each month (her ﬁxed costs). a

Calculate the revenue when Helene sells 6 bowls in a month.

If Helene breaks even when she sells 6 bowls in a month, determine the ﬁxed part of the cost.

Find the proﬁt made if Helene sells 20 bowls.

The revenue and cost for a small business are given by R = 110x and C = 50x + 390 respectively, where x is the number of items sold. a

Graph the revenue (R) and cost (C) lines on the same Cartesian plane.

Find the proﬁt if 10 items are sold.

Find the proﬁt if 10 items are sold, as a percentage of the revenue. Round your answer to one decimal place.

Give an equation for the proﬁt, in terms of x.

Give an expression for the proﬁt as a percentage of the revenue, in terms of x.

6.02

Income

After this lesson, you will be able to… • understand the concepts of salary, wage, and commission as different ways of earning income. • calculate and compare salaries using time conversions. • model the income from wages using linear equations. • model the income from commission using linear equations. • apply the understanding of salary, wage, and commission to real-world scenarios.

Salary Income is the amount of money a person earns. One common way to earn income is through a salary. A salary is a set amount of money you get over a certain time period, usually a year. When earning a salary, you get paid the same amount no matter how many hours you work (unless there are special arrangements like overtime or bonuses). For example, a teacher is paid the same annual (yearly) salary, regardless of how much time they spend preparing lessons and marking assessments. We can calculate and compare income using these time conversions:

1 year = 12 months 1 year = 26 fortnights 1 year = 52 weeks

We avoid using 1 month ≈ 4 weeks as it’s not accurate. If converting between months and weeks, convert to an annual value ﬁrst. 314

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Example 1 Derek is considering three different job offers and wants to choose the one that will pay him the greatest salary: • Offer 1: $2917 per month • Offer 2: $955.73 per week • Offer 3: $52 170 per year a Find the annual income for Offer 1.

Create a strategy Convert the monthly salary to its annual equivalent.

Apply the idea Annual income = 2917 × 12

Multiply the monthly salary by 12

= $35 004

Evaluate

Reﬂect and check Verify the calculation by dividing the annual income by 12. $35 004 ÷ 12 = $2917 The value matches the monthly salary, so it is correct. b Find the annual income for Offer 2.

Create a strategy Convert the weekly salary to its annual equivalent.

Apply the idea Annual income = 955.73 × 52 = $49 697.96

Multiply the weekly salary by 52 Evaluate

Reﬂect and check Verify the calculation by dividing the annual income by 52. $49 697.96 ÷ 52 = $955.73 The value matches the weekly salary, so it is correct. c Which job offer should Derek accept?

Create a strategy Compare the annual incomes and choose the greatest.

Apply the idea

• Offer 1: $35 004 per year • Offer 2: $49 697.96 per year • Offer 3: $52 170 per year

Offer 3 has the highest annual income, so Derek should choose Offer 3. 6.02 Income mathspace.co

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Idea summary A salary means earning a set amount of income (regardless of the hours worked). To calculate and compare income, use the conversions: 1 year = 12 months = 26 fortnights = 52 weeks

Wage Another way to earn income is with a wage. A wage is based on how much money you earn per hour (or another time period). Since your income depends on how long you work, we can model this using a linear equation: Income = Rate of pay × Time worked

y = kx y is the income k is the rate of pay x is the time worked Sometimes, the rate of pay can increase under certain conditions. For example, if you work on a public holiday, you might be paid 1.5 times your normal rate (this is called time-and-a-half) or even double your normal rate (double-time). Wages and salaries are paid differently because they account for different work conditions. For example: • A job that is paid by salary usually comes with extra benefits like sick leave or holiday leave. • Someone with a wage might have a higher hourly rate but doesn’t get paid leave benefits.

Example 2 Naziha earns $385 per hour as a lawyer. a Calculate Naziha’s weekly wage if she works 30 hours in a week.

Create a strategy Multiply the hourly rate of pay by the number of hours.

Apply the idea Income = 385 × 30 = $11 550

Write the multiplication Evaluate

Reﬂect and check Verify the calculation by dividing the weekly wage by 30. $11 550 ÷ 30 = $385 The value matches the weekly wage, so it is correct.

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b Write an equation for Naziha’s weekly wage ( y) if she works x hours per week.

Create a strategy Multiply the hourly rate by x.

Apply the idea y = 385x

Write the equation

Reﬂect and check Check by substituting known values, x = 30 and y = 11 550 to see if the result is true. 11 550 = 385 × 30 11 550 = 11 550 This equation is true, so our equation is correct. c Determine the number of hours that Naziha worked in a week if she earned $12 801.25

Create a strategy

Apply the idea

Use the equation from part (b).

y = 385x 12 801.25 = 385x

Write the equation Substitute the income

x = 33.25 hours Divide both sides by 385 to solve for x Naziha worked 33.25 hours, or 33 hours and 15 minutes.

Example 3 Waqas worked a total of 18 hours last fortnight at his normal hourly rate of pay. He was paid a wage of $390.96 a What is his normal hourly rate of pay?

Create a strategy Divide the income by the number of hours worked.

Apply the idea Divide income by hours worked

Evaluate

Reﬂect and check The rate of pay is approximately $20 per hour. Multiplying by the hours worked gives the income: 20 × 18 = $360, which is close to $390.96

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b Over the next fortnight, Waqas works for 14 hours at the normal rate and 4 hours at double time. Calculate his total pay for the fortnight.

Create a strategy Multiply each pay rate by the hours worked, then add these.

Apply the idea Normal income = 21.72 × 14 = $304.08 Double time pay rate = 2 × 21.72 = $43.44 per hour Double time income = 43.44 × 4 = $173.76 Income = 304.08 + 173.76 = $477.84

Multiply the pay rate by hours worked Evaluate Double the normal pay rate Evaluate Multiply by hours worked at double time Evaluate Add each part of the income Evaluate

Reﬂect and check Working for 4 hours at double time is equivalent to working for 8 hours at the normal rate. Multiplying the normal rate by the total equivalent hours: 21.72 × (14 + 8) = $477.84

Idea summary A wage means that income is based on the number of hours worked.

y = kx y is the income k is the rate of pay x is the time worked

Commission Another way to earn income is through a commission. Being paid on commission means that you earn money based on the total value of items that you sell. Usually, commission is a percentage of the total value of what you sell. For example, a real estate agent might earn 2% of the selling price of each house that they sell. Employees paid by commission may also be paid a retainer. This is a set amount of money paid over a given time period. A retainer ensures that an employee still earns some money, even if they don’t make any sales.

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Interactive exploration

Discover this concept in action online

mathspace.co

Income from commission can be modelled using a linear equation: Income = Commission × Value of items sold + Retainer

y = mx + c y is the income m is the commission rate (as a decimal) x is the total value of items sold c is the retainer

Example 4 Wentao is an art dealer. The gallery where he works pays him a retainer of $1100 per fortnight. He also earns a commission of 20% of the total value of art that he sells. a Determine Wentao’s income for a fortnight in which he sells art worth $8500.

Create a strategy Multiply the commission rate by the total value of art sold, and add the retainer.

Apply the idea Income = 20% × 8500 + 1100

Write the equation

= 0.2 × 8500 + 1100

Convert the percentage to a decimal

= $2800

Evaluate

b Write an equation for Wentao’s fortnightly income ( y) based on the total value of art sold (x).

Create a strategy Multiply the commission rate by x, and add the retainer.

Apply the idea y = 0.2x + 1100

Write the equation

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c Calculate the total value of art that Wentao must sell to make $4000 in a fortnight.

Create a strategy Use the equation from part (b).

Apply the idea y = 0.2x + 1100

Write the equation

4000 = 0.2x + 1100

Substitute the income

2900 = 0.2x

Subtract 1100 from both sides

x = $14 500

Divide both sides by 0.2 to solve for x

Reﬂect and check Substitute the answer back into the equation to check: y = 0.2 × 14 500 + 1100

Substitute x = 14 500

= $4000

Evaluate

Example 5 Jenina works at a car dealership. The monthly income that she earns ( y) is shown in the graph, where x is the total value of her car sales for the month. y 7000 6000 5000 4000 3000 2000

y = 0.05x + 2000

1000 x 20 000 40 000 60 000 80 000

a Find Jenina’s monthly retainer, and her commission as a percentage of the total value of car sales.

Create a strategy

Apply the idea

The coeﬃcient of x (the slope of the line) is the commission rate and the y-intercept is the retainer.

The coeﬃcient of x is 0.05, so the commission is 5% of the value of car sales.

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The y-intercept is 2000, so the retainer is $2000 per month.

b Find Jenina’s monthly income if she sells cars which have a total value of $80 000.

Create a strategy Use the graph to ﬁnd the y value when x = 80 000.

Apply the idea When x = 80 000, y = 6000.

So Jenina’s income is $6 000.

7000 6000 5000 4000 3000 2000

y = 0.05x + 2000

1000 x 20 000 40 000 60 000 80 000

Reflect and check Alternatively, we could have used the equation: y = 0.05x + 2000

Write the equation

= 0.05 × 80 000 + 2000

Substitute x = 80 000

= $6000

Evaluate

Idea summary Commission means that income is based on the total value of items sold.

y = mx + c y m x c

is the income is the commission rate (as a decimal) is the total value of items sold is the retainer

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6.02 Practice questions Understanding

1–3

Explorer

1–3

Adventurer

What is the difference between a wage and a salary?

True or false:

Ex 1

5 Ex 2

Trailblazer

Commission means that income is based on the number of items sold.

Double time means that your normal hourly rate of pay is doubled.

A retainer is a set amount of money earned, even if an employee doesn’t make any sales.

Income from commission can be modelled using a linear equation: y = mx + c. What is the correct description for the variables? A

y is the income, m is the retainer, x is the total value of items sold and c is the commission rate

y is the income, m is the commission rate, x is the retainer and c is the total value of items sold

y is the income, m is the commission rate, x is the total value of items sold and c is the retainer

Fluency 4

1–3

4–7, 9, 10, 12, 14, 15, 17, 18, 21

4, 5, 8–11, 13, 14, 16, 19, 20

Find the annual salary of a worker whose weekly pay is $332. Assume there are 52 weeks in a year. Christa earns $21 per hour working as a call centre operator. If she works 14 hours per week, how much is her weekly wage?

Roald’s annual salary is $100 482. Calculate his monthly income.

Bianca’s monthly salary is $5841.50. Calculate her fortnightly pay rate.

Yousaf is considering three different job offers and wants to choose the one that will pay him the most. • Offer 1: $4570 per month • Offer 2: $1020 per week • Offer 3: $53 450 per year

322

Find the annual income for Offer 1.

Find the annual income for Offer 2.

Which job offer should Yousaf accept?

James has a job that pays him $24 per hour and he works 38 hour weeks. Calculate to the nearest dollar his: a

Weekly income

Fortnightly income

Annual income

Monthly income

Katarina has a job that pays a salary of $76 000 per year. Calculate each to the nearest dollar: a

Monthly income

Fortnightly income

Weekly income

Hourly rate for a 38 hour week

Mathspace Victorian Curriculum – Year 9 mathspace.co

Pauline works at a tutoring company for 34 hours a week at a rate of $35 per hour. How much more will she earn per week if her wage is increased to $43 per hour?

Gurdip earns $52 per hour as a chef.

14 Ex 3

17 Ex 4

Write an equation for Gurdip’s weekly income, y, if he works x hours in a week.

How many hours does Gurdip need to work in a week to earn $2158?

Jmiaa works as a gardener and earns $36 per hour. a

Write an equation for Jmiaa’s monthly income, y, if she works x hours in a month.

Calculate Jmiaa’s annual income if she works 150 hours each month.

Ouafaa earned a total of $627 after working a total of 22 hours as a waitress. a

Calculate her hourly rate of pay.

Calculate Ouafaa’s weekly income if she works for 18 hours at the normal rate and 4 hours at double time.

Norik works as a trade assistant and earns $25 per hour. a

Calculate Norik’s annual income if he works for 38 hours each week at his normal rate of pay.

Calculate Norik’s annual income if each week he works for 32 hours at the normal rate, and 6 hours at time-and-a-half.

Neil normally works from 6:00 a.m. to 5:00 p.m. Monday to Friday, and from 10:00 a.m. to 3:00 p.m. on weekends. His normal weekly income is $1602. a

How many hours does he work per week?

Calculate his normal hourly rate of pay.

One week he works 60 hours at his normal rate, and 5 hours at double time. Calculate his income for the week.

Xiaoyi is a real estate agent. She earns a retainer of $502 per week and a commission of 1% of the total value of her property sales. Determine Xiaoyi’s income for a week in which she sells a property worth $540 000.

This year, Irene has made sales with a total value of $571 130. Her annual retainer is $39 500. If Irene earns a commission of 3.5% of the value of her sales, calculate her annual income.

Sahil earns a retainer of $1700 per month plus a commission of 2% of the total value of his sales.

Write an equation for Sahil’s monthly income, y, based on the total value of his sales, x.

Calculate Sahil’s monthly income if he makes sales worth $248 900.

Calculate the total value of sales that Sahil must make in a month to earn $5200 in income.

Cythia works as a ﬁnancial adviser and earns a commission of 0.5% of the value of her clients’ proﬁts. Her retainer is $800 per week. a

Write an equation for Cythia’s weekly income, y, based on the total value of her clients’ proﬁts, x.

Calculate Cythia’s weekly income if her clients were to make a total proﬁt of $85 000 each week.

What would her clients’ total weekly proﬁts need to be for Cythia to earn an annual income of $70 200? 6.02 Income mathspace.co

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21 Ex 5

Jarvis is a travel agent. His monthly income ( y) is shown in the graph, where x is the total value of bookings he takes.

3500

Find the value of the monthly retainer that Jarvis earns.

Find the commission as a percentage of the total value of bookings he takes.

2000

Find Jarvis’ monthly income if he takes bookings with a total value of $5000.

1000

What is the total value of bookings that Jarvis needs to take to earn a monthly income of $4000? 22, 23

4000

Reasoning 22

4500

3000 2500

y = 0.1x + 2500

1500

23–25

500

x 5000

10 000

15 000

23–25

Mbacke has two part-time jobs and is comparing the income from each. His job as a software developer pays $2850 per month, while his position as a manager at a computing store pays $680 per week. He multiplies his weekly income as a manager by 4 to get a monthly income of $2720 and concludes that he earns more as a software developer. Is Mbacke correct? Justify your answer.

Elna has two job offers, both of which pay a retainer and commission. • Offer A pays a weekly retainer of $500 and a commission of 10% of the total value of sales. • Offer B pays a weekly retainer of $1000 and a commission of 5% of the total value of sales. Elna thinks that she can make about $4000 per week in sales in either role. Explain which offer she should choose.

An employee works for a hours at their normal hourly rate of pay and b hours at double time. If the rate of pay is $20 per hour, give an expression for:

Weekly income in terms of a and b

Weekly income in terms of a, if

Weekly income in terms of b, if a = 3b

Compare these two jobs and discuss situations where one job may be more beneﬁcial than the other. • Job A: Wage of $28.50 per hour, for 30 hours per week, 40 weeks of the year. • Job B: A salary of $40 000 a year, which includes 4 weeks paid holidays.

Problem-solving 26, 27 26

324

26, 28, 29, 31

26–31

Nico earned $43 100 in one year and worked an average of 30 hours per week. a

What was his hourly rate of pay?

If his hourly rate of pay increased by $5, how much would his annual income increase by?

Calculate the percentage increase in his income, rounded to one decimal place.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Asmaa earns $24 per hour as a sales assistant. If she works on a Sunday, she is paid timeand-a-half. For Asmaa to earn $540 per week,

How many hours would she need to work at her normal rate?

How many hours would she need to work on a Sunday only?

How many hours would she need to work on a Sunday, if she also works for 12 hours at her normal rate?

Lavern is considering two job offers. • Offer A pays $35 per hour. • Offer B pays an annual salary of $65 520. For how many hours each week would Lavern need to work under Offer A to to earn the same or higher income than Offer B?

Arona earns a commission of 3.5% of the total value of her sales, plus a weekly retainer. In a week where she sold items with a total value of $25 200, her weekly income was $1122. Determine the value of Arona’s weekly retainer.

Luigi is paid a retainer of $2100 per month plus a commission on his mobile phone sales. Last month Luigi sold mobile phones with a total value of $13 680 and earned a monthly income of $2373.60. Determine his commission rate as a percentage of his total sales value.

Missy bakes cookies and sells them in their canteen. The graph shows the cost, revenue and the profit where x is the number of cookies sold. Each line has been labelled.

200

Match lines A, B and C to the profit, cost or revenue.

100

How many cookies must be sold to reach the break-even point?

-100

6.03

400

y (dollars)

300

A B 2

C x (cookies) 8 10 12 14 16 18 20

-200

Simple interest

After this lesson, you will be able to… • define what simple interest is and how it is calculated. • convert different time periods for calculating interest. • understand loan repayments and the formula used to calculate them. • solve problems involving calculating simple interest and determining loan repayments.

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Simple interest Simple interest is a way to calculate how much extra money you will earn (if you invest) or pay (if you borrow) over time. It is a percentage of the starting amount, called the principal. Simple interest = Principal amount × Interest rate × Time

I = PRT I is the interest (the extra money you earn or owe) P is the principal (the amount you start with) R is the rate of interest per time period, as a decimal T is the number of time periods (the duration of the loan/investment)

The units of time for R and T must be the same. For example, if an annual interest rate is used, then the time must be in years. Note that “p.a.” is an abbreviation of “per annum” and means per year. We can use the following conversions: 1 year = 4 quarters = 12 months = 26 fortnights = 52 weeks = 365 days

When you have borrowed money, you must pay back both the principal and the interest. When you have invested money, you own both the principal and the interest. In both cases, the total amount is: A=P+I

Example 1 $906 is invested at 5% p.a. simple interest. Dave wants to know the number of years it will take the investment to grow to $1132.50 a Calculate the interest that will be earned on the investment.

Create a strategy Use the formula for the total amount.

Apply the idea A=P+I 1132.50 = 906 + I

326

Write the formula Substitute the values of A and P

I = 1132.50 – 906

Subtract 906 from both sides to solve for I

= $226.50

Evaluate

Mathspace Victorian Curriculum – Year 9 mathspace.co

b Calculate the number of years it will take the investment to grow to $1132.50

Create a strategy Use the simple interest formula.

Apply the idea Write the formula

Substitute I, P, and R

Convert the percentage to a decimal

Evaluate the multiplication

Divide both sides by 45.3 to solve for T

Evaluate

Reﬂect and check Each year the investment of approximately $900 is earning 5% interest. 10% of $900 is $90, so 5% is $45. After 5 years the investment has earned 5 × 45 = $225 interest, bringing the total amount to about $1125, which is close to $1132.50

Example 2 Calculate the simple interest earned on an investment of $6050 at 0.7% per quarter for 3 years.

Create a strategy To use the simple interest formula the interest rate and time period both need to be in the same unit. In this case, quarters. Three years has 3 × 4 = 12 quarters. So now the interest rate of per quarter matches the time period.

Apply the idea I = PRT Write the simple interest formula = 6050 × 0.7% × 12

Substitute the values

= 6050 × 0.007 × 12 Convert the percentage to a decimal = $508.20

Evaluate

Reﬂect and check We could have left the time in years and converted the quarterly rate to an annual rate. To ﬁnd the annual rate, we multiply the quarterly rate of 0.7% by 4 to get 2.8%. Then I = 6050 × 2.8% × 3 which gives $508.20 again.

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Example 3 The simple interest on a loan of $6600 over 33 months is $1252.35 Find the annual interest rate as a percentage, rounded to one decimal place.

Create a strategy Use the simple interest formula.

Apply the idea Since we are looking for the annual interest rate, we should convert the time to years (by dividing by 12 months). Write the formula Substitute I, P, and T (in years) Evaluate the multiplication Divide both sides by 18 150 Evaluate the division Convert to a percentage

Reflect and check We could have left the time in months. Then the rate found would be a monthly value which we could convert to an annual rate by multiplying by 12: 1252.35 = 6600 × R × 33 R = 0.575% per month = 6.9% p.a.

Idea summary We can calculate the simple interest for a loan or investment using the formula:

I = PRT I P R T

is the interest is the principal (initial amount) is the rate of interest per time period (as a decimal) is the number of time periods

The time periods for R and T must be consistent. The total amount owed or in the account is: A=P+I

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Loan repayments Loans are usually repaid in a number of instalments (repayments), rather than all at once. We can use the following formula to calculate loan repayments:

Repayment amount =

A is the total amount (principal plus interest) n is the number of repayments

Example 4 Katrina takes out a loan to purchase a surround sound system. The total amount to be repaid is $95 000. a If Katrina makes 19 equal loan repayments, what is the value of each repayment?

Create a strategy Use the loan repayments formula.

Apply the idea Write the formula

Substitute the values

Evaluate

b If Katrina pays the loan back in 2 years with repayments of $11 875, how often did she make repayments?

Create a strategy Use the loan repayments formula.

Apply the idea Write the formula Substitute the values Multiply both sides by n Divide both sides by 11 875 Evaluate Divide by 2 years to ﬁnd repayments per year Katrina made quarterly repayments (1 repayment every 3 months).

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Idea summary

A n

Repayment amount = A n

is the total amount (principal plus interest) is the number of repayments

6.03 Practice questions Understanding 1

Explorer

1–3

Adventurer

1–3

Trailblazer

Calculate the simple interest on each scenario: a

An investment of $45 000 at 6% p.a. for 2 years

A 3-year loan of $3000 at 12%. p.a.

How can we convert: a

A time period from months to years?

A monthly interest rate to an annual rate?

True or false: a

A higher interest rate will result in more interest being paid on a loan.

A higher principal will result in less interest being paid on a loan.

A larger number of repayments over the same time period will result in a lower repayment amount.

Fluency 4

1–3

4, 6, 7, 9–11, 12ace, 13, 15, 16a, 17, 18

5–7, 9–11, 12bdf, 14, 15, 16b, 17, 18

5, 6, 8–11, 12bdf, 14, 15, 16c, 17, 19

Homer borrowed $7600 from a bank at a simple interest rate of 4.5% for one year. a

How much interest will Homer need to pay after 1 year?

Calculate the total amount Homer will have to repay the bank at the end of 1 year.

Find the total amount to be paid on a 4-year, $7000 loan at 11% p.a. ﬂat interest.

Toshko invests $12 500 at 4% p.a. simple interest and wants to know the number of years it will take for the investment to grow to $15 000.

Ex 1

7 8

330

Calculate the interest that will be earned on the investment.

Calculate the number of years it will take the investment to grow to $15 000.

For how many years does $7050 need to be invested for it to grow to $7950, if simple interest is paid at 7% p.a.? Round your answer correct to two decimal places. Ajaz invests $5680 at a simple interest rate of 5% p.a. and earns $426 in interest. a

Calculate the number of years taken to earn the interest.

Find the number of quarters taken to earn the interest.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Find the equivalent annual simple interest rate. a

Ex 2

0.03% per day

1.96% per quarter

The simple interest rate for an investment is 6% p.a. Find the percentage rate, rounded to two decimal places where necessary, used when calculating interest: a

0.63% per month

Quarterly

Monthly

Weekly

Calculate the simple interest earned on these investments, rounded to two decimal places. a

$5720 is invested at 10% p.a. for 63 weeks.

$9780 is invested at a rate of 5% p.a. for 17 months.

$9020 is invested at 1% per month for 2 years.

$5440 is invested at 0.3% per fortnight for 566 days.

Calculate the ﬁnal value of these investments over the given time period rounded to two decimal places: a

$4570 at 4% p.a. for 28 months

$9980 at 7% p.a. for 26 weeks

$1430 at 8% p.a. for 130 days

$6170 at 1.2% per month for 3 years

$9060 at 1% per quarter for 4 years

$4620 at 0.1% per fortnight for 3 years

Maria is given three investment account options by her bank: • Option 1: Interest is earned weekly at a simple interest rate of 0.14% per week. • Option 2: Interest is earned monthly at a simple interest rate of 0.63% per month. • Option 3: Interest is earned yearly at a simple interest rate of 7.3% p.a. Which option should Maria choose if she wants to earn as much interest as possible?

Sean is offered three different loan plans when borrowing money from the bank: • Option 1: Interest is earned daily at a simple interest rate of 0.02% per day. • Option 2: Interest is earned fortnightly at a simple interest rate of 0.34% per fortnight. • Option 3: Interest is earned quarterly at a simple interest rate of 1.98% per quarter. Which option should Sean go for if he wants to pay as little interest as possible?

For his investment into government bonds, Buzz was paid simple interest of 9% p.a. Calculate the size of Buzz’s initial investment if he earned $294.12 interest after 24 months.

Find the annual interest rate as a percentage. Round your answer to the nearest whole percent.

Ex 3

17 Ex 4

The simple interest on an investment of $5670 over 36 months is $1360.80

The simple interest on a loan of $25 100 over 39 weeks is $1694.25

The simple interest on a loan of $7500 over 50 months is $1250

Gitte takes out a loan to purchase a new fridge. She makes 16 equal loan repayments. The total loan amount paid is $1280. Find the value of each repayment.

Iain takes out a loan to purchase a jet ski. He makes 15 equal loan repayments of $5239. Calculate the total amount paid back on the loan.

Summer borrows money to pay for a vet bill and wants to pay it back in a year. If the total loan amount is $2470 and her repayments are $95, how often must Summer make repayments?

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Reasoning

20, 21

20–22

Yarel borrows $1200 at a simple interest rate of 8% p.a. To calculate the interest he will owe over 22 months, Yarel uses the calculation I = 1200 × 0.08 × 22. Explain why Yarel’s calculation is not correct, and how he can ﬁnd the correct value.

Ariela wants to invest her savings. She is considering two options: • Option 1: Simple interest rate of 0.5% per month for two years • Option 2: Annual rate of 2.5% simple interest, for 42 months Which option will result in Ariela earning the most interest? Justify your choice.

Explain why a shorter loan term (T value) will result in less interest paid, even though the monthly repayments may be higher.

Problem-solving 23–26 23

24, 26–30

Tara takes out a car loan of $6000 at a simple interest rate of 2% per quarter. She plans to repay the loan over 2 years through regular monthly repayments. a

Calculate the total interest that Tara will accrue over the duration of the repayment.

Calculate the value of each repayment of the loan.

The graph shows the amount of simple interest charged each year by a particular bank, on a 8-year loan. a

Find the total amount of simple interest charged on a loan of $5000.

Calculate the simple interest rate per annum charged by the bank on these 8-year loans. Give your answer as a percentage.

350

Interest ($)

300 250 200 150 100 50

Mohcin and Emily are looking to invest $11 600 and $18 300 respectively for 6 years. The simple interest rate applied is determined by the following table: Principal Between $4800 and $14 800 More than $14 800 to $24 800 Greater than $24 800

332

24–27

Principal ($) 1000 2000 3000 4000 5000

Quarterly rate 1% 1.25% 1.5%

How much interest will Mohcin earn on his investment?

How much interest will Emily earn on her investment?

How much will they earn if they invest their combined principal amounts?

How much more interest will they earn by combining their principal amounts?

Mathspace Victorian Curriculum – Year 9 mathspace.co

The graph shows the amount of money in a savings account earning simple interest. a

Find the principal amount.

Calculate the interest earned after 10 months.

Calculate the monthly rate of simple interest.

1400

Amount ($)

1200 1000 800 600 400 200

Time (months)

An amount is invested at 0.5% per quarter, simple interest, for 8 years. What loan term would earn the same amount of interest if the interest rate was 0.8% per quarter?

A borrower received a loan of $5500 which was repaid in monthly instalments of $269 over 36 months. Calculate the flat rate of interest per annum as a percentage, rounded to two decimal places.

At the start of a month, Lana had $7000. She invested some of it in a friend’s startup and the rest in a savings account with a simple interest rate of 3% per month. By the end of the month, her friend’s startup gave her a 10% return on her investment and her total worth was $7410. Use equations to calculate how much she invested in her friend’s startup.

Shawn is going to purchase a car. He has two options:

• Option 1: Using dealer finance which uses simple interest of 5.6% over 3 years for the full purchase amount of $5599. • Option 2: By paying an initial deposit of $1000, and then financing the rest at a higher rate of 7.2% for 3 years. Which option should Shawn take if he wants to pay the least amount in total for the car?

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6.04

e-investing with simple R interest

After this lesson, you will be able to… • use tables to calculate simple interest over different time periods. • compare the effects of charging interest on the total amount versus only on the principal amount. • apply these concepts to real-world scenarios involving loans and investments.

Using tables for simple interest We have used the formula I = PRT to calculate the total interest accrued at the end of the term of a loan or investment. An alternative approach is to use a table which shows how interest is calculated each time period and how repayments affect the balance. The opening balance is the amount owed, (or in the account), at the start of each time period. The closing balance is the amount left at the end of the time period after repayments. The table shows the monthly balances for a loan of $20 000 at a simple interest rate of 1% per month, with monthly repayments of $1500. Month 1 2 3 4

Opening balance 20 000 18 700 17 400 16 100

Interest 200 200 200 200

Repayment 1500 1500 1500 1500

Closing balance 18 700 17 400 16 100 14 800

The interest is calculated based on the opening balance, so even though the balance decreases each month, the interest stays the same at $200. Each month, the interest is added to the opening balance, and the repayment is subtracted to give the closing balance. The closing balance then becomes the opening balance for the next month. If we continue the table until month 16, we get: Month 13 14 15 16

Opening Balance 4400 3100 1800 500

Interest 200 200 200 200

Repayment 1500 1500 1500 700

Closing Balance 3100 1800 500 0

In month 16, the repayment is only $700, because the opening balance plus interest is less than the normal repayment amount. This brings the closing balance to zero. When using a table to calculate the interest on an investment, there will be no repayments and the ﬁnal balance will not be zero (as the balance is always increasing).

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Exploration Using a spreadsheet tool, see if you can design a spreadsheet of the table mentioned. As an extra challenge, can you design it so that you can explore different interest rates, different opening balances, or different loan terms?

Example 1 Tirsa invests $425 000 in an account that pays 1.5% simple interest per quarter. Construct a table showing the opening balance, interest and closing balance each quarter for the ﬁrst year.

Create a strategy Each quarter, add the interest to the opening balance to get the closing balance, then copy this into the next row as the opening balance of the next quarter.

Apply the idea One year is four quarters. Quarter 1 will have an opening balance of $425 000. Each quarter the interest will be: I = 1.5% × 425 000

Write an expression for quarterly interest

= 0.015 × 425 000

Convert the percentage to a decimal

= $6 375

Evaluate

Quarter 1 2 3 4

Opening Balance 425 000 431 375 437 750 444 125

Interest 6 375 6 375 6 375 6 375

Closing Balance 431 375 437 750 444 125 450 500

Reﬂect and check We can use the simple interest formula (with T = 4 quarters) to check our table: I = PRT = 425 000 × 1.5% × 4 = $25 500 A=P+I = 425 000 + 25 500 = $450 500

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Example 2 Inas takes out a short-term loan of $1000 at a simple interest rate of 2% per month and makes monthly repayments of $220. a Create a table showing the opening balance, interest, repayment and closing balance for each month. Continue the table until the closing balance is zero.

Create a strategy Each month, add the interest and subtract the repayment to get the closing balance, then copy this into the next row as the opening balance.

Apply the idea Month 1 will have an opening balance of $1000. Each month the interest will be: I = 2% × 1000

Write an expression for monthly interest

= 0.02 × 1000

Convert the percentage to a decimal

= $20

Evaluate

Month 1 2 3 4 5

Opening Balance 1000 800 600 400 200

Interest 20 20 20 20 20

Repayment 220 220 220 220 220

Closing Balance 800 600 400 200 0

b How long did it take Inas to pay off the loan and what was the total amount repaid?

Create a strategy Use the table from (a).

Apply the idea The last row of the table (with a closing balance of zero) was month 5, so it took Inas 5 months to pay off the loan. Inas made 5 monthly repayments of $220 so the total amount repaid was 5 × 220 = $1100.

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c What was the total interest paid?

Create a strategy Use the table from (a).

Apply the idea It took Inas 5 months to pay off the loan and the interest was $20 each month. The interest paid was 5 × 20 = $100.

Reﬂect and check From (b) we know that the total amount repaid was $1100. The principal was $1000 so the interest paid was 1100 − 1000 = $100. We could also use the simple interest formula to check our table: I = PRT = 1000 × 2% × 5 = $100

Example 3 Danita takes out a $30 000 car loan where simple interest is charged quarterly. The last quarters of the loan balances are shown in the table. Quarter 15 16 17 18

Opening Balance 5500 3750 2000

Interest 750 750 750

Repayment 2500 2500 2500

Closing Balance 3750 2000 250

a Complete the last row of the table for quarter 18.

Create a strategy The opening balance for quarter 18 will be the closing balance for quarter 17. The interest will be the same as previous rows. Add the interest to the opening balance to calculate the ﬁnal repayment.

Apply the idea The opening balance plus the interest is 250 + 750 = $1000. Since this is less than the normal repayment amount, the last repayment will be reduced to $1000, to bring the closing balance to zero. Quarter 15 16 17 18

Opening Balance 5500 3750 2000 250

Interest 750 750 750 750

Repayment 2500 2500 2500 1000

Closing Balance 3750 2000 250 0

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b How many years did it take Danita to pay off the loan?

Create a strategy Use the table from (b).

Apply the idea The last row of the table (with a closing balance of zero) was quarter 18, so it took Danita 18 quarters to pay off the loan. Divide the number of quarters by 4 to ﬁnd the time in years

c Calculate the quarterly simple interest rate charged on this loan.

Create a strategy Find the interest as a percentage of the principal

Apply the idea Write the expression for $750 as a percentage of $30 000 Evaluate

Idea summary We can use tables to ﬁnd the opening and closing balances for a loan or investment each time period. • The opening balance is the amount owed or in the account at the start of each time period. This is either the principal (for the ﬁrst row) or the closing balance of the previous row. • The closing balance is the amount owed or in the account at the end of the time period. The closing balance is calculated by adding the interest accrued and subtracting any repayments.

Interest on the total amount We are familiar with simple interest, where interest is calculated as a percentage of the principal. Under simple interest, the interest amount is always the same. A more common (but more complex) situation is where interest was calculated as a percentage of the total amount in the loan or investment each time period. This means that the interest can change each time period.

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Exploration Complete the table for a loan of $20 000, with monthly repayments of $1500. For each row, calculate the interest by ﬁnding 1% of the opening balance. Round values to two decimal places. Month 1 2 3 4 … 12 13 14 15

Opening Balance 20 000 18 700

Interest 200 187

Repayment 1500 1500

Closing Balance 18 700 17 387

… 4963.11

… 49.63

… 1500

… 3512.74

What happened to the interest when it was calculated as a percentage of the opening balance, instead of a percentage of the principal (simple interest)?

Was the total interest more or less than it would have been under simple interest?

If there were no repayments (if this was an investment rather than a loan), how would your answers to the above change?

Example 4 Saﬁa invests $15 000 in an account that pays interest of 0.5% of the total balance per month. a Construct a table showing the opening balance, interest and closing balance for the ﬁrst four months. Round values to two decimal places.

Create a strategy Month 1 will have an opening balance of $15 000. Each month the interest will be 0.5% of the opening balance. Add the interest to get the closing balance, then copy this into the next row as the opening balance.

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Apply the idea Month 1 Interest = 0.5% × 15 000

Multiply the opening balance by the interest rate

= 0.005 × 15 000

Convert the percentage to a decimal

= $75

Evaluate

Month 1 Closing Balance = 15 000 + 75 = $15 075 Month 2 Interest = 0.005 × 15 075 = $75.38 Month 2 Closing Balance = 15 075 + 75.38 = $15 150.38

Add the interest to the opening balance Evaluate Multiply the opening balance by the interest rate Evaluate Add the interest to the opening balance Evaluate

Continuing this process for the next two months gives: Month 1 2 3 4

Opening Balance 15 000 15 075 15 150.38 15 226.13

Interest 75 75.38 75.75 76.13

Closing Balance 15 075 15 150.38 15 226.13 15 302.26

b How much interest has Saﬁa earned on her investment in four months?

Create a strategy The closing balance is the total amount (principal plus interest from each month), so the total interest earned is the closing balance minus the principal.

Apply the idea I = 15 302.26 − 15 000 Write the subtraction = $302.26

Evaluate

c Compare your answer to (b) with the interest that Saﬁa would have earned with 0.5% simple interest per month.

Create a strategy Use the simple interest formula.

Apply the idea I = PRT Write the formula = 15 000 × 0.5% × 4

Substitute the values

= 15 000 × 0.005 × 4 Convert the percentage to a decimal = $300

Evaluate

When the interest was calculated as a percentage of the total amount, Saﬁa earned $302.26 interest. This is $2.26 more than the simple interest where interest was calculated as a percentage of the principal only. 340

Mathspace Victorian Curriculum – Year 9 mathspace.co

Example 5 Emeric borrowed $340 000 to purchase a house. The table shows the opening balance, interest, repayment and closing balance for each month at the end of the loan term. Month 189 190 191 192

Opening Balance 9288.53 6349.26 3392.36 417.71

Interest 55.73 38.10 20.35 2.51

Repayment 2995 2995 2995 420.22

Closing Balance 6349.26 3392.36 417.71 0

a Determine the monthly interest rate charged on this loan. Write your answer as a percentage rounded to one decimal place.

Create a strategy Find the interest as a percentage of the opening balance

Apply the idea Write the expression for interest as a percentage of the opening balance for month 189 Evaluate and round to one decimal place

Reﬂect and check We could have used values from a different row of the table. Below is the calculation using the interest and opening balance of month 190:

b How many years did it take Emeric to pay off the loan?

Create a strategy Use the last row of the table.

Apply the idea The last row of the table (with a closing balance of zero) was month 192, so Emeric took 192 months to pay off the loan. = 16 years

Divide the number of months by 12 to ﬁnd the time in years

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c What was the total value of Emeric’s repayments?

Create a strategy

Apply the idea

Add all of the repayments from the table.

Emeric made 191 monthly repayments of $2995 then a ﬁnal repayment of $420.22, so the total amount repaid was 191 × 2995 + 420.22 = $572 465.22.

d Calculate the interest Emeric paid.

Create a strategy Subtract the principal from the total repayments.

Apply the idea I = 572 465.22 − 340 000

Subtract the principal from the total repayments

= $232 465.22

Evaluate

Idea summary Simple interest is calculated as a percentage of the principal, so the interest amount is always the same. Interest can also be calculated as a percentage of the total amount (opening balance) each time period, which means that the interest amount changes.

6.04 Practice questions Understanding 1 2

342

1–2

Explorer

1–2

Adventurer

1–2

Trailblazer

How is the closing balance of a loan calculated each time period? True or false: a

The amount of simple interest each time period is always the same.

The repayment amount each time period is always the same.

The opening balance of an account is the closing balance of the previous time period.

Simple interest is when interest is calculated as a percentage of the total amount.

For an investment, calculating interest as a percentage of the total amount results in more interest accruing.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Fluency 3 Ex 1

5 Ex 2

7 Ex 3

3, 5, 7, 9–11

4, 6, 8, 10, 12

Lifeng invests $73 000 in an account that pays 0.1% simple interest per month. Construct a table showing the opening balance, interest and closing balance for the ﬁrst six months. Vidas has $44 000 in a bank account that earns 2% simple interest per quarter. a

Construct a table showing the opening balance, interest and closing balance each quarter for the ﬁrst year.

How long does it take for the investment to grow to $46 640?

How much interest is earned in total over the year?

Lexie takes out a loan of $2500 at a simple interest rate of 3% per month and makes monthly repayments of $700. a

Create a table showing the opening balance, interest, repayment and closing balance for each month. Continue the table until the closing balance is zero.

How many months does it take Lexie to pay back the loan?

What was the total interest paid?

Abilio borrows $27 800 at a simple interest rate of 1.5% per quarter and makes quarterly repayments of $3200. a

Create a table showing the opening balance, interest, repayment and closing balance for each quarter of the ﬁrst year.

What was Abilio’s total repayments for the ﬁrst year?

Yousra takes out a home loan for $315 000 where simple interest is charged monthly. The last months of the loan balances are shown in the table. Month 177 178 179 180

Opening Balance 7000 5250 3500 1750

Interest 1575 1575 1575 1575

Repayment 3325 3325 3325 3325

How many years did it take Yousra to pay off the loan?

Calculate the simple interest rate charged on this loan.

What was the total amount repaid?

Find the total amount of interest Yousra paid.

Closing Balance 5250 3500 1750 0

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Elijah borrows $8000 where simple interest is charged fortnightly. The table shows the loan balances for the duration of the loan. Fortnight 1 2 3 4 5

9 Ex 4

11 Ex 5

Repayment 1700 1700 1700 1700

Calculate the ﬁnal repayment.

How many weeks did it take Elijah to pay off the loan?

What simple interest rate was charged on the loan?

Calculate the total amount that Elijah repaid.

How much interest did Elijah pay over the loan term?

Closing Balance 6380 4760 3140 1520 0

Page invests $9400 in an account that pays interest of 0.5% of the total balance per month. a

Complete the table showing the opening balance, interest and closing balance for the ﬁrst four months. Round values to two decimal places. Opening Balance 9400 9447

Interest 47 47.24

Closing Balance 9447 9494.24

How much interest did Page earn on her investment in four months?

Compare your answer to (b) with the interest that Saﬁa would have earned with 0.5% simple interest per month.

The opening balance of an investment is $8720 and the interest accrued that month is $34.88 a

What is the closing balance for the month?

What is the monthly rate of interest, as a percentage of the opening balance?

Susana takes out a car loan where interest is charged on the total amount each month. Three of the last four months of Susana’s repayments are shown in the table. Month 69 70 71

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Interest 80 80 80 80 80

Month 1 2 3 4

Opening Balance 8000 6380 4760 3140 1520

Opening Balance 868.48 622.82 375.93

Interest 4.34 3.11 1.88

Repayment 250 250 250

Closing Balance 622.82 375.93 127.81

Calculate the monthly interest rate charged on this loan rounded to one decimal place.

Complete the next row of the table for month 72, rounding your answers to two decimal places.

How many years did it take for her to pay off the loan?

Calculate her total repayments.

If her loan was for $15 000, calculate the total interest paid on the loan.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Mr and Mrs Maximilian have a mortgage. Some of the ﬁnal months of their repayments are shown in the table. Month 146 147 148 149 150

Opening Balance 26 452.84 21 651.24 16 813.62 11 939.72 7029.27

Interest 198.40 162.38 126.10 89.55 52.72

Repayment 5000 5000 5000 5000 5000

Closing Balance 21 651.24 16 813.62 11 939.72 7029.27 2081.99

Calculate the monthly interest rate charged on this loan. Write your answer as a percentage rounded to two decimal places.

Complete the next row of the table for month 151, rounding your answers to two decimal places.

How long, in years and months, did it take for them to pay off the loan?

Calculate the total repayments.

If they paid $302 097.60 in interest, how much did they initially borrow?

Reasoning

13, 14

13–15

Explain why the ﬁnal repayment on a loan is sometimes less than the normal repayment amount.

Ibryam is considering two different options for an investment. • Option A pays simple interest at a rate of 9% p.a. • Option B pays interest on the total amount at a rate of 0.75% per month Which option would earn Ibryam the most interest? Justify your answer.

When paying off a loan, you can usually choose to make higher monthly repayments. Discuss the beneﬁts of making higher monthly repayments when being charged simple interest, and when interest is calculated as a percentage of the total amount.

Problem-solving 16, 17 16

16–18

Angela has $100 to invest. a

Calculate the interest she would earn each year for the ﬁrst four years at a simple interest rate of 10% p.a.

Find the interest she would earn each year for the ﬁrst four years if annual interest were calculated as 10% of the total amount in the account.

Graph your answers from (a) and (b) as points connected by lines on the same Cartesian plane where x is time in years and y is interest earned each year.

Comment on the differences between the graphs of the two types of interest.

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Teodor is taking out a loan of $3000 at a monthly interest rate of 0.5% of the total amount. a

Use tables to calculate the time it takes Teodor to repay the loan, and the total amount of interest paid, if he makes monthly repayments of: i

b 18

$600

$800

How much money does Teodor save by making monthly repayments of $800 instead of $600?

Ashly is considering two different options for a loan of $5600. She can afford to make monthly repayments of $1200. • Option A charges simple interest at a rate of 0.4% per month. • Option B charges interest on the total amount at a rate of 0.6% per month. a

Calculate the total interest paid with: i

346

Option A

Option B

On the same Cartesian plane, graph both Option A and Option B, using x as the time in months and y as the amount owing on the loan.

Which option should Ashly choose and why?

Mathspace Victorian Curriculum – Year 9 mathspace.co

6 Chapter review 1

Calculate the proﬁt (or loss) for each scenario: a

The revenue from selling handmade crafts is $7590 and the cost of materials and labor is $1500.

The cost of producing organic juices is $320 and the revenue from selling them is $1100.

The revenue from a tech workshop is $14 000 but the expenses for the venue, speakers, and materials amounted to $14 500.

The cost of setting up a pop-up store is $820 but the revenue from the sales was only $340.

Maria sells artisanal cheeses to local cafes for $50 per wheel. Each wheel of cheese costs her $30 to produce, and the cost to advertise and deliver the cheeses to cafes each week is $150. In a particular week, Maria sells 4 wheels. a

Determine the revenue for the week.

Determine the total cost for the week.

Does Maria make a proﬁt or a loss?

Calculate the proﬁt (or loss) that Maria makes that week.

Lila earned $38 500 in one year, and worked an average of 25 hours per week. a

What was her hourly rate of pay?

If her hourly rate of pay increased by $4, how much would her annual income increase by?

What is this increase as a percentage of her original income? Round your answer to two decimal places.

Mayca is a livestock sales agent and her weekly income ( y) is shown in the graph, where x is the total value of sales that she makes. a

Find Mayca’s weekly retainer (in dollars) and commission (as a percentage).

Calculate Mayca’s annual income if she makes sales worth $10 000 each week.

Mayca wants to make a fortnightly income of $3200. What is the total value of sales she must make each week?

1800

1600 1400 1200 1000 800 600

y = 0.05x + 600

400 200

x 5000 10 000 15 000 20 000

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Determine the simple interest earned on the following investments, rounded to two decimal places. a

$6200 is invested at 8% p.a. for 50 weeks.

$10 500 is invested at a rate of 6% p.a. for 15 months.

$8300 is invested at 1.5% per month for 3 years.

$5800 is invested at 0.4% per fortnight for 480 days.

Shin took out a loan of $8500 from a credit union at a simple interest rate of 5% p.a. for two years. a

How much interest will Shin need to pay after 2 years?

Calculate the total amount Shin will have to repay the credit union at the end of 2 years.

Juana invests $85 000 in a savings account which pays 0.15% simple interest per month. Construct a table showing the opening balance, interest, and closing balance for the ﬁrst six months.

Elise decides to invest her inheritance of $30 000 in a premium account which offers an interest of 2% of the total balance per month. a

Construct a table showing the opening balance, interest, and closing balance for the ﬁrst four months. Round values to two decimal places.

How much interest did Elise earn on her investment in the four months?

What monthly rate of simple interest would give the same interest as your answer to (b)?

Give your answer as a percentage rounded to two decimal places. 9

Carlos takes out a personal loan where interest is charged on the total amount each month. Three of the last four months of Carlos’s repayments are shown in the table. Month 57

Opening Balance 945.60

Interest 4.73

Repayment 280

Closing Balance 670.33

670.33

3.35

280

393.68

1.97

280

115.65

Calculate the monthly interest rate charged on this loan to one decimal place.

Complete the next row of the table for month 60, rounding your answers to two decimal places.

How many years did it take for him to pay off the loan?

Calculate his total repayments.

If his loan was for $14 000, calculate the total interest paid on the loan.

This graph shows the cost, the revenue and the profit from making and selling x units of a certain good. Each line has been labelled. a

What letter corresponds to the graph of: i

the profit?

iii

the revenue?

the cost?

How many units must be sold to reach the break-even point?

60 50 40 30 20 10 −10 −20 −30

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y (thousand $)

A B 20 C

x units 100

Marco is planning to start a coffee shop. He estimates his fixed costs to be $10 000 each month. The cost to produce each cup of coffee is expected to be $5. a

Write an equation for Marco’s monthly cost, C, if he sells x cups of coffee.

Determine the number of cups of coffee that will result in a monthly cost of $15 000.

How much revenue per cup is needed for Marco’s coffee shop to break even if the monthly cost is $15 000?

Aria is a freelance graphic designer and a part-time tutor. They are comparing their income from both roles. As a graphic designer, they earn $3150 per month, while tutoring pays her $750 per week. Aria multiplies their weekly income from tutoring by 4 to get a monthly income of $3000 and concludes that they earn more as a graphic designer. Is Aria correct? Justify your answer.

Leo has two consulting offers, both of which pay a base salary and a bonus for each project completed: • The Strategist package provides a monthly base salary of $1200 and a bonus of 15% for each project’s value. • The Executive package provides a monthly base salary of $1800 and a bonus of 8% for each project’s value. Leo believes he can complete projects worth $5000 each month in either role. Explain which offer he should choose.

Jaden borrows $1500 at a simple interest rate of 6% p.a. To calculate the interest he will owe over 18 months, Jaden uses the calculation I = 1500 × 0.06 × 18. Explain why Jaden’s calculation is not correct, and how he can ﬁnd the correct value.

Choi has $150 to invest.

Calculate the interest he would earn each year for the ﬁrst four years at a simple interest rate of 8% p.a.

Use a table to ﬁnd the interest he would earn each year for the ﬁrst four years, if annual interest were calculated as 8% of the total amount in the account.

Graph your answers from (a) and (b) as points connected by lines on the same Cartesian plane where x is time in years and y is interest earned each year.

Which of the two scenarios returned the larger investment and by how much?

Nutu is evaluating two different options for a loan of $6500. He plans to make monthly repayments of $1400. • Option 1 charges simple interest at a rate of 0.5% per month. • Option 2 charges interest on the total amount at a rate of 0.7% per month. a

Calculate the total interest paid with: i

Option 1

Option 2

On the same Cartesian plane, graph both Option A and Option B, using x as the time in months and y as the amount owing on the loan.

Which option should Nutu choose to minimise the amount of interest paid?

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Big ideas Understanding similar triangles and scale factors is key to making accurate measurements and representations in math and the real world.

Curriculum Outcomes VC2M9SP01 recognise the constancy of the sine, cosine and tangent ratios for a given angle in right-angled triangles using properties of similarity VC2M9SP02 apply the enlargement transformation to shapes and objects using dynamic geometry software as appropriate; identify and explain, using language of similarity, ratio and scale, aspects that remain the same and those that change VC2M9SP03 design, test and refine algorithms involving a sequence of steps and decisions based on geometric constructions and theorems; discuss and evaluate refinements VC2M9M03 solve spatial problems, applying angle properties, scale, similarity, ratio, Pythagoras’ theorem and trigonometry in right-angled triangles VC2M9M05 use mathematical modelling to solve practical problems involving direct proportion, rates, ratio and scale, including financial contexts; formulate the problems and interpret solutions in terms of the situation; evaluate the model and report methods and findings

7 Scale and similarity Chapter outline 7.01 7.02 7.03 7.04

Investigation: Algorithms in geometry Similar right-angled triangles Scale factors Investigation: Guessing distance Scale and area Scale and volume Investigation: Scale diagrams Chapter 7 review

352 355 364 375 377 382 389 392

Bonsai trees are miniaturised versions of full-sized trees, just like scale models in mathematics!

INVESTIGATION

Algorithms in geometry Investigate the process of developing and testing an algorithm to geometrically bisect an angle.

Objectives • To understand and practise the steps in constructing an angle bisector using a compass and straight edge • To design an algorithm that systematically details the construction of an angle bisector • To test and evaluate the developed algorithm through practical geometric constructions and peer reviews

Materials • Pencil

• Compass

• Straight edge

Develop an algorithm We are going to use an animation of a geometric construction for an angle bisector to write an algorithm. The algorithm, like a recipe, will be a step-by-step process for how to bisect any angle using a compass and straight edge.

Interactive investigation Discover this concept in action online

mathspace.co

Investigate

352

After observing the animation in the applet, list down, in your own words, the sequential steps you noticed. How do these steps help in bisecting an angle? Make sure to describe the role of each tool used in the process.

Based on your observations, draft a clear and concise algorithm for constructing an angle bisector using a compass and straight edge. Think of this algorithm as a recipe someone else could follow even if they haven’t seen the applet.

Using your drafted algorithm, try to recreate the angle bisector on a piece of paper. Did your construction match the applet’s result? If there were any differences, what do you think caused them? How could you reﬁne your algorithm to account for these differences?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Flow chart algorithm Here are the steps for constructing an angle bisector: • Identify the angle to be bisected. • With the compass point on the vertex of the angle, draw an arc intersecting both lines. • Label the intersections with points. • With the compass point on one of the labelled points and its setting the same, draw an arc halfway through the angle’s interior. • With the compass point on the other labelled point and its setting the same, draw another arc halfway through the angle’s interior, intersecting the previous arc. • Label the arc intersection point. • Connect the vertex and the new intersection point with a straight line. Your task is to convert the instructions into a ﬂow chart algorithm using rectangles, diamonds, parallelograms, and arrows:

The rounded rectangle is used for start and end points (e.g., start with an angle). Diamonds are used when a decision needs to be made (e.g., does the arc intersect both legs?). Parallelograms are used for input (e.g., if yes, then… or if no, then…). Arrows indicate the direction of the diagram. After creating your ﬂow chart, follow your algorithm to construct an angle bisector for ∠PNO: P

Swap ﬂow charts with a partner. Without referring to the original instructions, your partner will attempt to bisect ∠PNO using only your ﬂow chart as a guide.

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Investigate 1.

Which steps, if any, did your partner ﬁnd confusing or not straightforward when using your ﬂow chart?

How can you modify your ﬂow chart to make the construction process clearer or more eﬃcient?

Discuss with your partner: Were there differences in interpretation of the ﬂow chart steps? If so, what might have caused them?

Discussion

354

Research how to construct a copy of an angle using a compass and straight edge. Can you convert these steps into a ﬂow chart or pseudocode, effectively turning them into a structured algorithm?

Use the algorithm you’ve created to replicate an angle. Were you successful? Were there any challenges?

Find a classmate who used different steps to construct a copy of an angle. Compare your algorithms.

Mathspace Victorian Curriculum – Year 9 mathspace.co

7.01 Similar right-angled triangles After this lesson, you will be able to… • understand the conditions that make right-angled triangles similar. • identify similar right-angled triangles by comparing angles and side ratios. • identify corresponding sides and angles in similar right-angled triangles. • use ratios to confirm that two right-angled triangles are similar. • solve for missing side lengths or angles in similar right-angled triangles using ratios or Pythagoras’ theorem. • apply knowledge of similar right-angled triangles to real-world problems.

Similar right-angled triangles In similar triangles, all corresponding angles are equal and all corresponding sides are proportional, or in the same ratio. Here are two similar, right-angled triangles, △UST and △CAB. We can see that: • ∠U corresponds to ∠C • ∠S corresponds to ∠A • ∠T corresponds to ∠B • TU corresponds to BC • SU corresponds to AC • ST corresponds to AB

S 8 52°

38° 10

T A

52°

38° 25

Each pair of angles are equal: ∠U = ∠C = 90°

∠S = ∠A = 52°

∠T = ∠B = 38°

The ratios of the corresponding sides are equal:

The symbol for similarity is ‘∼’, so we would write △UST ∼ △CAB.

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Exploration These four triangles are similar:

45° 5

50 45°

Triangle A

Triangle B

Triangle C

Triangle D

How could we show that Triangle A and Triangle B are similar?

How could we show that Triangle A and Triangle D are similar?

How could we show that Triangle B and Triangle C are similar?

What information do we need to show that two right-angled triangles are similar?

For two right-angled triangles, they are similar if: 1. They have an acute angle in common, let’s call it θ - AA or AAA similarity. • This means that all angles are the same since they both have a right angle, have one acute angle in common, the other angle will be (180 − 90 − θ ) ° A S

θ U

UST

CAB

2. They have two or three pairs of corresponding sides that are in the same ratio - RHS (right angle- hypotenuse-side) or SAS (side-angle-side) similarity • If we know two pairs of sides are corresponding, then we can use Pythagoras’ theorem to show that the third pair is also in the same ratio.

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A S c ck

UST

CAB

If we can show two triangles are similar, then we can solve for missing sides and angles.

Example 1 For these two triangles:

4 2

60°

30°

a Are the triangles similar?

Create a strategy Determine whether all the corresponding angles are equal or whether all the corresponding sides are proportional.

Apply the idea

Reﬂect and check

From the diagram, we can see that the Since the hypotenuse of the larger triangle is 4 and corresponding angles in both triangles are equal. the smaller triangle is 2, the larger triangle will have By AAA similarity, these triangles are similar. all lengths double those of the smaller triangle.

b Find x.

Create a strategy Since we know they are similar, we can set up a proportional ratio and solve for the unknown.

Apply the idea Ratio of hypotenuses to sides opposite 30° angle

Simplify

Reﬂect and check The scale factor is 2, so we knew that this side length would be double the corresponding length on the smaller triangle. 7.01 Similar right-angled triangles mathspace.co

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Example 2 A

These two triangles are similar:

1 Q

a°

45° C

a Find the value of a.

P B

Create a strategy Since these two triangles are similar, this means that corresponding angles are equal. We may also use properties of right-angled triangles.

Apply the idea Since △ABC is a right-angled triangle, this means: ∠A + ∠B + 90° = 180° Sum of interior angles in a triangle ∠A + ∠B = 90°

Subtract 90° from both sides

∠A + 45° = 90°

Substitute ∠B = 45°

∠A = 45°

Subtract 45° from both sides

Since the triangles are similar, we know that corresponding angles are equal. a can correspond to ∠A or ∠B since they are equal. a = 45°

Reﬂect and check Another strategy is to use that △PQR is isosceles since it has two sides that have length 1. This means that ∠Q = ∠P. Since △PQR is a right-angled triangle, this means ∠Q + ∠P = 90° So ∠P = ∠Q = 45° a = 45° b Find the value of b.

Create a strategy Use the ratios of corresponding sides to set up an equation and solve for b.

Apply the idea The corresponding sides are: • AB and PQ • CA and RP • BC and QR

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Side BC corresponds to side QR. Corresponding sides are proportional Substitute known values Take the reciprocal of both sides

Reﬂect and check Alternatively, we could use that they are both isosceles triangles to say that b = 4.

Idea summary For two right-angled triangles, they are similar if: • They have an acute angle in common - AA or AAA similarity as we already know they have a right angle in common. • They have two or three pairs of corresponding sides that are in the same ratio - RHS (right angle-hypotenuse-side) or SAS (side-angle-side) similarity If we can show two triangles are similar, then we can solve for missing sides and angles.

7.01 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

The two triangles shown are similar: a

Identify the angle corresponding to: i

∠D

∠B

DC corresponds to which side in △RPQ?

CB corresponds to which side in △RPQ?

iii

60°

∠C

R 30° Q

In the diagram, △ECD ∼ △LMN.

What must be true about each pair of corresponding angles?

What must be true about each pair of corresponding sides?

N C

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For two right-angled triangles △ABC and △XYZ, is each statement true or false? a

Since they are both right-angled triangles, they are similar.

If they have one acute angle in common, they are similar.

If they have the same hypotenuse length, they are similar.

If △ABC ∼ △XYZ, then

Fluency 4

4, 5, 6abcf, 7

If △ABC ∼ △XYZ, then

4, 5, 6acef, 7, 8

4, 5, 6c–f, 7, 8

For each set, identify which two triangles are similar. a

20 11

C 9

D 7 10

10 5

Determine whether each pair of triangles is similar. a

30°

45°

3 45°

41.4° 6

48.6°

360

2 5

4 Ex 1a

D 5

Mathspace Victorian Curriculum – Year 9 mathspace.co

30°

60°

2 2 30°

60°

Ex 1b

Find the missing length y of each pair of similar right-angled triangles. a

7 cm

2 cm

17.5 mm

32° 42 mm 12 mm

y mm

y cm

5 cm

58°

3 cm

14 m 7m

3 2 cm

30°

45°

9 2 cm

ym y cm

60° 3m

D 3m ym A

A 9m

11.4 m C 2m

E ym

3.7 m

7 Ex 2

Find the value of the variables in each pair of similar, right-angled triangles: a

s°

8 55°

r°

b 1.6

m 3 n

3 17

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2 y° y cm

16 cm

5 cm x cm

8 cm

Find the value of the variables in each triangle. Express your answer in exact form. a

x cm

30° 30° 3 cm

y cm

45°

x cm

45° y cm

60°

60° 3 cm

Reasoning 9

9, 10

9–11

Is the information given enough to show

△ABD ∼ △DEF ? Justify your answer. a

∠B = ∠E

Explain how you know this pair of triangles is similar: 5

34° 9

22.45

27 56°

Explain why all the triangles are similar. J

H D x° A

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Problem-solving 12, 13 12

12–14

13–15

Jenny wants to find the height of her school’s flag pole. During recess, she measures the length of the flag pole’s shadow to be 335 cm. Her friend then measures her own shadow which turns out to be 95 cm. The triangles formed by casting these shadows are similar. If Jenny is 160 cm tall, what is the height of the flag pole? Round your answer to the nearest whole centimetre.

160 335

A ﬂagpole that stands 4.9 metres high casts a shadow of 4.5 metres. At the same time, the shadow of a nearby building falls at the same point S. The shadow cast by the building measures 13.5 metres. Find h, the height of the building.

4.9 m

4.5 m

13.5 m

The design for a slide at a water park is shown. Assume the two triangles created are similar. How tall is the slide? Round your answer to two decimal places. h

2.0 2m

1.2

Solve for the height of the roof. Round your answer to two decimal places.

.91

8.2 m

5.1 m

9.66 m

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7.02 Scale factors After this lesson, you will be able to… • understand and explain the concept of a scale factor. • calculate the scale factor between two similar figures. • use the scale factor to find unknown lengths. • solve problems using scale factors in practical situations.

Scale factor The scale factor is the ratio of corresponding sides in similar ﬁgures. An enlargement is an increase in size without changing the shape. This means a scale factor greater than 1.

A reduction is a decrease in size without changing the shape. This means a scale factor between 0 and 1.

Interactive exploration Discover this concept in action online

mathspace.co

Corresponding angles of similar figures are equal. Corresponding sides of similar figures are in the same ratio. If we know the scale factor between two similar figures, then any length in one figure can be used to calculate the corresponding length in the other figure.

Example 1 These two shapes are similar.

61°

6 16

68°

7 104°

127° 14

364

x°

Mathspace Victorian Curriculum – Year 9 mathspace.co

a What is the scale factor from the larger to the smaller quadrilateral?

Create a strategy Since this is a reduction, it should be between 0 and 1. We can divide the corresponding side lengths of the smaller shape by the larger shape.

Apply the idea We are told the shapes are similar, so only need to check one pair, but you can pick any pair. We’ll show all pairs from shortest to longest. Shortest sides =

= Second shortest sides =

= Second longest sides =

= Longest sides =

Smaller shape divided by larger shape Simplify Smaller shape divided by larger shape Simplify Smaller shape divided by larger shape Simplify Smaller shape divided by larger shape Simplify

The scale factor for the reduction is .

Reﬂect and check If any of these had been different, then the shapes would not be similar.

b What is the size of the angle marked x?

Create a strategy Identify the sides that correspond to the sides marked 6 and 7, and ﬁnd the measure of the angle between those sides.

Apply the idea The sides in the large shape that correspond to the sides of length 6 and 7 in the small shape are 12 and 14 in the large shape. The angle between them is 127°. x = 127

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Example 2 The triangles in this diagram are similar: A′

6 5 4 3 2 1

−6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6

B′ B x 1 2 3 4 5 6 C

C′

What is the scale factor from △ABC to △A′B′C′?

Create a strategy Determine the ratio between two corresponding points’ distances from the origin.

Apply the idea Points A and A′ correspond to each other. Point A is 3 units away from the origin and point A′ is 6 units away from the origin. Divide 6 by 3 Evaluate

Reﬂect and check This is an enlargement, so the scale factor should be greater than 1.

Example 3 z

Y 65°

30°

85° 6

30° 12

B Z

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a Is △ABC similar to △XYZ?

Create a strategy To check if two triangles are similar, we can use the angle-angle-angle or side-angle-side similarity tests. We can’t use RHS since this is not a right-angled triangle.

Apply the idea The interior angles of ABC are C = 30°, A = 85° and B = 180 − 30 − 85 = 65°. The interior angles of XYZ are Z = 30°, Y = 65° and X = 180 − 30 − 65 = 85°. Since all corresponding angles are equal, we have that △ABC ∼ △XYZ.

b What is the scale factor from ABC to XYZ?

Create a strategy Set up a ratio using corresponding sides with the side of XYZ in the numerator and the side of ABC in the denominator.

Apply the idea Ratio of known corresponding sides

Substitute known values

Simplify the scale factor

Reﬂect and check The scale factor should be greater than 1 since it is an enlargement.

c What is the value of z?

Create a strategy Set up a proportional equation using corresponding sides.

Apply the idea Ratio of corresponding sides are equal

Substitute using labels on triangles

Simplify the scale factor

Multiply both sides by 6

Evaluate

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Reflect and check Since z is on the larger triangle, we should expect that it is longer than the corresponding side on the smaller triangle. Another way to do this would be to first find the scale factor, and then multiply the length of the corresponding side by it. z = AB × scale factor = 6 × 1.5 =9

Example 4 Adonis has a poster on A4 paper that he wants to enlarge to be on A0 paper. 297 mm 210 mm

126 mm

1189 mm h 841 mm

a What is the enlargement factor? Round your answer to the nearest whole number.

Create a strategy Divide one side of the large poster by the corresponding side in the small poster.

Apply the idea

Reﬂect and check

Dividing the horizontal length of the large poster by We would have achieved the same answer if we the horizontal length of the small poster, we get: had used the vertical lengths instead:

The enlargement factor is 4.

which rounds to 4.

b What is the height of the tallest mountain in the A0 size poster?

Create a strategy Multiply the height of the tallest mountain in the small poster by the enlargement factor from part (a).

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Apply the idea 126 × 4 = 504 mm

Idea summary The scale factor between two similar ﬁgures is the ratio of lengths between the larger and smaller ﬁgure. An enlargement will have a scale factor greater than 1. A reduction will have a scale factor between 0 and 1.

7.02 Practice questions Understanding 1

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

For the two similar triangles: a

Which side in △LMN corresponds to the side DE?

Which side in △LMN corresponds to the side DC?

C D

These two quadrilaterals are similar: a

Which side in the larger shape corresponds to side AB?

Which angle in the larger shape corresponds to ∠ADC?

T B

D C

Is each statement true or false? a

All squares are similar.

All rectangles are similar.

All isosceles triangles are similar.

All right-angled isosceles triangles are similar.

The sum of the interior angles of a triangle is 180°.

The sum of the interior angles of a quadrilateral is 360°.

For two similar shapes, the longest side in one shape must correspond to the longest side in the other shape.

If a shape is dilated by a scale factor of , will the new shape be larger or smaller than the original?

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5, 6aceg, 7a–c, 8ac, 9–11

Fluency Ex 1

5, 6bdfg, 7a–c, 8ac, 10, 11

5c–f, 6bdfh, 7b–d, 8bd, 10, 11

Find the value of the unknown angles for each of the similar ﬁgures: a

I G

28°

10 b°

64°

x°

14 M

A 71°

d 112°

q°

R Q

p°

F k° j° 8 6

W X

80°

O 108°

31°

y° M

a°

63°

44°

s° r°

3 P

150° 6 90° 117° 75° M

x° 10

E y°

Find the value of the unknown sides for each of the similar ﬁgures: a

S 18

15 f

g 6 B

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c d 12

T X

25 20 Z

B k

C 5

C 16

j 33

13 V

e P

U 22

N S

E 10

R 24

S X

8 Z

g R

18 B

21 35

15 Q

10.5

C 12

Ex 2

y S

Find the scale factor for each of the similar shapes: a

10 8 6 4 2

D A −10−8 8− −6 6− −4 4− −2 2 − 2 −2 − 4 −4 −6 −8 −10

X E B

2 4 6 8 10 C

10 y 9 8 7 6 5 4 B 3 2 1

−4 −3 3 − −2 2 − −11− −11 A − −2 2 −3 3 W −4

C x 1

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9 8 7 6 5 E 4 3 2 1

y 10 B 8 6 C 4 Y 2

D C

−8 −6 −4 −2 −2 −4 −6 −8 −10

F −8 −6 −4 −2 −1 Ex 3

Z 4

Find the scale factor.

Find the value of the variable.

b x cm

a cm

4 cm 6 cm

24 cm 28 cm

12 cm

21 cm

P 24

16 B

8 C

8 R

m Q

Determine the scale factor between the two circles: 12

Ex 4

A 4 in × 6 in (10 cm × 15 cm) picture of a sun is enlarged. a

What is the enlargement factor?

If the diameter of the sun in the 4 × 6 photo is 6.35 cm, what is the diameter of the sun in the 8 in × 12 in photo?

2.5 in

d 6 in

4 in 8 in

372

For each of the similar triangles: i

Mathspace Victorian Curriculum – Year 9 mathspace.co

12 in

Alistair and Abigail are rearranging their room. They measured their room and are trying to make a scale model to see what might ﬁt in different places. 4m 1m 12 cm

closet

0.6 m

What is the scale factor from the room to the drawing?

Draw the scale model with all sides labelled.

Reasoning 12

13, 14

12–14

A square has side length 3 cm and a rectangle has dimensions 18 × 6 cm. Nila said that because all of the angles are equal, the AAA similarity tests means they are similar with a scale factor

= 3.

Is she correct? Justify your answer. 13

Brady took a picture of a lighthouse in the ratio 1 : 1, then posted it online. Who is right, Brady or Dorothy? Explain. 100 x 300 px

Brady Look how tall this lighthouse is Dorothy What did you do to the picture? It is not that tall in real life. Brady I just enlarge it by 3 times, I didn’t distort it or anything

A ﬁgure was enlarged by a scale factor of 3. What scale factor could be used to get it back its original size? Explain.

Problem-solving 15–17 15

16–18

17–19

An equilateral triangle of side length 6 cm is to be enlarged by a factor of 5. a

What will be the side length of the resulting triangle?

What will be the size of each angle in the resulting triangle?

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Mae wants to insert a picture into a document. She enlarges it by a factor of 12 but it becomes too blurry, so she reduces the resulting picture by a scale factor of 4. Find the overall scale factor from the original to the ﬁnal size.

A 1.1 m high fence casts a shadow of length 1.5 m. At the same time, a lamp post of height 3.9 metres casts a shadow of length L metres. Find the value of L, rounding to one decimal place.

A swimming pool of length 40 m and width 20 m is to be represented on a scale drawing with a scale of 1 : 2000. Find the scale dimensions of the pool in centimetres. a

Find the length in centimetres.

Find the width in centimetres.

Giuseppe took a picture of a tall waterfall, which is in the ratio 2 : 3 then He visits a photo centre that does prints from digital photos. This is their pricing guide:

$0.10

$0.35

4 × 6” (15 × 10 cm) Digital Photo Print

5 × 7” (12 × 17 cm) Digital Photo Print

8 × 10” (20 × 25 cm) Digital Photo Print

$3.50

$12

$13

8 × 12” (20 × 30 cm) Digital Photo Print

374

$2.50

12 × 16” (30.5 × 40.6 cm) Digital Photo Print

12 × 18” (30.5 × 45.7 cm) Digital Photo Print

For each option, would the picture be distorted or cropped, or would it display correctly? i

4 × 6″ (15 × 10 cm)

5 × 7″ (12 × 17.5 cm)

iii

8 × 10″ (20 × 25 cm)

8 × 12″ (20 × 30 cm)

12 × 16″ (30.5 × 40.6 cm)

12 × 18″ (30.5 × 45.7 cm)

He bought a frame which holds 8 × 10″ (20 × 25 cm) pictures. i

If he scales the picture down using the shorter side, what area of the picture will get cropped off?

If he doesn’t maintain the original ratio, and distorts the photo, will the waterfall look taller or wider than in reality?

Mathspace Victorian Curriculum – Year 9 mathspace.co

INVESTIGATION

Guessing distance Investigate the use of similar triangles to estimate the distance from yourself to a distant object.

Objectives •

To determine the reasons two triangles might be similar

•

To apply the principles of similar triangles to real-world situations

Guessing distance Did you know? You can use similar triangles to estimate the distance from yourself to a distant object. In this investigation, you’ll discover how to achieve this. Similar triangles have historically been employed by land surveyors to measure distances indirectly. A good example is the use of a theodolite, which measures angles. By taking angular readings and using a baseline of known length, the distances to faraway points can be determined through the properties of similar triangles. Shown is how to create triangles between yourself and a distant object: 1.

Choose an object (for example, a bag) and measure its height (label this as H).

Position the object at an undetermined distance (label this as D) - perhaps on the opposite side of the room.

Shut one eye and stretch out your arm with your thumb pointing up.

Align your thumb with the distant object.

Note the part of your thumb where the object starts and ends (e.g., the top of the bag might be at your thumb’s tip while the bottom might align with the thumb’s base. This might only span a few centimetres. Label this measurement as T).

Gauge the distance from your eye to your thumb (label this as L).

Refer to the provided diagram. We’ve illustrated two triangles, one from our eye to our thumb and the other from our eye to the distant object. H Eye

T L D

Investigation: Guessing distance mathspace.co

375

After following the example, ﬁnd an object to create your own triangles and answer these questions: 1.

Are the triangles that you created with your thumb and object similar? Can you explain why or why not?

If they are similar, which rule or proof regarding triangle similarity can be applied to these two triangles? Why?

Utilising this triangle similarity rule, can you estimate the distance from your eye to the object?

If possible, directly measure the distance. How accurate was your estimation?

3,474 km

If the diameter of the Moon is 3474 kilometres, using the methods of this investigation, estimate how far away it is from the Earth. How close is your answer to the accepted answer?

Investigate 1.

Team up with a classmate and measure each other’s height. Then, walk an undetermined distance apart. Employ the method of similar triangles to estimate the distance between you two.

Record the actual distance. Which of you came closer to the accurate measurement?

Discussion

376

If you were standing at a certain distance from the Sydney Opera House, how could you use the methods learned in this investigation to estimate the height of one of its sails? What data would you need to know beforehand?

Imagine you’re at a beach and you see a tall cliff in the distance. If you had a friend stand at the base of the cliff and you walked a distance away, could you estimate the height of that cliff using the methods you’ve learned? What challenges might the landscape present?

Mathspace Victorian Curriculum – Year 9 mathspace.co

7.03 Scale and area After this lesson, you will be able to… • explain how the scale factor impacts the area of similar figures. • apply understanding of scale factors to calculate the area of similar figures. • interpret and solve problems involving areas of similar figures and scale factors.

Scale and area Since area is a two-dimensional measurement, the calculation of area always involves two lengths multiplied together. For example: Square Square Rectangle

Rectangle

Triangle

h bh b h hb

Triangle

Parallelogram

Trapezium

Circle Circle

A = s2

s s

b a a h h b b r

A=l×w

A= 1 ×b×h 2 A=b×h

A = 1 × h × (a + b) 2 2

AA == πrπ2 r

When we multiply two lengths with the same units, the unit for the area is the length unit squared. For example, cm × cm = cm2. This means that if the side lengths of a shape are multiplied by a scale factor, the area of that shape will be multiplied by that scale factor twice (once for each dimension in the formula).

Interactive exploration Discover this concept in action online

mathspace.co

7.03 Scale and area mathspace.co

377

The ratio of areas in similar figures is equal to the square of the scale factor. For a scale factor of k, if the smaller figure has an area of A, then the area of the larger figure would be k2 × A.

Example 1 The two rectangles shown in the diagram are similar.

x cm2 15 cm2 12 cm

Find the value of x.

6 cm

Create a strategy Multiply the area of the smaller rectangle by the scale factor squared.

Apply the idea Find the ratio of the side lengths Evaluate

Multiply the area by the scale factor squared

Evaluate

Example 2 A triangle has side lengths of 6 cm, 12 cm and 16 cm. A second similar triangle has an area that is 9 times larger than the ﬁrst. a Find the scale factor between the two triangles.

Create a strategy Take the square root of the area scale factor.

Apply the idea Scale factor = =3

378

Take the square root Evaluate

Mathspace Victorian Curriculum – Year 9 mathspace.co

b What is the relationship between the perimeters of the two triangles? A

The larger triangle’s perimeter is the same as the smaller triangle’s perimeter.

The larger triangle’s perimeter is 9 times the smaller triangle’s perimeter.

The larger triangle’s perimeter is 3 times the smaller triangle’s perimeter.

Create a strategy Use the scale factor from part (a).

Apply the idea Since the scale factor between the two triangles is 3, then each side length will be 3 times longer in the larger triangle. This means that the larger triangle’s perimeter is 3 times the smaller triangle’s perimeter, so the correct answer is option C.

Idea summary The ratio of areas between two similar ﬁgures is equal to the square of the scale factor.

7.03 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

True or false? a

Area calculations involve the multiplication of two lengths.

If the side lengths of a shape are multiplied by a scale factor, the area of that shape will be multiplied by that scale factor three times.

The ratio of areas in similar ﬁgures is equal to the square of the scale factor.

A square has sides of 5 cm. Give the side length of the new square if: a

The square was enlarged by a factor of 12.

A scale factor of 0.2 was applied to the square.

A scale model of a building has a height of 32 cm. The actual building is 8 m tall. Determine the scale factor used when creating the model based on the actual building.

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Fluency Ex 1

4–6, 8, 9

4, 6, 8–10

4, 6–8, 10

Find the value of x for each pair of similar rectangles. a

320 cm2

x cm2

x cm2 5 cm

20 cm

7 cm 9 cm

27 cm

c 10 cm

x cm2 6 cm

12 cm

The corresponding sides of two similar triangles are 6 cm and 12 cm. The area of the smaller triangle is 18 cm2. Find the area of the larger triangle.

A square has a side length of 8 cm. Another square is created by scaling it up by a factor of 7. Find the area of the larger square.

Two rectangles are similar, with matching sides in the ratio 6 : 7. The smaller rectangle has an area of 576 cm2 and its shortest side has length 8 cm. Find the length of the longest side of the bigger rectangle.

A triangle has side lengths of 3 cm, 12 cm and 14 cm. A second similar triangle has an area that is 25 times larger than the ﬁrst.

Ex 2

Find the scale factor between the two triangles.

Describe the relationship between the perimeters of the two triangles.

A shape with an area of 36 cm2 is enlarged using the scale factor 3. Find the area of the enlarged shape.

If a square with an area of 25 m2 is dilated by a factor of 0.4, what is the side length of the dilated square?

Reasoning

12, 13

11–13

A rectangle has been enlarged with a scale factor of k. Describe the relationship between the area of the two rectangles.

A circle has an area of approximately 78.5 cm2. Explain why a circle with a radius twice as large has an area of approximately 314 cm2.

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Consider two similar ceilings: the ﬁrst with dimensions 5 m by 4 m, and the second with dimensions 20 m by 16 m. a

Find the length scale factor.

Find the area scale factor.

The smaller ceiling took 1.5 L of paint to cover it. How many litres of paint would be required to paint the second ceiling?

Justify why the amount of paint needed for the second ceiling is not simply four times the amount needed for the ﬁrst ceiling.

Problem-solving 15, 16, 18–20 14

14–16, 18–20, 22

All dimensions of the miniature stadium are

14, 16, 17, 19–22

of the actual stadium sizes. If the pitch area

of the actual stadium is 7200 m2, ﬁnd the area of the pitch in the miniature version. 15

The figure shows the field on which a game of tag is being played during a physical education class. To make it more of a fitness challenge, the teacher dilates the boundaries of the field by a factor of 1.5 What is the area of the new field?

12.2 m

A rectangular sign has a side length of 1.5 m. In a photograph, the corresponding side has a length of 12.5 cm. a

Find the scale factor from the photograph to the sign.

The area of the sign in the photograph is 75 cm2. Determine its area of the sign in real life.

A rectangular prism has dimensions of 2 cm, 4 cm and 5 cm. a

Calculate the surface area of the prism.

If the prism is scaled up by a factor of 3, give the dimensions of the enlarged prism.

Calculate the surface area of the enlarged prism.

By what factor has the surface area increased by?

If a three-dimensional shape is enlarged with a scale factor of k, by what factor will the surface area increase by?

Maria runs a business where she makes and sells soft toys. Her most popular item is a teddy bear that is 14 cm long. For a special promotion, she decides to make a larger version of the bear that is 56 cm long. a

Find the scale factor between the two bears.

Maria uses 224 cm2 of felt to make the small bear. How much felt will she need to make the big bear?

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A model of the Eiffel tower is made in the ratio 1 : 12 960

The height of the Eiffel tower is 324 m. Find the height of the model in centimetres.

The model requires 15 mL of paint to cover its entire surface area. How much paint, in litres, would be required to cover the entire surface area of the real Eiffel tower?

The Dominican Republic is shown on the map. According to the scale given, estimate the actual area of the Dominican Republic. Dominican Republic Santo Domingo 0 40 80 km

A square has sides of length x cm. Determine the area of the new square if the original is dilated by a factor of:

a 22

The ﬁgures shown are similar. What is the value of k?

k cm 6 cm Area = 300 cm2 Area = 75 cm2

7.04 Scale and volume After this lesson, you will be able to… • explain how the scale factor impacts the volume of similar objects. • identify and determine the scale factor based on the volumes of similar objects. • apply knowledge of scale factors and volume to find unknown sides in similar shapes. • solve problems involving volume of similar objects and scale factors.

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Scale and volume Since volume is a three-dimensional measurement, the calculation of volume involves three lengths multiplied together. For example: Vrectangular prism = length × width × height

V=l×w×h

Height Width

Length

Vcube = length3 V = s3

s s

Vtriangular prism = Area of triangle × prism height Height

This means that if the dimensions of an object are multiplied by a scale factor, the volume of that object will be multiplied by that scale factor three times (once for each dimension in the formula).

Exploration ×2

×2

Each side length of the cube in the image has been scaled by a factor of 2. 1.

If the original volume was 1 unit3, what is the volume of the new cube?

Did the volume also double? If not, how did it change?

If the original volume was 7 units3, what would the volume of the new cube be?

×2

This means that the ratio of volumes in similar objects is equal to the cube of the scale factor. For a scale factor of k, if the smaller ﬁgure has a volume of V, then the volume of the larger ﬁgure would be k3 × V.

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Example 1 The two rectangular prisms shown in the diagram are similar.

V = 11 cm3 2 cm V = x cm3 8 cm

Find the value of x.

Create a strategy Multiply the volume of the smaller rectangular prism by the scale factor cubed.

Apply the idea

Find the ratio of the side lengths

Evaluate

Multiply the volume by the scale factor cubed

704

Evaluate

Example 2 Spheres are always similar since they have only one deﬁning measurement (the radius). Two spheres have volumes of 40 cm3 and 625 cm3. Find the scale factor between the two spheres.

Create a strategy Find the volume scale factor, then take the cube root.

Apply the idea Find the ratio of the volumes

Evaluate

Take the cube root of the volume scale factor

Evaluate

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Idea summary The ratio of volumes in similar objects is equal to the cube of the scale factor.

7.04 Practice questions Understanding 1

Ex 1

Adventurer 1–3

Trailblazer 1–3

Is each statement true or false? a

Volume calculations involve the multiplication of four lengths.

If the side lengths of a shape are multiplied by a scale factor, the area of that shape will be multiplied by that scale factor three times.

The ratio of volumes in similar objects is equal to the cube root of the scale factor.

A cube has a side length of 2 cm. a

What is the volume of the cube?

What is the volume of the cube if we double the original side length to 4 cm?

What is the volume of the cube if we double the side length again to 8 cm?

What is the effect of doubling the side length on the volume?

Joe wanted to build a 5 m × 10 m pool in his backyard. He then decided he would increase the size of the pool by 50% but keep the depth the same. How much more water would he need to fill the pool with the new dimensions?

Fluency 4

Explorer 1–3

4–6, 8–10

4–6, 8, 9, 11

4, 5, 7, 8, 10, 11

Find the value of x for each pair of similar objects. a

V = 20 cm3

3 cm

V = x mm3 V = 36 mm3

3.5 m 1.4 m

V = x cm

12 cm

V = 62.5 m3

V = x m3

4 cm

4 mm

Two similar cones have bases with radius 7 cm and 28 cm respectively. a

Find the scale factor of the height of the smaller cone to the height of the larger cone.

Find the scale factor of the volume of the smaller cone to the volume of the larger cone.

Calculate the volume of the larger cone, if the volume of the smaller cone is 852 cm3.

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For these two similar trapezoidal prisms: a

Find the length scale factor from the smaller prism to the larger prism.

Find the volume scale factor from the smaller prism to the larger prism.

20 cm

5 cm

The corresponding sides of two similar prisms are 5 cm and 15 cm. The volume of the smaller prism is 40 cm3. Find the volume of the larger prism.

Two similar prisms have volumes of 24 cm3 and 3 000 cm3.

Ex 2

Find the scale factor between the two prisms. The two solids shown are similar. Find the value of x. x cm V = 193.7 cm3

41 cm

V = 24 212.5 cm3

The volume of two similar objects are 125 cm3 and 1331 cm3. Find the ratio of the larger side to the smaller side.

The volume of a soup can is reduced from 125 cm3 to 91.125 cm3. What is the scale factor?

SOUP

Reasoning 12

12, 13

13, 14

The length l, width w, and thickness h, are the dimensions of a cement slab. If these dimensions are tripled, describe the effect on: a

The surface area of the cement slab

The volume of the cement slab

Basili’s old esky has a capacity of 20 L. Basili wants a new one with a capacity of 160 L, so he plans to buy one with dimensions 8 times the old one’s dimensions. Explain why Basili doesn’t need to buy such a large esky.

A game company makes six-sided dice in two sizes: a standard size with side lengths of s cm and a jumbo size with side lengths of j cm.

386

A six-sided die is a cube. Explain why a cube is always similar to another cube.

Give an expression for the scale factor for the two dice in terms of s and j.

Give an expression for the volume scale factor for the two dice in terms of s and j.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Problem-solving 15–19 15

16–20

18–22

For these two rectangular prisms: a

Find the length scale factor.

Find the surface area scale factor.

Find the volume scale factor.

Find the volume scale factor if the measurements of the smaller prism are doubled.

2.3 cm 5.1 cm

3.22 cm 2.66 cm 7.14 cm

The surface areas of two similar triangular prisms are in the ratio 64 : 49. Give, as improper fractions: a

The scale factor.

The volume scale factor.

Consider the two spheres shown, where the smaller has radius 3 cm and the larger has a radius of 12 cm. a

Write the ratio (in the form m : n) of the radius of Sphere A to Sphere B.

Write the ratio of the volume of Sphere A to Sphere B.

If the matching dimensions of two similar ﬁgures are in the ratio m : n, what ratio are their volumes in?

1.9 cm

Sphere B Sphere A 12 cm 3 cm

Two similar cylinders have volumes of 5760 cm3and 90 cm3: a

What is the volume scale factor for the two cylinders?

How many times taller is the larger cylinder?

Find the value of h.

h cm 20 cm V = 5760 cm3 V = 90 cm3

The radii of two spherical balloons are 12 cm and 6 cm respectively. a

What is the scale factor?

How many times greater is the surface area of the larger balloon?

How many times greater is the volume of the larger balloon?

If half the air is released from the smaller balloon, how many times more air does the larger now hold?

Valerie was making a trial birthday cake for her son. The trial cake dimensions are 10 cm for the length, 7 cm for the width and 3 cm for the height. The actual cake has dimensions 20 cm for the length, 14 cm for the width and 6 cm for the height. a

Find the ratio of the dimensions of the trial cake to those of the actual cake.

Deduce the ratio of the volume of the trial cake to that of the actual cake.

To make the actual cake, by what number must Valerie multiply each quantity of ingredients?

7.04 Scale and volume mathspace.co

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388

A beehive consists of regular hexagonal cells with side length 3.2 mm and depth of 3.8 mm. A plastic container is built which is modelled on these hexagonal cells, with a side length of 6.4 cm. a

Find the scale factor between the cells of the beehive and the container.

What is the depth of the container?

A beehive with an estimated 70 000 cells holds approximately 45 000 cm3 of honey. What is the volume of a single cell? Round your answer to four decimal places.

What is the volume of the container? Round your answer to the nearest cm3.

A geologist found two pieces of pyrite, or fool’s gold. She found that the volume of Specimen B is 3.375 times greater than the volume of Specimen A. a

What is the scale factor?

How many times greater is the area the base of Specimen B than Specimen A?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Specimen A

Specimen B

INVESTIGATION

Scale diagrams Investigate images of proportional relationships to estimate actual measurements.

Objectives To understand scale models To calculate real-life measurements based on a proportional relationship To create a ﬂoor plan

Materials •

Grid paper

•

Sticky notes

•

Ruler

•

Pen or pencil

Scale diagrams Scale drawings An architectural ﬂoor plan is an example of a scale drawing. A scale drawing is a diagram of an object in which the dimensions are in proportion to the actual dimensions of the object. 3.00 m

4.00 m 2.60 m

3.20 m

The scale tells us how the drawing’s dimensions and the real-life dimensions are related. For example, the scale 1 cm : 2 m tells us that 1 centimetre in the ﬂoor plan drawing represents 2 metres in the actual building. If there are no units mentioned, then we have to assume that the units are the same, so we represent this as 1 : 200, where 1 cm : 200 cm.

Problem 1: A swimming pool of length 8 m and width 4 m is to be depicted in a drawing with a scale of 1 : 2000. Find the scale dimensions of the pool in millimetres.

Problem 2: Shown is a 1 : 200 floor plan of a house. The homeowner wishes to add a 140 cm long dining room table at the spot marked with an ‘x’ on the floor plan. What length should the table be drawn to in the floor plan?

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Problem 3: Use the floor plan to answer the questions. • According to the scale of the diagram, 1 cm on the diagram represents how many metres in the house? • Using a ruler, Neville measures the length of Bed 1 on the plan and finds it to be 4 cm. How many metres does this represent? • Using a ruler, Neville measures the width of Bed 1 on the plan and finds it to be 2 cm. How many metres does this represent? • Neville wants to tile the floor of Bed 1. If each tile is 20 cm by 10 cm, how many tiles would Neville need? Scale 1 : 100

Kitchen

Laundry

Patio

Bedroom 2 Dining Bath

Living Bedroom 1

Investigate 1.

Why is it important to understand and use the correct scale when creating ﬂoor plans or maps?

How does the representation of real-life objects on a scale drawing (like the swimming pool or house) aid in practical applications such as construction or design?

If you had a room that was shaped irregularly (not a perfect rectangle or square), how might you approach creating a scale diagram for it?

How do the dimensions of various rooms or objects in the ﬂoor plan inﬂuence your perception of the actual space or object size?

Design a floor plan In this activity you will be designing your dream room. This could be a bedroom, a restaurant, anything that you’d like. You will need to think about the size and shape of the room, as well as the furniture that will be inside it.

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Steps for creating a ﬂoor plan: • Determine the shape of your room and draw the outline, or where the walls will be, on your piece of grid paper. Ensure that all walls are straight and that you have enough space inside the outline for drawing furniture. • Think of 5 pieces of furniture that you want in your room and add them to the ﬂoor plan. Consider the size of the furniture so that when scaled up, they are appropriately sized. You may want to research the dimensions of the items you would like to add to your room. • You now need to create the scale to show how big the room and furniture will be. Measure your room and come up with an appropriate scale.

Investigate Answer these questions after completing the procedure. 1.

When drawing your ﬂoor plan, were there any pieces of furniture that you had to resize or adjust to ﬁt within the scale of the room? Explain.

Why is it essential to consider the size of furniture when creating a room’s ﬂoor plan, and how does that inﬂuence the room’s functionality?

How would the inclusion of windows and doors in your ﬂoor plan affect your decisions on furniture placement?

If you wanted to replicate your ﬂoor plan in a real-world setting, what challenges might you encounter that you didn’t on paper?

Think about different rooms in a house (kitchen, bathroom, living room). How might the considerations and challenges differ when creating scale diagrams for each?

Discussion 1.

How might advancements in technology, like augmented reality or 3D modelling, change or inﬂuence the way we approach creating and visualising ﬂoor plans in the future?

Imagine a situation where a ﬂoor plan’s scale was miscalculated. What potential realworld implications could this have, and how might it impact a project?

Explore the history of architectural drawings and ﬂoor plans. How have these evolved over time, and what tools or techniques were used in the past compared to now?

Consider other areas, besides architecture, where understanding and applying scale is crucial. What are some of these ﬁelds, and how is scale utilised in them?

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7 Chapter review 1

Determine whether each pair of triangles is similar. a

60°

30° 60°

30°

7.5

4 3

Find the missing length y of each pair of similar, right-angled triangles: a

13 cm

6 mm

y mm

12 cm

21 mm

18 mm y cm

18 cm

Find the value of the variables in each pair of similar, right-angled triangles: a

s° 45 7 62°

r°

a°

10.5 cm

b cm

3 cm 65°

7 cm

A person that is 1.6 m tall casts a shadow of 1.8 metres. At the same time, the shadow of a nearby tree falls at the same point S. The shadow cast by the building measures 5.4 metres. a

Identify two similar triangles. Explain how you know they are similar.

Find the height of the tree.

1.6 m S 1.8 m 5.4 m

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The design for a ramp at a skateboard park is shown. The two triangles created are similar. How tall is the ramp?

pm 2.25 m 1.5 m 18 m

For each of the similar triangles: i

Find the scale factor.

Find the value of z.

b 30 cm

z cm

35 mm 21 mm

30 cm

25 cm

10 mm

z mm

Find the value of the variables in each of the similar figures: a

g°

a° 117°

83°

b° f° 29°

Find the value of the variables in each of the similar figures: a

24 9

32 8

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A rectangle has the dimensions 4 × 7. Jay dilates the rectangle vertically by a scale factor of 2 to get a rectangle with dimensions 8 × 7. Jay said that because he dilated the first rectangle and all angles are equal, the rectangles are similar with scale factor

7 4

4 7

Is Jay correct? Justify your answer. 10

Lyor wants to print a photograph for a frame. He magnifies it by a factor of 8 but it doesn’t fit the frame, so he shrinks the enlarged photo by a scale factor of 2. Determine the overall scale factor from the original to the final size.

A football ﬁeld of length 105 m and width 68 m is to be depicted on a scale diagram with a scale of 1 : 500.

Find the length on the scale diagram in centimetres.

Find the width on the scale diagram in centimetres.

Find the value of y for each pair of similar rectangles. a

450 m2 y cm2

32.5 cm

30 m y m2 10 m 13 cm 10 cm

The corresponding sides of two similar triangles are 8 cm and 16 cm. The area of the smaller triangle is 32 cm2. Determine the area of the larger triangle.

If a square has an area of 49 m2 and it is dilated by a factor of , how long is the side of the resulting square?

A triangle has been enlarged with a scale factor of p. How does the area of the dilated triangle relate to the original triangle’s area?

All dimensions of a toy house are 2

of the actual house sizes. If the living room area of the

actual house is 48 m , what is the area of the living room in the toy version?

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Find the value of x for the pair of similar figures. a

6.5 cm V = 14.5 cm3 4.2 mm 2.8 mm 13 cm

V = 243 mm3

V = x mm3

V = x cm3

Two similar cones have a radius length of 5 cm and 20 cm respectively. a

State the scale factor of the height of the smaller cone to the height of the larger cone.

State the scale factor of the volume of the smaller cone to the volume of the larger cone.

Calculate the volume of the larger cone, if the volume of the smaller cone is 250 cm3.

Two similar prisms have volumes of 27 cm3 and 729 cm3. If two, corresponding sides in the prisms have lengths of x cm and 9 cm, where x represents the length in the smaller prism, find the value of x.

An office fish tank has a capacity of 400 L. The manager wants to downgrade to a new fish tank with a capacity of 50 L. The manager estimates that the dimensions of the new fish tank should be half of the old tank’s dimensions. Is the manager correct? Justify your answer.

For these two rectangular prisms: a

Are the two rectangular prisms similar? Justify your answer.

Find the scale factor of the length.

Find the scale factor of the surface area.

Find scale factor of the volume.

Find the volume scale factor if the measurements of the larger prism are halved.

1.8 m 6.4 m

2.4 m

2.88 m 3.84 m 10.24 m

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Big ideas Pythagoras’ theorem and trigonometric ratios provide the tools for calculating unknown sides and angles with practical applications in the real world. Curriculum Outcomes VC2M9M03 solve spatial problems, applying angle properties, scale, similarity, ratio, Pythagoras’ theorem and trigonometry in right-angled triangles VC2M9SP01 recognise the constancy of the sine, cosine and tangent ratios for a given angle in right-angled triangles using properties of similarity VC2M9SP03 design, test and refine algorithms involving a sequence of steps and decisions based on geometric constructions and theorems; discuss and evaluate refinements

Chapter outline

Trigonometry Chapter outline 8.01

8.02 8.03 8.04 8.05

Pythagoras’ theorem Investigation: The triangle inequality Investigation: Ratios in triangles Trigonometric ratios Side lengths Angles Applications of trigonometry Chapter 8 review

398 406 409 411 423 433 443 450

The Eiffel Tower in Paris relies on trigonometric principles in its design to make it strong!

8.01

Pythagoras’ theorem After this lesson, you will be able to… • apply Pythagoras’ theorem to find the length of the hypotenuse in a rightangled triangle. • use Pythagoras’ theorem to find the lengths of the other sides in a rightangled triangle. • explore the derivation of Pythagoras’ theorem using squares and triangles. • define the altitude in a right-angled triangle and its role in dividing the triangle into two similar ones. • use the theorem to solve real-world problems involving right-angled triangles.

Pythagoras’ theorem Exploration For the squares shown in the diagram: b

What is the area of the inside square, where length is c?

What is the area of the four triangles, where height is a and base is b?

What is the area of the outside square, where length is (a + b) ?

Equate the area of the inside square to the area of the outside square minus the area of the four triangles. Can you derive Pythagoras’ theorem?

The hypotenuse is labelled c and the two smaller sides are a and b. a2 + b2 = c2

Pythagoras’ theorem states that in a right-angled triangle, the sum of the squares of the lengths of the two shorter sides is equal to the square of the length of the hypotenuse.

We can use this theorem to find the unknown side lengths of a right-angled triangle.

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Example 1 Find the length of the unknown side c in the triangle.

Create a strategy We can use the Pythagoras’ theorem: c2 = a2 + b2.

Apply the idea Write the formula

Substitute a and b

Evaluate the indices

Evaluate the sum

Take the square root of both sides

Evaluate the square root

Example 2 Find the length of the unknown side x in the triangle. 15

Create a strategy Use Pythagoras’ theorem: a2 + b2 = c2.

Apply the idea We can substitute the following a = x, b = 12, and c = 15 into the formula: Write the formula

Substitute the values

Evaluate the indices

Subtract 144 from both sides

Evaluate the difference

Square root both sides

Evaluate the square root

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399

Idea summary c

a2 + b2 = c2

Pythagoras’ theorem relates the three sides of a right-angled triangle. a and b are the two shorter sides, and c is the longest side, called the hypotenuse.

8.01 Practice questions Understanding 1

Explorer 1–3

Adventurer

Trailblazer

1–3

Are these statements true or false? a

Pythagoras’ theorem can only be applied to a right-angled (90°) triangle.

The side directly across from the right angle is known as the opposite.

To ﬁnd the unknown length we must know the length of the other two sides.

The hypotenuse is the longest side.

For this right-angled triangle, write the equation that can be used to ﬁnd x. 19

Using the diagram: a

400

Identify:

Three different right-angled triangles

The hypotenuse of △ABC

iii

The shortest side of △ABC

The perpendicular height of △ABC from B

The hypotenuse of △ABD

Can we ﬁnd DC given: i

AD and BC?

AD and BD?

iii

BD and BC?

AB, AD, and BC?

Mathspace Victorian Curriculum – Year 9 mathspace.co

4ac, 5aceg, 6ac, 7ace, 8, 9

Fluency 4 Ex 1

4ac, 5aceg, 6ac, 7ace, 9, 10

4bd, 5bdfh, 6bd, 7bdf, 9–11

For each triangle, ﬁnd the length of the unknown side c. a

21 m

c cm

20 m

48 cm

14 cm

For each triangle, ﬁnd the length of the hypotenuse. Give your answer as an exact value. a

17 cm

c cm

9 cm

3 cm

c cm

d 4 c 21

The two smallest side lengths of a right-angled triangle are 6 and 9.

The two smallest side lengths of a right-angled triangle are 14 and 17.

The two smallest side lengths of a right-angled triangle are 6 and 7.

The two smallest side lengths of a right-angled triangle are 17 and 28.

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6 Ex 2

For each triangle, ﬁnd the length of the unknown side: a

y 12

29 m

d 82 m

80 m

20 m

For each triangle, ﬁnd the length of the unknown side as an exact value. a

14 m

b cm

17 m

15 cm

23 cm

d 42

31 m

A right-angled triangle q, has a hypotenuse of length 27 and another side of length 9.

A right-angled triangle r, has a hypotenuse of length 39 and another side of length 13.

Violet parks her car in a parking lot and walks 6 km north to get to a river. She walks 8 km east along the riverbank. If she is now h km directly from her car, ﬁnd the value of h.

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8 km 6 km

h km

TVs are measured in inches along their diagonal. A 65 in TV has dimensions of 144 cm by 81 cm. Use this to ﬁnd the value for 65 inches to the nearest cm.

144 cm 81 cm

65 in

In designing a diamond-shaped picture frame, the vertical length is 48 cm and the horizontal length is 14 cm. How long will each diagonal side of the diamond be? Round your answer to two decimal places.

48 cm

14 cm

Yasmina is standing in the northwest corner of a rectangular field. A walking path runs along the diagonal of the field. If she walks 250 m along the diagonal path, then walks due west 240 m, she will reach the southwest corner of the field. If the distance between the northwest and southwest corners of the field is x m, determine the value of x.

Reasoning 12

12–14

Ella tried to solved for a on the Pythagoras’ formula c2 = a2 + b2. Her working is shown: c2 = a2 + b2

12–15

c2 – b2 = a2

c–b=a

What was her mistake? Justify your answer.

Explain why she did not include the negative square root in step 3 .

Solve the equation correctly.

Using Pythagoras’ theorem: a

Can a triangle with sides of length 5 cm, 7 cm and 9 cm be a right-angled triangle? Justify your answer.

Can a triangle with sides of length 7.5 cm, 10 cm and 12.5 cm be a right-angled triangle? Justify your answer.

Given a rectangle with side lengths of 16 cm and 32 cm, what is the length of the diagonal rounded to two decimal places? Explain why the two diagonals in a rectangle are equal.

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A triangle with sides 3, 4, and 5 units is a right-angled triangle. If the triangle is enlarged by a factor of 2, does Pythagoras’ theorem still hold? Justify your answer.

Problem-solving 16, 17, 21 16

17, 18, 20, 21

18–23

A surveyor measures three sides of a piece of land in the shape of a quadrilateral, as shown in the diagram. a

Find the value of d.

Find the area of the land.

If one acre is 4046.86 m2, how many acres is this plot of land? Round your answer to one decimal place.

111 m 167 m 119 m

Using the given diagram, ﬁnd the value of x rounded to one decimal place. 6.4 m

xm 3m 3.9 m

AB is perpendicular to the radius of the circle with centre O.

21.2 cm

Using the diagram, ﬁnd:

The radius of the circle, rounded to one decimal place.

The length of BC, rounded to 1 decimal place.

A farmer wants to build a fence around the entire perimeter of his land, as shown in the diagram. The fencing costs $37 per metre. a

Find the length x, rounded to two decimal places.

Find the length y, rounded to two decimal places.

How many metres of fencing does the farmer require, if fencing is sold by the metre?

404

At $37 per metre of fencing, how much will it cost him to build the fence along the entire perimeter of the land?

Mathspace Victorian Curriculum – Year 9 mathspace.co

24 cm

A 1m F x

5m 7m B

E2 m

G y

In the diagram, △ABE ∼ △ACD, with BE = 3.2, DE = 9 and CD = 8. a

Find the length of EA.

Find x, the length of AB.

A x E

3.2

9 D

A chocolate box in the shape of a triangular prism has a base of width 6 cm, a height of 8 cm and a length of 16 cm. a

Find the length x.

Every edge of the chocolate box is to be sealed with special gold strips. What length of gold strip is required?

A ramp is being built in a shopping centre to make it more accessible. The building is 68 cm off the ground and the ramp is to start 1.2 m away from the doorway. Calculate the length of the ramp, rounded to the nearest mm.

Jesmond and Bilal are meeting each other for a mountain top picnic. Bilal took a wrong turn and is walking on a trail that is parallel to the trail that Jesmond is on. Bilal is at a height of 1870 m and Jesmond is at a height of 2340 m. The horizontal distance between them is 1080 m.

8 cm

x 16 cm 6 cm

What is the direct distance between the two friends? Round your answer to the nearest metre.

Going off trail, how long will it take for Bilal to walk directly to Jesmond at a speed of 1.42 m/s? Round your answer to the nearest minute.

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INVESTIGATION

The triangle inequality Investigate the properties and constraints of the triangle inequality and create a structured approach to test these properties algorithmically.

Objectives •

To understand and explore the triangle inequality through a hands-on activity

• To design a pseudocode algorithm that determines if given side lengths can form a triangle based on the triangle inequality.

Materials • Colored straws (at least seven different colors) • Scissors

• •

Paper Pencil

The triangle inequality Special requirements for triangles The triangle inequality speciﬁes a general rule for the side lengths of a triangle. Gather your materials to complete the procedure: 1.

Measure and cut straws into pieces so that you have one piece each of length 2 cm, 3 cm, 4 cm, 5 cm, 6 cm, 8 cm, 9 cm, and two pieces of length 10 cm.

Use the pieces of straw that you have cut out to create the triangles and complete the table:

Attempt 1 Attempt 2 Attempt 3 Attempt 4 Attempt 5 Attempt 6 Attempt 7 Attempt 8

Side 1 4 cm 2 cm 3 cm 4 cm 2 cm 10 cm 10 cm 10 cm

Side 2 5 cm 4 cm 4 cm 5 cm 3 cm 6 cm 6 cm 6 cm

Side 3 8 cm 5 cm 8 cm 7 cm 5 cm 10 cm 4 cm 8 cm

Which attempts resulted in the creation of a triangle?

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Does it create a triangle?

For each of the attempts ﬁnd the sum of the lengths of: • Sides 2 and 3 • Sides 1 and 2 • Sides 1 and 3 For the attempts that created triangles, compare each of the sums you found in the previous question to the remaining side not included in that sum. For example if I am looking at the sum of the lengths of sides 2 and 3 I would compare this number to the length of side 1. What do you notice? For the attempts that did not create triangles from your lists of sums, what do you notice?

Investigate 1.

How are the sums of two sides compared with the third side different between the attempts that did create triangles and the attempts that did not create triangles?

Based on your observations, what is a general rule you can make for the sides of a triangle?

Compare your rule with a classmate. Did they come to the same conclusion?

Pseudocode to apply the triangle inequality What is pseudocode? Pseudocode is essentially a way to outline a solution to a problem using plain English without worrying about formal programming syntax. Pseudocoding is a tool to help us think through a problem before diving into detailed solutions. Elements of pseudocode • Variables: These are like containers or boxes that can hold values • Operators: Actions like adding, subtracting, multiplying, etc. • Conditions: “If-then” type statements to make decisions • Loops: Repeating actions a certain number of times • Input and output: Getting input and showing output Pseudocode should be understandable even to someone who doesn’t know programming. It’s like giving a list of steps to follow to solve a problem. It often uses indentation to show structure (like steps within steps). There’s no strict format, but it should be clear and organised. Shown is an example of pseudocode for the deﬁnition of an even number: A number is even if it can be divided by 2 without leaving a remainder. Otherwise, it’s odd.

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Here is the breakdown of the example: 1.

We’ve deﬁned a clear goal: Determine if N is even or odd.

We’re getting a result: “the remainder when N is divided by 2”.

We’re using a condition: “If the remainder is 0”.

Create your own pseudocode for determining if given side lengths satisfy the triangle inequality theorem, based on what you discovered in the ﬁrst activity. Use the questions to help guide your creation. • Based on the triangle inequality theorem, what conditions must the three side lengths meet to form a triangle? • When outlining your pseudocode, how will you account for all possible combinations of the three side lengths? • How will you structure your pseudocode to handle the input (side lengths) and produce a clear output (whether it forms a triangle or not)? • What variables will be necessary for your pseudocode, and how do you plan to utilise them?

Investigate 1.

Exchange your pseudocode with a classmate. Follow their pseudocode to determine if the side lengths 7 cm, 18 cm, and 10 cm satisfy the triangle inequality theorem.

Follow your partner’s pseudocode again to determine if the side lengths 7 cm, 11 cm, and 9 cm satisfy the triangle inequality theorem.

Determine if their pseudocode is understandable and provide feedback to your partner on their code. Make improvements to your pseudocode based on your partner’s feedback.

Discussion

408

With a classmate, determine more combinations of side lengths that could create a triangle.

With the foundational understanding of pseudocode and the triangle inequality, how do you think you can extend this knowledge? For instance, could you foresee creating pseudocode for other geometric theorems or properties? Give examples.

Mathspace Victorian Curriculum – Year 9 mathspace.co

INVESTIGATION

Ratios in triangles Investigate the patterns that emerge when creating nested similar right-angled triangles on a coordinate plane.

Objectives To analyse the relationships between the sides of nested triangles drawn on a coordinate plane To understand the constancy of ratios of corresponding sides based on given angles

Materials • Pencil

•

Graph paper

•

Straight edge

Ratios in triangles Using the concept of similar triangles, we will construct a series of nested right-angled triangles on a coordinate plane, each sharing a common hypotenuse. Use the diagram to help you create your own nested triangles on graph paper: 1.

Start by graphing any diagonal line on the coordinate grid. This line will serve as the hypotenuse for your series of right-angled triangles. Label the y-intercept C.

Identify a point on the line and label it A. This will be one of the vertices of your ﬁrst right-angled triangle.

From point A, draw a vertical line segment downwards until it meets the horizontal line through C, forming a right angle. Label the point where it meets this point as point B.

10 9 8 7 6 5 4 3 C 2 1 0

2 3 4 5 6 7 8 9 10

You have now formed your ﬁrst right-angled triangle, ABC.

Next, select another point on line segment AC and label it D. This will be one of the vertices of your next right triangle.

Repeat the process to form another right-angled triangle, DEC, with D as one of its vertices.

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Continue the process to form as many nested triangles as you like on your coordinate plane.

Investigate 1.

What is the relationship between triangle ABC and the other triangles you created? Explain.

Measure the lengths of the sides of each triangle. What patterns or ratios do you notice between the corresponding sides of the nested triangles?

As you move further along the hypotenuse and create more nested triangles, what do you predict will happen to the ratios of the corresponding sides? Why?

Relate the slopes of the lines to the ratios of the sides of your triangles. How does this help to conﬁrm they are similar?

Given that all your triangles have one angle in common, do you think there’s a relationship between the angle of a triangle and the side ratios?

Discussion 1.

Think about other shapes besides triangles. How might the concept of consistent ratios apply if we were working with nested rectangles or parallelograms on the coordinate plane?

Besides roofs, where else in architecture or building design might the concept of nested similar triangles be useful? Discuss the aesthetic and functional implications with a classmate.

Did you know?

Trigonometry is used in everyday life, from navigation to construction! By measuring angles and distances, we can calculate the height of buildings, the distance across rivers, or even guide airplanes. For instance, surveyors use trigonometry to determine land boundaries, while architects rely on it to design ramps, roofs, and bridges. 410

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8.02

Trigonometric ratios

After this lesson, you will be able to… • identify the hypotenuse, opposite side, and adjacent side in a right-angled triangle. • define and use the trigonometric ratios sine, cosine, and tangent. • express the trigonometric ratios as fractions of side lengths. • solve problems involving trigonometric ratios.

Ratios in right-angled triangles Hypotenuse

Opposite

θ Adjacent

AC is the hypotenuse in this right-angled triangle as it is opposite the 90° angle. BC is opposite the angle θ, so we can refer to it as the opposite side. AB is next to the angle θ, so we can refer to it as the adjacent side.

Interactive exploration Discover this concept in action online

mathspace.co

For similar right-angled triangles (triangles which have the same angles), the ratio of any pair of sides will always be the same. Trigonometry has special ratios of side lengths in right-angled triangles. These are the ratios we will explore:

Example 1 For this right-angled triangle:

θ B

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a Which side is the hypotenuse?

Create a strategy The hypotenuse is the side opposite the right angle.

Apply the idea

Reﬂect and check

The hypotenuse is side AB.

We will often label sides that are opposite a vertex with the same letter as the label of the vertex but lower case, so we can also call this side c.

b For the angle θ, which side is the opposite?

Create a strategy The opposite side across from the angle.

Apply the idea

Reﬂect and check A

Notice that the opposite side is the one side that does not have an endpoint at the vertex where the angle is located.

α C

θ B

Side AC is opposite the angle θ.

c For angle α, which side is the opposite?

Create a strategy Starting from angle α we look across the triangle.

Apply the idea

Reﬂect and check

Side BC is opposite angle α.

Notice that in the same triangle, we can have different sides labelled as the opposite side depending on which angle we are starting from.

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d For angle α, which side is the adjacent side?

Create a strategy The adjacent side is one of the sides next to the angle, but is not the hypotenuse.

Apply the idea

Reﬂect and check

Side AC is adjacent to angle α.

We can check that this side is not the hypotenuse and not the opposite side.

Example 2 For the angle θ, what is the value of the ratio

15 17

Create a strategy First we can identify the opposite and adjacent sides from angle θ, then we can ﬁnd their ratio.

Apply the idea Starting from angle θ, side AC is the opposite and side BC is the adjacent side.

A Opposite 15 17 Hypotenuse

Now we can ﬁnd the ratio: C

8 Adjacent

We cannot simplify this any further.

Reﬂect and check It can help to label the hypotenuse as well to check that we have labelled the other sides correctly.

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Hypotenuse

θ A

Adjacent

AC is opposite the right angle, so we can call it the hypotenuse.

Opposite

Idea summary

AB is next to the angle θ, and is not the hypotenuse, so we call it the adjacent side.

BC is opposite the angle θ, so we can call it the opposite side.

For any triangle, we can ﬁnd these ratios: Opposite Opposite Adjacent , , Adjacent Hypotenuse Hypotenuse If we change the side lengths, but keep the angles the same, to make similar triangles, then the ratios do not change.

Trigonometric ratios We can define the ratios we have seen as the trigonometric ratios.

If we are given one of the angles in a right-angled triangle and all of the side lengths, we can write the trigonometric ratios as fractions of the side lengths.

Interactive exploration Discover this concept in action online

mathspace.co

The ratios are often abbreviated to H

θ A

Where H is the hypotenuse, O is the opposite side and A is the adjacent side. Note that we can also abbreviate sine, cosine and tangent to sin, cos and tan.

A helpful technique to remember these ratios is to use the mnemonic SOH-CAH-TOA.

Example 3 Which trigonometric ratio relates the given sides and the angle θ ? 8

θ 15 414

Mathspace Victorian Curriculum – Year 9 mathspace.co

Create a strategy Remember the mnemonic SOH-CAH-TOA. Consider the relationship between the given sides and the angle θ.

Apply the idea Based on the given triangle, the triangle has side lengths that are opposite and adjacent to the angle θ. This means that the trigonometric ratio for this triangle is tangent

Example 4 Write the expressions for sine, cosine and tangent for the angle θ.

A 25

θ 7 C

Create a strategy Label the triangle for the hypotenuse, opposite and adjacent sides. Identify the ratios.

Apply the idea With respect to the given angle θ, the opposite side is AC, the adjacent side is BC, and the hypotenuse is AB, that is O = 24, A = 7 and H = 25.

A H

Use the sine ratio

Substitute the values

θ A

Use the cosine ratio

Substitute the values Use the tangent ratio

Substitute the values

Reﬂect and check Note that it is customary in trigonometry to leave fractions as improper fractions.

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Example 5 Find the value of tan θ in △ABC.

A 10

θ B

Create a strategy Label the sides of the triangle with reference to angle θ. Label the missing side length x. Find the missing side length (x) using Pythagoras’ theorem. Use the trigonometric ratio for tangent.

Apply the idea From the diagram, the opposite side of the given angle θ, is the missing side length x, so O = x, the adjacent side is A = 8, and the hypotenuse is H = 10.

A x (O) B

(H)

θ 8 ( A)

. Use Pythagoras’ theorem to ﬁnd the value of x. Use Pythagoras’ theorem

Substitute the values and x

Subtract 82 from both sides

Evaluate

Take the square root of both sides

Evaluate

The length of the opposite side x is 6 and we can substitute this length into the trigonometric ratio for tangent. Use the tangent ratio

Substitute the values

Simplify

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Idea summary

Hypotenuse

θ A

For a right-angled triangle, the trigonometric ratios are given by:

Opposite

Opposite Hypotenuse Adjacent cos θ = Hypotenuse Opposite tan θ = Adjacent sin θ =

Adjacent

A helpful technique to remember these ratios is to use the mnemonic SOH-CAH-TOA.

8.02 Practice questions Understanding 1

Explorer 1–4

Adventurer

Trailblazer

1–4

Which side is the hypotenuse? a

b Z

Z Y

Z X

Which angle is opposite the hypotenuse? a

A B

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Fill in the blanks. a

Determine if the statement is true or false based on the given triangles. a

These triangles are similar triangles.

The angles in each triangle have the same values.

One triangle is twice as large as the other.

The value for sin(45) in the ﬁrst triangle will have the same value in the second triangle. 5–7, 8aceg, 9ab, 10, 11

Fluency

Ex 1

c 2 45°

45°

45° 2

4 45° B

5, 6, 7ace, 8bdfh, 9bc, 12a

5, 6, 7bdf, 8bdfh, 9cd, 12

Identify the adjacent and opposite sides to the angle θ. a

C C

θ B

Z Z

θ Y

Ex 2

For the angle θ : a

Find the length of the hypotenuse.

Find the length of the opposite side.

Find the length of the adjacent side.

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What is the value of the ratio

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C 48 cm

52 cm

Ex 3

Which trigonometric ratio relates to the given sides and the angle θ? a

b 7

θ θ

12 B

9 31

Ex 4

Write the ratios for each triangle: i

sin θ

cos θ

iii

tan θ

A b c

C 61 20

34 B

B 7

h B 17

A 15 8

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9 Ex 5

For each of the right-angled triangles: i

Find the value of x.

Write the ratio for tan θ.

B 8

B 29

In terms of the side lengths AB, AC and BC, write the ratios for: a

sin θ

sin α

cos θ

cos α

tan θ

tan α

For each right-angled triangle: i

Identify the ratio that relates the given angle and sides.

Write the trigonometric ratio.

b 6

θ 13

d w

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25°

7.5

For each triangle: i

Find the value of sin θ.

Find the value of cos θ. A

Reasoning

13–15

A right-angled triangle is adjusted so that the length of the side opposite an angle is doubled while the adjacent side length is unchanged. By doing this the hypotenuse and the angle will change. How do the sine, cosine and tangent ratios change?

Using the diagram:

What is sin θ?

What is cos α?

Is the value of the ratio for sin θ and cos α the same?

Is this true for all right-angled triangles? Explain your answer.

Using the diagram shown, show that

4 3 C

= tan θ.

A a° 13

Problem-solving 16–18 16

16–19, 21

θ°

18–22

Find the value of tan θ.

19 15

θ° 28

If sin θ = : a

Which angle is represented by θ?

Find cos θ.

Find tan θ.

5 4 C

8.02 Trigonometric ratios mathspace.co

A 421

If tan θ =

Which angle is represented by θ?

Find cos θ.

Find sin θ.

If cos θ = a

Find sin θ.

Find tan θ.

△PQR is a right-angled triangle and S is the midpoint of QR as shown.

If ∠SPQ = θ, show that QR = 2a tan θ.

Using the diagram: T 56.64 m 57 m

28.32 m

19 m B

422

42 m

A 21 m

Using the triangle created by the building, ﬁnd the ratio of the opposite side to the adjacent side of ∠TAB.

The tree has a height of 19 metres. For the smaller triangle created by the tree, ﬁnd the ratio of the opposite side to the adjacent side of ∠HAC.

Using the triangle created by the tree, ﬁnd the ratio of the adjacent side to the hypotenuse for ∠HAC.

Using the triangle created by the building, ﬁnd the ratio of the opposite side to the hypotenuse for ∠TAB.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Beth is building a house. She wants the roof height to be between 3.5 m and 5 m. She must decide the angle measure to use for the slant of the roof when the slant height is d m.

Roof height

d x°

Write an inequality that Beth can use in order to ensure that her roof height will be within the desired range. Also find an appropriate value for the slant height d.

8.03

50 metres

Side lengths

After this lesson, you will be able to… • evaluate trigonometric ratio expressions using technology. • apply trigonometric ratios to find unknown angles and side lengths in rightangled triangles. • solve problems that require the application of trigonometric ratios to find a missing side length.

Evaluate trigonometric ratios We can evaluate trigonometric expressions like sin 42° or cos 71° using a calculator. When evaluating trigonometric expressions, make sure that the calculator is in degrees mode.

Exploration Use sin 30° = 0.5 and sin 60° = 0.866 (rounded to 3 decimal places) to answer these questions. 1.

What is the value of (sin 30°) × 2?

What is the value of sin(30° × 2)?

Are they the same?

What is the value of (sin 60°) ÷ 2?

What is the value of sin(60° ÷ 2)?

Are they the same?

Why do you think this is?

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Example 1 Evaluate these using a calculator, round your answer to two decimal places. a sin (45)

Create a strategy Use a calculator and make sure it is in degrees mode.

Apply the idea Using a calculator we get: sin 45 = 0.707 106 781 2 = 0.71

Round the answer

Reﬂect and check If your calculator is in radian mode, your answer would have been sin 45 = 0.8509. Change it to degree mode, then evaluate sin 45 again.

b cos (45)

Create a strategy Use a calculator and make sure it is in degrees mode.

Apply the idea Using a calculator we get: cos 45 = 0.707 106 781 2 = 0.71

Round the answer

c tan (45)

Create a strategy Use a calculator and make sure it is in degrees mode.

Apply the idea tan 45 = 1

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Reﬂect and check We can use a right isosceles triangle with side lengths of 1 to check our answers to the previous parts. Using Pythagoras’ theorem to ﬁnd the length of the hypotenuse:

45° 1 45° 1

Use the sine ratio

Substitute known values

Evaluate with a calculator Use the cosine ratio

Substitute known values

Evaluate with a calculator Use the tangent ratio

Substitute known values

Simplify

Example 2 Evaluate 7 cos 77° to two decimal places.

Create a strategy Use a calculator and make sure it is in degrees mode.

Apply the idea Enter this into the calculator: 7 cos 77° = 1.574 657 38 = 1.57

Round the answer

Reﬂect and check A way to check to make sure your calculator is in degrees is to evaluate sin (30°) and check that you get 0.5.

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Idea summary Evaluate trigonometric expressions like sin 42° and cos 71° using the trigonometric functions on a calculator and entering the desired angle. When evaluating trigonometric expressions, make sure that the calculator is in degrees mode.

Find the side Hypotenuse

θ Adjacent

Trigonometric ratios for a right-angled triangle:

Opposite

Steps to ﬁnd a missing side in a right-angled triangle: 1. Identify the reference angle. 2. Label the side lengths as hypotenuse, opposite, and adjacent. 3. Choose the ratio which has one of the required sides given and the missing side: • Sine: opposite and hypotenuse • Cosine: adjacent and hypotenuse • Tangent: opposite and adjacent 4. Substitute the information you know into the chosen trigonometric ratio. 5. Solve for the unknown.

Example 3 f

Find the value of f rounded to two decimal places.

53°

Create a strategy Use the trigonometric ratio that relates the two sides and substitute the given values. In this case: • Label the sides for opposite, adjacent and hypotenuse. • Write an equation using the cosine ratio as we have the angle, the adjacent side and the hypotenuse. • Solve for f.

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Apply the idea Use the cosine ratio

Substitute the values and f

Multiply both sides by 12

Evaluate using a calculator

Reﬂect and check To check our answer, we can evaluate both sides of the initial equation with the cosine ratio. Use the cosine ratio

Substitute the values

Evaluate both sides on a calculator

They aren’t the exact same because we rounded to only two decimal places, but they are close enough that we can tell our answer is correct.

Example 4 Find the value of h rounded to two decimal places.

21° hm

16 m

Create a strategy Use the trigonometric ratio that relates the two sides and substitute the given values. In this case: • Label the sides for opposite, adjacent and hypotenuse. • Write an equation using the sine ratio as we have the angle, the opposite side and the hypotenuse. • Solve for h.

Apply the idea Use the sine ratio

Substitute the values and h

Multiply both sides by h

Divide both sides by sin 21°

Evaluate using a calculator

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Reﬂect and check The hypotenuse should always be longer than any other side. It is helpful to check this when ﬁnding or using the hypotenuse.

Example 5 A sailor looks up at a 13° angle to the top of 50 m tall lighthouse. Find the distance d metres from the sailor’s boat to the base of the lighthouse.

50 m A

13° dm

Create a strategy Use the trigonometric ratio that relates the two sides and substitute the given values. In this case: • Label the sides for opposite, adjacent and hypotenuse. • Write an equation using the tangent ratio as we have the angle, the opposite side, and the adjacent side. • Let d represent the length AB, the distance that we need to ﬁnd. Solve for d.

Apply the idea Use the tangent ratio Substitute the values and d Multiply both sides by d Divide both sides by tan 13° Evaluate using a calculator

Idea summary Using the trigonometric ratios sin(θ ) =

Opposite Adjacent Opposite cos(θ ) = tan(θ ) = Hypotenuse Hypotenuse Adjacent

we can solve for any of the side lengths in a right-angled triangle. Maintain equality by performing the same operation to both sides of the equation, and use a calculator to evaluate the trigonometric expressions.

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8.03 Practice questions Explorer

Understanding

1–3

Adventurer

Trailblazer 2, 3

1–3

In a right-angled triangle, the side opposite the right angle is called the hypotenuse. What are the other two sides called?

For each diagram, is there enough information to ﬁnd the missing length? a

b bm

27°

θ°

12 m

12 cm

44°

25° g cm

16.68 cm a cm

What is the ratio that will help ﬁnd the missing side? Do not ﬁnd the value of the missing side: a

If the angle is 45 degrees and the adjacent side is 7 cm, what is the length of the opposite side?

If the angle is 30 degrees and the hypotenuse is 10 cm, what is the length of the opposite side?

If the angle is 60 degrees and the adjacent side is 5 cm, what is the length of the hypotenuse?

Fluency 4 Ex 1

5 Ex 2

4a–c, 5a–c, 6a–c, 7, 8, 12

4c–f, 5c–f, 6cdf, 7a–c, 8a–c, 9–12

4f–h, 5dgh, 6egh, 7cdf, 8d–f, 9–12

Evaluate and round to three decimal places: a

sin 82°

cos 28°

tan 84°

cos 70°

tan 47°

sin 39°

cos 54°

tan 5°

Evaluate and round to three decimal places: a

9 sin 63°

12 cos 7°

8 tan 28°

5 cos 35°

1.8 tan 56°

4.2 sin 23°

2.1 sin 56.5°

0.75 tan 1.22°

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7 Ex 3

Evaluate and round to three decimal places: a

Find the value of the variable, rounded to three decimal places: a

f mm

33°

56°

6 cm

y cm

11 m 25°

11 mm

44°

12 mm

15 m 15.75 cm

58°

x cm

42°

y mm

8 Ex 4

Find the value of the variable, rounded to three decimal places: a

4 cm

c 41°

b mm

25° g cm

hm 63° 3.75 mm

10 m

43 m

xm 32°

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11.5 m

42°

f mm 23.2 mm 26°

Find the length of AC, rounded to one decimal place.

C 63°

Find the value of a, rounded to the nearest centimetre. 12 cm

44°

16.68 cm a cm

Find the value of the variable, rounded to two decimal places: a

54°

16 cm 42°

26 mm

f cm h mm

12 Ex 5

Ambrose’s home is located at point A, and Ambrose is standing at point B.

A 2313 m

If d is the distance between Ambrose and their home, ﬁnd the value of d rounded to two decimal places.

55°

Reasoning 13

13, 14

13–15

Given the value of tan θ = , ﬁnd the value of b, rounded to two decimal places. Explain your thinking. bm

q° 10 m

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Jin asks his friend Ethan if he can construct a right-angled triangle with an angle of 30° and side lengths of 10 cm and 20 cm. Ethan says it is not a right-angled triangle. Look at Ethan’s work and describe the error.

Pythagoras’ theorem

Evaluate: i

sin 80° and cos 10°

sin 67° and cos 23°

iii

sin 45° and cos 45°

sin 32° and cos 58°

What is the pattern developed in part (a)?

If sin (θ ) = x and cos (α ) = x, what is the relationship between θ and α?

Problem-solving 16–18 16

17–19

Find a rounded to two decimal places.

Find b rounded to one decimal place.

Find d rounded to the nearest whole number.

18–21

b cm 37°

37° 5 cm

d cm

Find the length of AD, rounded to two decimal places.

Find the length of BD, rounded to two decimal places.

Find the length of AB, rounded to two decimal places.

C 9 B

60°

35°

Two ﬂag posts of height 12 m and 17 m are erected 20 m apart. Find the length, l, of the string needed to join the tops of the two posts. Round your answer to one decimal place.

a cm

D lm

17 m

12 m 20 m

A is a point on the circumference of the circle with centre O.

OB is 24 cm long and cuts the circle at C. a b

432

62°

Find the radius of the circle, r, rounded to one decimal place. Find the length of BC rounded to one decimal place.

Mathspace Victorian Curriculum – Year 9 mathspace.co

24 cm

For the following diagrams: i

Find y, rounded to two decimal places.

Find w, rounded to two decimal places.

iii

Find the value of x, rounded to one decimal place.

y w 39°

67°

68 63 37°

46° w

x y

A safety fence is constructed to protect tourists from the danger of a 9 metre tall eroding castle toppling down. A surveyor measures the angle from a distance to the top of the castle as 10°. She then walks towards the fence and takes a second measurement to the top of the tower of 22°. How far did she walk between taking the two measurements? Round your answer to the nearest metre. 9m 10°

8.04

22°

Angles

After this lesson, you will be able to… • differentiate between trigonometric ratios and their inverses. • apply inverse trigonometric ratios to find missing angles in right-angled triangles. • understand the relationship between the tangent ratio and the gradient of a line. • solve problems involving the calculation of missing angles in right-angled triangles and the determination of the gradient of a line.

Inverse trigonometric ratio To ﬁnd a missing angle in a right-angled triangle, we can calculate the inverse of the trigonometric ratio. 8.04 Angles mathspace.co

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The inverse effectively undoes the trigonometric ratio. If sin θ = x, then we undo the sine by using the inverse operation sin−1. sin θ = x sin−1(sin θ ) = sin−1 x

θ = sin−1 x We use the inverse function on the calculator to evaluate the expressions. When evaluating inverse trigonometric functions, make sure that your calculator is in degrees mode.

Exploration Evaluate the following, where θ = 0.5: 1.

sin−1 θ and

. Are they the same?

cos−1 θ and

. Are they the same?

tan−1 θ and

. Are they the same?

While we may represent the inverse trigonometric functions using an index of −1, they are not reciprocals of the original functions,

To ﬁnd missing angles in right-angled triangles, we require at least two known sides in order to use trigonometric ratios.

Hypotenuse

θ Adjacent

Trigonometric ratios for a right-angled triangle:

Opposite

To determine which trigonometric ratio we can use to ﬁnd the unknown angle, we must label each of the two known side lengths of the triangle as either opposite, adjacent or the hypotenuse. We then solve for the angle by applying the inverse of that ratio.

Example 1 If cos θ = 0.256, ﬁnd θ. Round your answer to the nearest degree.

Create a strategy Use the inverse cosine to ﬁnd the angle θ.

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Apply the idea cos θ = 0.256 −1

θ = cos (0.256)

Write the cosine Use the inverse cosine to solve for θ

= 75.167 152…°

Evaluate using a calculator

= 75°

Round to the nearest degree

Example 2 Find the value of θ to the nearest degree.

Create a strategy Use the trigonometric ratio that relates the two sides then apply the inverse ratio.

Apply the idea With respect to the given angle of θ, the side AC = 8 is the opposite side and the side AB = 10 is the hypotenuse so we can use the sine ratio. Write the sine ratio

Substitute the values

Use the inverse sine to solve for θ

Evaluate and round to the nearest degree

Example 3 The person in the picture sights a pigeon above him. If the angle the person is looking at is α, ﬁnd α in degrees. 19 m a 13 m

Round your answer to two decimal places.

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Create a strategy Use the trigonometric ratio that relates the two sides then apply the inverse ratio.

Apply the idea With respect to the given angle of α, the side length of 13 m is adjacent and the side length of 19 m is opposite, so we can use the tangent ratio. Write the tangent ratio Substitute the values Use the inverse tangent to solve for α Evaluate and round to two decimal places

Idea summary To ﬁnd an unknown angle in a right angled triangle: • Label the sides opposite, adjacent and hypotenuse. • Write an equation using the trigonometric ratio that relates to the angle and the two given sides. • Solve for the angle using the inverse ratio.

Tangent and gradient A special relationship exists between the tangent ratio and the gradient of a line. If we wanted to ﬁnd the gradient (m) of this straight line, we

would use m = A

In this case the rise AB is opposite to the angle θ and the run CB is adjacent to θ. x

θ C

This means that

The gradient of a line is equal to tan θ, where θ is the angle between the line and the positive direction of the x-axis: m = tan θ

Example 4 A line has a gradient of 6. Find the angle between the line and the positive direction of the x-axis. Write your answer to the nearest degree.

Create a strategy Use the relationship between the gradient and tan θ then apply the inverse tangent. 436

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Apply the idea m = tan θ

Write the equation relating gradient and tan θ

6 = tan θ

Substitute m = 6

−1

θ = tan (6)

Use the inverse tangent to solve for θ

= 80.537 677…°

Evaluate using a calculator

= 81°

Round to the nearest degree

Example 5 Point T (25, 6) lies on the given line passing through the origin:

y 7 6

5 4 3 2 1

x 5

Find θ, the angle between the line and the positive x-axis. Round your answer to two decimal places.

Create a strategy Use the relationship between the gradient and tan θ then apply the inverse tangent.

Apply the idea Write the equation relating gradient and tan θ

Substitute m =

Substitute the values

Use the inverse tangent to solve for θ

Evaluate and round to two decimal places

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Idea summary The gradient of a line is equal to tan θ, where θ is the angle between the line and the positive direction of the x-axis: m = tan θ

8.04 Practice questions Explorer 1, 2a–c, 3

Understanding 1

Adventurer 1, 2b–d, 3

Name the opposite side to angle θ.

Name the adjacent side to angle θ.

Name the opposite side to angle α.

Name the adjacent side to angle α.

Name the angle that is opposite the hypotenuse.

Trailblazer 2d–f, 3 A α

θ B

With reference to the angle θ, ﬁnd the value of these ratios: i

iii b

A 34

θ B

13 B

5 C

θ θ

25 A

Fill in the missing information. ⬚ trigonometric functions (sin−1, cos−1, ⬚) help you ﬁnd the unknown ⬚ in a right-angled triangle when you know the side lengths. On the other hand, regular trigonometric functions (sin, ⬚, tan) are used to work out the ratios of the side lengths of a ⬚ triangle when one of the angles is already known.

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4a–c, 5a–dm, 6, 7a–e, 8, 10a–c, 11ab, 12, 13

Fluency 4

5 Ex 1

sin−1(0.7035)

cos−1(0.4447)

tan−1(0.5085)

sin−1(0.2734)

cos−1(0.6484)

tan−1(8.1519)

sin−1(0.8176)

cos−1(0.7035)

tan θ = 1.739

Find θ rounded to the nearest degree: a

sin θ = 0

cos θ = 1

sin θ = 0.210

cos θ = 0

tan θ = 1

cos θ = 0.296

tan θ = 0.869

tan θ = 0

cos θ = 0.5

cos θ = 0.899

tan θ = 2.2

Find θ, rounded to the nearest hundredth of a degree: a

7 Ex 2a

4f–h, 5i–lo, 6, 7d–j, 8, 9, 10d–f, 11bd, 13, 14

Evaluate, rounded to the nearest degree:

m 6

4c–e, 5e–hn, 6, 7c–h, 8, 10a–c, 11bc, 13, 14

sin θ = 0.8613

cos θ = 0.4825

tan θ = 8.958

Find the unknown angles, marked by a variable, rounded to the nearest degree: a b c B 7.9 cm

27 cm

39 cm

24 C

x A x

e A

i 63.6 cm

68.5 cm

15 cm

θ 10

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x A

8 Ex 3

The person in the ﬁgure sights a pigeon above him. If the person is looking at an angle of θ above horizontal, ﬁnd θ, rounded to the nearest minute.

17 m

θ 12 m

Write an expression for the gradient of the line that is inclined at an angle of 33° to the positive x-axis.

Find the gradient of a line, rounded to one decimal place, that is inclined to the positive x-axis, at the angle:

11 Ex 4

10°

30°

50°

20°30′

40°15′

60°45′

For each given gradient, ﬁnd the angle between the line and the positive x-axis. Round your answer to the nearest degree. a

Consider the given line which has an angle of θ with the positive x-axis: a

Find the gradient of the line.

Find θ. Round your answer to two decimal places.

x 18

13 Ex 5

Point P (5.2, 5.72) lies on the given line passing through the origin: Find θ, the angle between the line and the positive x-axis. Round your answer to two decimal places.

8 7 6

5 4 3 2 1

x 1

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2 3 4 5 6 7 8 9

Point P (15, 7) lies on the given line passing through the origin: Find θ, the angle between the line and the positive x-axis. Round your answer to two decimal places.

y 7

6 5 4 3 2 1

x 5

Reasoning 15

15, 18

15, 16, 18

15–18

An isosceles triangle has equal side lengths of 10 cm and a base of 8 cm as shown. a

Calculate the size of angle A rounded to one decimal place.

Explain why the cosine function is the most appropriate choice in this case.

10 cm

A°

A° 8 cm

When sin θ = cos θ, what angle will θ be? Explain your answer.

In a right-angled triangle, if the sine ratio of an angle is greater than the cosine ratio of the same angle, what can you conclude about the angle? Explain your reasoning.

If a gradient is doubled then the angle between the line and the positive x-axis is also doubled. Is this statement true or false? Explain your reasoning. (Hint: Questions 10 or 11 above might help.)

Problem-solving 19–21 19

20–23, 25

20, 22–25

Find x, y and z, rounded to two decimal places. 10

z x 30

Find x and y rounded to one decimal place.

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Consider the graph of the line

and the triangle

whose vertices are the origin, the x-intercept and the y-intercept.

3 2

The angle θ is the acute angle in the triangle that the line makes with the x-axis. a b

1 x

State the gradient of the line.

−3 −2

Find the value of tan θ.

−1

−1 −2

a b

ind the size of the angle x in degrees, rounded to two F decimal places. Find the size of the angle y in degrees, rounded to two decimal places.

During rare parts of Mercury and Venus’ orbit, the angle from the Sun to Mercury to Venus is a right angle.

19 cm

x 33 cm Venus

115 000 000 km

The distance from Mercury to the Sun is 60 000 000 km. The distance from Venus to the Sun is 115 000 000 km.

8 cm

Mercury

60 000 000 km

What is the angle θ, from Venus to the Sun to Mercury? Round your answer to the nearest degree. 24

During a large-scale renovation of a historical town, city planners decided to connect the bell tower of an old church to the base of the new city library with a glass walkway. If the top of the bell tower stands at a height of 80 metres from the ground and the base of the library is 200 metres away from the tower’s base, ﬁnd the angle θ, at which the walkway should be built to connect them. Round your answer to two decimal places.

△PQR and △QRS are both right-angled triangles as shown. a

Find the length of QR. Round your answer to two decimal places.

Find the value of θ. Round your answer to two decimal places.

Q P 7 cm 12 cm

θ R

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15 cm

8.05

Applications of trigonometry

After this lesson, you will be able to… • translate real-world scenarios into right-angled triangle problems. • solve real-world problems involving trigonometry. • interpret the results of trigonometric calculations in the context of real-world scenarios.

Applications of trigonometry Remember that the trigonometric ratios for a right-angled triangle are:

To ﬁnd unknown sides in right-angled triangles, we require at least one angle (other than the right angle) and one other side length. To ﬁnd missing angles in right-angled triangles, we require at least two known sides and we use the inverse trigonometric ratios. To solve trigonometry problems, in a real-world scenario, we can follow these steps: 1. Draw the diagram as a right-angled triangle and label the hypotenuse, opposite and adjacent. 2. Label the given side lengths and angle measures. 3. Choose the trigonometric ratio that applies to the given information. 4. Substitute the known values into the trigonometric ratio and solve for the unknown. 5. Give the answer in the context of the situation. This section will need you to choose the correct trigonometric relationship and method to solve problems.

Example 1 Ilhana is sitting 3.2 metres from the wall of her room. She spots a spider in the lower corner of the wall. If Ilhana is looking down at an angle of 19°, what is the height, x m, of Ilhana’s line of sight above the ground? Round your answer to one decimal place. 3.2 m 19° xm

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Create a strategy Use the trigonometric ratio that relates the two sides and substitute the given values.

Apply the idea With respect to the given angle of 19°, the side of length x is opposite and the side of length 3.2 m is adjacent, so we can use the tangent ratio. Write the tangent ratio

Substitute the values and x

Multiply both sides by 3.2

Evaluate using a calculator

Ilhana’s sightline is 1.1 m above the spider.

Example 2 A tree of lights is formed by 9 strings of lights that are each 350 cm long. The strings are all joined at one point to form the top of a tree, and the other end of each string is attached to a point on the ground.

θ° 350 cm

305 cm

A 305 cm pole stands in the centre, from the top of the tree to the ground. Find the angle, θ, one string of lights makes with the centre pole, rounded to two decimal places.

Create a strategy Identify the relevant trigonometric ratio, then solve for the angle θ.

q 350 cm 305 cm

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Apply the idea With respect to the given angle θ, 305 is the length of the adjacent side and 350 is the length of the hypotenuse, so we can use the cosine ratio. Write the cosine ratio Substitute the values Apply the inverse to solve for θ Evaluate using a calculator

Idea summary The trigonometric ratios for a right-angled triangle are: sinθ =

Opposite Adjacent Opposite cosθ = tanθ = Hypotenuse Hypotenuse Adjacent

To ﬁnd missing sides in right-angled triangles, we require at least one acute angle and one other side length. To ﬁnd missing angles in right-angled triangles, we require at least two known sides and we use the inverse trigonometric ratios.

8.05 Practice questions Understanding 1

Explorer 2–4

Adventurer

Trailblazer

1–4

A man standing at point C, is looking at the top of a tree at point A.

If you were to use θ σ

to solve for an angle, which angle would you be solving for? 2

A lifeguard standing at the top of the observation deck at point B, is looking down at the shore at point C.

Which angle is equivalent to 90 − α?

Which two labelled angles are equal?

B B α

θ A

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Rayna is standing 20 m from a tree and measures the angle from the ground to the top of the tree to be 43°. If she wants to ﬁnd the height of the tree, which trigonometric ratio does she need to use?

h 43° 20 m

A ladder is leaning against a wall. The angle between the ladder and the ground is 44° and the ladder reaches to the top of the wall, which is 1.36 m high. a

Sketch a diagram of the situation.

If you want to solve for the length of the ladder, which trigonometric ratio should you use?

Fluency 5

5–10, 12, 14

5–7, 9–11, 13, 14

6, 7, 10, 11, 13–15

Find the height of the tree, h, rounded to two decimal places. 23 m

29°

6 Ex 1

A helicopter is ﬂying at an altitude of 198 m. The angle from the helicopter to the landing pad is 44°.

44°

What is the distance, d, between the helicopter and the landing pad? Round your answer to the nearest whole number.

198 m

Landing Pad

7 Ex 2

A boy ﬂying his kite releases the entire length of his string which measures 27 m, so that the kite is 18 m above him as shown in the diagram. Find θ rounded to two decimal places.

27 m

18 m

A slide casts a shadow 5.66 m along the ground. The distance between the end of the shadow and the top of the slide is 7.84 m. Find θ rounded to two decimal places.

7.84 m

θ 5.66 m

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A ladder measuring 2.36 m in length is leaning against a wall. If the angle the ladder makes with the ground is y°, ﬁnd y rounded to two decimal places. 2.36 m

1.25 m

A wooden support which is 2.79 m long is preventing a fence from falling over. The angle that the wooden support makes with the ground is θ. a

Sketch a diagram of the situation.

Find θ rounded to two decimal places.

A suspension bridge is being built. The top of the concrete tower is 35.5 m above the bridge and the connection point for the cable is 65.9 m from the tower. Assume that the concrete tower and the bridge are perpendicular to each other.

cable

Find the length of the cable to two decimal places.

Find the angle the cable makes with the bridge. Round your answer to two decimal places.

35.5 m 65.9 m

During a particular time of the day, a tree casts a shadow of length 24 m. The height of the tree is estimated to be 7 m. a

Sketch a diagram of the situation.

Find the angle θ formed by the length of the shadow and the line from the end of the shadow to the top of the tree. Round your answer to two decimal places.

The air traffic controller is communicating with a plane in flight approaching an airport for landing. The plane is 10 369 m above the ground and is still 23 444 m from the runway. If θ° is the angle at which the plane should approach, find θ rounded to one decimal place.

10 369 m

θ° 23 444 m

Georgia is riding her bike up a hill that has an incline of 5°. If she rides her bike 817 m up the hill, ﬁnd d, the horizontal distance from where she started, rounded to the nearest metre.

817 m 5°

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A rocky ledge is 261 m high. From the top of the ledge, a boat is at an angle of 8° below horizontal. If the boat is d m from the foot of the cliff, ﬁnd d rounded to two decimal places.

8° 261 m d

Reasoning 16

16, 17

16–18

18, 19

Luca says that he can use the tangent ratio to ﬁnd angle θ. Is Luca correct? Explain your answer. 30 m

θ 23 m

A diagonal path through a rectangular garden is 10 m long. The path makes an angle of 26°48′ with/ the longer side of the garden. a

If the length of the shorter side of the garden is y m, which trigonometric ratio should we use to solve for y? Explain your answer.

Find y rounded to two decimal places.

A ladder of length 3.46 m needs to make an angle with the wall of between 10° and 20° for it to be safe to stand on. At smaller angles, the ladder is at risk of toppling backwards when the climber leans away from it. At larger angles, the ladder may lose its grip on the ground. The base of the ladder is placed at 190 cm, 42 cm, or 86 cm from the wall. At which of these base lengths is the ladder safest to climb? Explain how you found your answer.

At a seaside resort, two sailors, Vern and Raz, are having a debate. Vern argues that as their ship drifts away from the anchor, the angle the anchor line makes with the water surface decreases. However, Raz believes that the angle increases. The ship dropped anchor off the coast and the anchor fell 83 m to the sea bed. During the next 4 hours, the ship drifted 105 m. Who is correct and what is the angle x between the anchor line and the surface of the water, rounded to the nearest degree?

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Problem-solving 20–22 20

20, 22–24

A jet takes off and leaves the runway at an angle of 34°. It continues to ﬂy in this direction for 7 min at a speed of 630 km/h before levelling out. a

20–22

T = 7 min V = 630 km/h

Find the distance in metres covered by the jet just before it levels out.

T=0

h 34°

If the jet reaches a height of h m just before levelling out, calculate h to the nearest metre.

The angle from an observer’s horizontal line of sight to the top of a building is 65°. The observer is 20 metres from the base of the building, and the distance from the ground to her eyes is 1.6 m. a

Sketch a diagram of the situation.

Find the height of the building rounded to two decimal places.

Milad is on a beach, standing 46 metres from the base of a lighthouse. The angle from the ground to the top of the lighthouse is 54°. θ°

54° 85 m

46 m

Find the height of the lighthouse, rounded to one decimal place.

Milad’s picnic spot is 85 metres from the base of the lighthouse. What is the angle below horizontal from the top of the lighthouse to the picnic spot, θ°? Round your answer to two decimal places.

In the diagram shown, OQ is parallel to ST. Angle TPQ is 36°. Angle OQR is 25°. If QT = 15 metres and RS = 18 metres, ﬁnd the size of angle SPR. Round your answer to two decimal places.

Liam is standing at a distance from a tall monument. The angle below horizontal from the top of the monument to the ground at Liam’s feet is 48°. Liam then walks a further 40 metres away from the monument, and from there, the angle from the ground to the top of the monument is 28°.

R Q

O 18 m

15 m

48°

If the height of the monument is 40.8 metres, ﬁnd the original distance between Liam and the base of the monument, rounded to one decimal place.

28° 40 m

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8 Chapter review 1

For each triangle, ﬁnd the length of the hypotenuse. Give your answer as an exact value. a

11 7

9 5 c

For each triangle, ﬁnd the length of the unknown side as an exact value. a

12 m

g cm

14 cm

23 cm 18 m

Explain why Pythagoras’ theorem cannot be used to ﬁnd the value of c in this triangle.

63°

28°

Rosalind and Natalia leave their hotel room and walk 400 m east to get to the main street in the city. Then, they walk due south for 1.8 km. If d is the shortest distance between the pair’s current location and their hotel, ﬁnd the value of d in metres. Round your answer to the nearest metre.

Suella is building a vegetable garden according to the given diagram. Suella wants the fence to go around the entire edge of the garden. Find the amount of fence that she needs to purchase.

400 m

270 cm

190 cm 300 cm

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1.8 km

On one particular night, the light from a lighthouse can be seen up to 19.3 kilometres away. A boat travelling due north sails past the lighthouse. The boat is closest to the lighthouse when it is due east of it, as shown in the diagram. At that point, the boat is 16.1 kilometres from the lighthouse.

19.3 km

Find the distance in which the light will be visible from the boat, d. Round your answer to one decimal place.

19.3 km

16.1 km d km

Which trigonometric ratio relates the given sides and the angle θ? a b θ 17

θ 15

Write the ratios for each triangle: i

sin θ

cos θ

iii

tan θ

b 48

For each of the right-angled triangles: i

Find the value of y.

Write the ratio for tan θ.

A 34

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If sin θ =

1 2

Describe the type of triangle this is and explain why.

Give examples of three different triangles that would result in the ratio sin θ =

1 2

Use technology to evaluate and round to two decimal places: a

sin 40°

tan 79°

2.5 tan 51°

0.85 cos 3.15°

cos 14°

8 sin 58°

Find the value of the variable, rounded to two decimal places: a

25 m

s cm

61°

vm 16 cm 27°

40°

ym g mm

48 m

30°

26°

1567 m

h cm

Find θ rounded to the nearest degree: a e

452

f 45 cm

48°

28.5 mm

sin θ = 0.5

cos θ = 0.7071

Mathspace Victorian Curriculum – Year 9 mathspace.co

tan θ = 0.577

tan θ = 0.8391

cos θ = 0.2588

sin θ = 0.8192

Find the unknown angles, marked by a variable, rounded to the nearest degree: a

z°

j° 29

w°

m°

A line has a gradient of 3. Find the angle between the line and the positive direction of the x-axis. Round your answer to one decimal place.

Point Q (6, 7) lies on the given line passing through the origin: Find θ, the angle the line makes with the positive direction of the x-axis. Round your answer to two decimal places.

8 7

6 5 4 3 2 1

x 1

Trusses are used for bridges and roofs, and they can have various designs. One design, shown in the diagram, is called a Howe truss. a

Find θ.

Find x rounded to two decimal places.

2 3 4 5 6 7 3m 3m

8 9

3m 3m

Gianna is relaxing on the beach outside of her hotel. She looks upward at an angle of 25° and sees the balcony of her hotel room. If she is 92 m from the hotel, ﬁnd H, the vertical height of her hotel room, rounded to the nearest metre. Hm 25° 92 m

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Alena starts her hike 3 km away from a mountain’s base. The mountain stands 800 m tall. After hiking towards the mountain, she is only 2.5 km from the base. Ezekiel believes that as Alena moves closer, the angle between the ground and the peak increases, while Celia thinks the angle decreases. Who is correct? Justify your answer.

The top 6 stories of a building are 25.8 m in height. From a point 70 m from the base of the building, the angle from the ground to the point where the top 6 stories begin is 22°. What is the angle from the ground to the top of the building, θ ? Round your answer to one decimal place.

25.8 m

θ 22° 70 m

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Geometry has two great treasures: one is the theorem of Pythagoras; the other, the division of a line into extreme and mean ratio. The first we may compare to a measure of gold, the second we may name a precious jewel. Johannes Kepler

Big ideas Calculating the area, surface area, and volume of various shapes like polygons, prisms, and cylinders involves applying the right formulas, breaking down complex shapes, and ensuring accurate measurements.

Curriculum Outcomes VC2M9N01 recognise that the real number system includes the rational numbers and the irrational numbers, and solve problems involving real numbers with and without using digital tools VC2M9A06 use mathematical modelling to solve applied problems involving change, including financial contexts involving simple interest; formulate problems, choosing to use either linear or quadratic functions or other simple variations; interpret solutions in terms of the context; evaluate the model and report methods and findings VC2M9M01 solve problems involving the volume and surface area of right prisms, cylinders and composite objects using appropriate units VC2M9M02 solve problems involving very small and very large measurements, timescales and intervals expressed in scientific notation VC2M9M04 calculate and interpret absolute, relative and percentage errors in measurements

9 Measurement Chapter outline 9.01 9.02 9.03 9.04 9.05 9.06 9.07 9.08

Area Area of composite shapes Nets of prisms Surface area of prisms Volume of prisms Investigation: Volume versus area Accuracy of measurement Surface area of cylinders Volume of cylinders Chapter 9 review

458 465 475 482 490 500 503 513 521 530

Graffiti artists can create 50m of 10cm-wide lines with just one 400mL spray can! Can you figure out how thick the paint will be?

9.01 Area After this lesson, you will be able to… • identify and use appropriate units for measuring area. • calculate the area of common polygons using given formulas. • understand and apply the formula for the area of a circle, including special cases like semicircles and quarter circles. • interpret and solve real-world problems related to area calculation.

Area of polygons Area is the amount of space enclosed inside a two-dimensional shape. From smallest to largest, the common units used for measuring area are: • square millimetres (mm2) • square centimetres (cm2) • square metres (m2) • square kilometres (km2) A hectare is a special unit of area often used to measure the area of a piece of land. 1 ha = 10 000 m2. A summary of the area formulas for common polygons: Rectangle

Square

Parallelogram

Rhombus

Kite

Triangle

x y

h b

Trapezium

a h b

Example 1 Find the area of the triangle. 10 m

Create a strategy Use the formula for the area of a triangle: Area =

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1 × b× h 2

Apply the idea For this triangle the base is 8 m and the height is 10 m. 1 × b× h 2 1 = × 8 × 10 2

Area =

Write the formula Substitute the values Evaluate

= 40 m2

Idea summary A summary of the area formulas for common polygons: Rectangle Square Parallelogram Rhombus

w l

Kite

x y

Triangle

x y

Trapezium

Area of circles We can calculate the area of a circle using the formula:

A = π r2 A is the area r is the radius of the circle

When you are required to give the exact value of the area of a circle, you will leave your answer in terms of π. To find the area of a semicircle (half a circle) we find the total area, then find a half of that by dividing by 2. To find the area of a quarter of a circle we find the total area, then find a quarter of that by dividing by 4.

Example 2 Find the area of the circle as an exact value.

3 cm

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Create a strategy Use the formula for the area of a circle: Area = π × r2

Apply the idea Area = π × r2

Write the formula

= π × 32

Substitute r = 3 into the formula 2

= 9π cm

Evaluate, leaving your answer as an exact value

Reﬂect and check It can be helpful when using exact values to ﬁnd an approximate answer. A rough estimate for π is 3, so a rough estimate for this area would be 9 × 3 = 27 cm2.

Example 3 If the diameter of a circle is 21 cm, ﬁnd its area rounded to one decimal place.

Create a strategy Find the radius by halving the diameter. Use the area of a circle formula to ﬁnd the area, and round the answer.

Apply the idea r = 21 ÷ 2

Halve the diameter

= 10.5 cm

Evaluate

Area = π r

Write the formula for area of a circle 2

= π × 10.5

= 346.4 cm

Substitute the radius 2

Evaluate and round to one decimal place

Reﬂect and check Since the diameter is 21 cm, a square drawn around the circle would have an area of just over 20 × 20 = 400 cm2. The area of the circle should be less than this, and it is.

Idea summary We can calculate the area of a circle using the formula:

A = π r2 A is the area r is the radius of the circle

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9.01 Practice questions Understanding 1

Trailblazer 1–3

A circle has a diameter of 10 km. i

Is each area value correct or incorrect?

Is each area value exact or approximate?

10π km2

78.54 km2

314.16 km2

7 mm to centimetres

50 000 m2 to hectares

25π km2

Convert to the unit indicated: c

3 m to centimetres 2

0.27 km to square metres

4a–h, 5, 8ab, 10ac, 11ac, 4e–l, 5, 6, 8cd, 10b–d, 12 11c, 13ab

Fluency

Ex 1

Adventurer 1–3

For the given diagram, is each statement true or false? a We must ﬁnd the lengths of all the sides of the triangle before we can ﬁnd its area. b The area of the triangle is half of the area of the rectangle. c An appropriate unit would be m2. d The height of the triangle is 9 m. e For the rectangle, it must be l = 9 m and w = 6 m, not l = 6 m and w = 9 m.

Explorer 1–3

4g–n, 5–7, 8bd, 9, 10bd, 11bd, 13cd

Find the area of these shapes: a

c 29 mm

8 cm

9 cm

12 cm

10 mm

6m 27 cm 9m 11 cm 57 cm

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i 14 cm

5m 30 cm 6m

10 m

18 m

9 cm 18 cm

4 cm 23 mm

12 mm

17 cm

25 mm

6 mm

4 mm 16 mm

9 mm

13 mm

Complete the table for bases, heights and areas of parallelograms. Base 5 mm 15 cm

Height 2 mm 5 cm

7.5 m

Area 105 cm2 50 cm2 37.5 m2

A harbour has a pier in the shape of a trapezium with a perpendicular height of 8 m. One base of the pier has a length of 9 m and the other has a length of 5 m. Using a labelled diagram, find the area of the pier.

Buzz used some scrap paper to make a birthday card in the shape of a parallelogram. The base of the card is 18 cm long, and the perpendicular height is 16 cm. Using a labelled diagram, find the area of the card.

Find the area of these triangles with its given side lengths to one decimal place:

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Base length of 5 cm and a perpendicular height of 8 cm

Base length of 3.9 cm and a perpendicular height of 1.7 cm

Base length of 6 cm and a perpendicular height of 9 cm

Base length of 17.4 cm and a perpendicular height of 9.2 cm

The faces on a 4-sided die are all triangular. Each face has a base length of 13 mm and a perpendicular height of 20 mm. Find the area of one face of the die.

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Ex 2

Find the exact area of these circles. a

b 22 mm

13 cm

d 8.5 m

11 cm

11 Ex 3

Find the area of these circles to one decimal place. a

19 cm

24 mm

32 cm 14 m

The radius of a circular baking tray is 10 cm. Find its area, rounding to two decimal places.

Find the radius, rounded to two decimal places, of a circle that has an area of: a

16π cm2

Reasoning 14

289π mm2

14, 15

812.25π cm2

14–16

10π m2

16–18

Explain how the areas of the following shapes are related to the area of a rectangle: a

Triangle

Parallelogram

Kite

Rhombus

Will doubling each side of the rectangle also double the area of the rectangle? Justify your answer.

6m 9m

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Antonio states that if the radius of a circle is halved then the area will also be halved. Nadine believes that the area will be a quarter of the original area. Who is correct? Explain your answer using an example.

A circle has a circumference of 16π. Explain how to ﬁnd the area of the circle.

A trapezium is put together with an identical trapezium to make a parallelogram: a

Find an expression for the base of the entire parallelogram in terms of a and b.

Find an expression for the area of the entire parallelogram in terms of a, b and h.

Find an expression for the area of one trapezium in terms of a, b and h.

Problem-solving 19–21 19

20–25

20, 22–26

Consider a rectangular park that is 500 m long and 300 m wide: a

Calculate the area of the park in square metres.

Calculate the area of the park in hectares. (Note: 10 000 m2 = 1 ha)

300 m

500 m

Five farms around Australia were measured and their dimensions, in metres, were recorded in the table: a

Find the area of each farm in square metres.

Which farms have an area of more than 1 ha?

Which farms have an area of less than 1 ha?

Which farm has an area of exactly 1 ha?

What is the average area of the farms?

Farm Length (m) Width (m) Area (m2) 1 200 100 2 350 21 3 100 24 4 400 40 5 100 100

Katrina is digging a rectangular garden bed to plant some new hedging. a

If the garden bed is 0.5 m wide and 8 m long, calculate the total area of the garden bed.

Each hedge plant ﬁlls an area that is the equivalent of 0.8 square metres. How many hedge plants are needed to ﬁll Katrina’s garden bed?

Sharon has purchased a rectangular piece of fabric that is 12 m long and 7 m wide. Find the area of the largest triangular piece she can cut out from it.

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An area measuring 2016 cm2 is to be paved with identical tiles in the shape of parallelograms. Each tile measures 14 cm along the base, and has a perpendicular height of 6 cm. a

Determine the area that each tile covers.

Calculate the number of tiles needed to cover the whole area.

Mathspace Victorian Curriculum – Year 9 mathspace.co

6 cm 14 cm

Some car parks require the cars to park at an angle as shown.

4.4 m

The dimensions of the car park are as given, where each individual parking space has a perpendicular length of 4.9 m and a width of 4.4 m.

4.9 m

Determine the area needed to create an angled carpark suitable for 7 cars. 25

A circular lake has a diameter of 50 metres. There’s a small circular island in the lake with a diameter of 10 metres. Calculate the area of the lake water, not including the island.

The design shown is made using a large circle and two smaller circles of diameter 3 cm. Find the area of the shaded region correct to one decimal place.

3 cm

9.02 Area of composite shapes After this lesson, you will be able to… • identify the basic shapes within a composite shape. • calculate the area of basic shapes within a composite shape. • apply the addition and subtraction methods to calculate the area of composite shapes.

Area of composite shapes There are two main methods for ﬁnding the area of a composite shape: the addition method and the subtraction method. The addition method involves identifying the basic shapes, calculating the area of the basic shapes and adding them all together to ﬁnd the total area of the composite shape. This composite shape is made up of two semicircles and a trapezium.

The subtraction method involves identifying the basic shapes, calculating the area of the basic shapes and subtracting the area of the smaller shapes (that were cut out) from the area of the larger shape to ﬁnd the total area of the composite shape. This composite shape is made up of a trapezium and a semicircle. We could use either of these two methods, and sometimes we may need to use both. 9.02 Area of composite shapes mathspace.co

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Interactive exploration Discover this concept in action online

mathspace.co

Example 1 24 mm

Consider the composite shape. 32 mm

16 mm

18 mm 56 mm

a List the basic shapes that make up the composite shape.

Apply the idea The composite shape is made up of a parallelogram, a triangle, and a rectangle.

b Find the total area of the figure.

Create a strategy

24 mm

Add the area of the simple shapes that make up the composite shape.

32 mm 16 mm

18 mm 56 mm

Apply the idea The composite shape is made up of a parallelogram, a triangle, and a rectangle so we need to find the area of each and add them. Area of parallelogram = bh

Start with the area formula of a parallelogram

= 24 × 32

Substitute the base and height 2

= 768 mm

Evaluate Start with the area formula of a triangle

Substitute the base and height

Evaluate

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Area of rectangle = lw

Start with the area formula of a rectangle

= 56 × 18

Substitute the height and width 2

= 1008 mm

Total area = 768 + 256 + 1008 2

= 2032 mm

Evaluate Add the areas Evaluate

Reﬂect and check Find the area of the nearest rectangle that would encompass the whole shape. Area of rectangle = lw

Start with the area formula of a rectangle

= 56 × 50

Substitute the length and total width 2

= 2800 mm

Evaluate

Given the area of the nearest rectangle is 2800 mm2, it is expected that the actual shape is less than this as we can see a part cut out. The actual area of the shape is 2032 mm2 which is less, so it’s a reasonable answer.

Example 2 Consider the composite shape.

7 cm 6 cm

4 cm

a Which basic shapes make up this composite shape? 12 cm

One trapezium and one rectangle minus two semicircles.

One trapezium and one rectangle minus two quarter-circles.

One trapezium and one parallelogram minus two quarter-circles.

One trapezium and one rectangle.

Create a strategy We can draw some lines on the composite shape to break it up into its components.

Apply the idea By drawing lines we can see a trapezium at the top, and a rectangle at the bottom with two quarter-circles cut out from each side.

The correct answer is option B: One trapezium and one rectangle minus two quarter-circles.

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b Find the area of the composite shape. Round your answer to two decimal places.

Create a strategy We need to add the areas of the trapezium and the rectangle together, then take away the areas of the two quarter-circles.

Apply the idea Start with the area formula of a trapezium Substitute a, b and h Evaluate Rectangle area = lw

Start with the area formula

= 12 × 4

Substitute l, w 2

= 48 cm

Evaluate

The area of the 2 quarter-circles is the same as the area of 1 semicircle. Start with the area formula Substitute r Evaluate and round Composite area = 57 + 48 − 25.13 = 79.87 cm

Combine the areas

Evaluate

Reﬂect and check Find the area of the nearest rectangle that would encompass the whole shape. Area of rectangle = lw

Start with the area formula of a rectangle

= 12 × 10

Substitute the length and total width 2

= 120 cm

Evaluate 2

Given the area of the nearest rectangle is 120 cm , it is expected that the actual shape is less than this as we can see parts cut out. The actual area of the shape is 79.87 cm2 which is less, so it’s a reasonable answer.

Idea summary There are two main methods for ﬁnding the area of a composite shape: • The addition method involves breaking the composite shape up into simple shapes and adding the areas together. • The subtraction method involves ﬁnding the area of the larger shape and subtracting the smaller areas that were cut out.

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9.02 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

Are these statements true or false? a

The area of a composite shape can be found by dividing it into smaller, basic shapes, ﬁnding the area of each of these shapes, and then adding these areas together.

The area of a composite shape can be found by subtracting the larger shape from the smaller shape.

The area of a composite shape is always expressed in square units.

If a composite shape is made up of two identical rectangles, the total area can be found by calculating the area of one rectangle and then doubling it.

ame the two geometric shapes that make up this N composite shape.

Explain how to ﬁnd the total area.

6 cm

8 cm

10 cm

ame the two geometric shapes that make up this N composite shape.

Explain how to ﬁnd the total area. 12 cm 4 cm

Fluency 4

4, 5a–e, 6, 8bc, 9ab, 10a 5a–ce, 6, 8, 9a–c, 10

Find the area of rectangle A.

Find the area of rectangle B.

Find the total area of the given shape.

5c–f, 6, 7, 8b–d, 9–11

6m 13 m

B 3m

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5 Ex 1

Find the total area: a

2 cm

b 4 mm

8 cm 14 cm

9 mm 13 mm

5 cm

6 mm

8 cm 3 cm

19 mm

6 cm 11 cm

8 mm

13 mm

20 cm

21 mm

7.7

6 cm 19 cm 15 cm

15 8

6 cm 14.6

Find the shaded area: a

7 cm

b 9 cm

6 cm 6 cm 6 cm

16 cm

17 cm

d 6 cm

12 cm 4 cm

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3 cm 14 cm

For each of these composite shapes: i

What basic shapes make up the composite shape?

Find the area, rounded to two decimal places.

16 cm 7 cm 18 cm

16 cm

Find the area of the composite shapes. Round your answer to one decimal place. a

27 cm

18 cm 16 cm

5 cm

6 cm

9 Ex 2

Find the shaded area in the diagrams. Round your answer to one decimal place. a

6 cm 7 cm 10 cm

d 36

20 cm

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Find the shaded area, rounded to two decimal places: a

6 cm

18 cm

16 cm

Eddie designs a plot of land which contains his house and garden. a

Find the total area of the plot of land.

Find the area of the garden.

18 m 11 m

Garden 23 m House

Reasoning 12

13, 14

12–14

12, 14, 15

A rectangle with dimensions width, w, and length, 2w, has a square with side length s cut out of it. a

Write an expression for the area, A, of the composite shape.

Does the area change if the position of the square changes?

A piece of origami paper, originally in the shape of a parallelogram, is folded along its shortest diagonal. Explain how to ﬁnd the total area covered by the folded paper.

18 cm

8 cm

Joshua and Lila are calculating the area of a composite shape that consists of a rectangle and a semicircle. The rectangle is 10 cm long and 6 cm wide. The diameter of the semicircle is the same as the width of the rectangle. Joshua calculates the area of the rectangle and the semicircle separately and then adds them together. Here’s his solution:

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Lila argues that they can calculate the area of the entire shape as a whole, without separating the rectangle and the semicircle. She calculates the area as: (10 cm × 6 cm) + π × (3 cm)2 Which student’s approach is correct and why? 15

Ava and Max are calculating the area of a composite shape that consists of a square of side lengths 5 m and a quarter-circle.

Ava’s approach is to calculate the area of the square and the quarter-circle separately. Here’s her calculation:

However, Max argues that the radius of the quarter-circle should be the same as the side of the square, and calculates the area of the quarter-circle as:

Who is correct in this situation and why?

Problem-solving 16, 18, 19

16–19

17, 18, 20, 21

A square has a quarter of a circle attached to one of its sides. The square has side length 9 cm and the quarter of a circle has radius equal to the sides of the square. Find the total area, rounded to two decimal places.

You are tiling a kitchen ﬂoor, which has a rectangular shape measuring 4.5 m in length and 3 m in width. Along the edges of the room, there is a border of square tiles, each measuring 15 cm on a side. The remaining interior of the ﬂoor will be covered with square tiles measuring 30 cm on a side. How many of each type of tile are needed?

A rectangular park is 30 m wide and 50 m long, inside the park there is a playground area. The playground consists of a square sandbox with sides measuring 3.2 m and a circular swing set section with a radius of 3 m. a

Draw a possible map of the playground.

What is the area of park excluding the playground areas? Round your answer to two decimal places.

The council is going to fertilise the lawn in the park (the area excluding the playground), for each square metre of lawn they need 250 g of fertiliser. How many kilograms of fertiliser is needed? Round your answer to two decimal places.

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The hotel lobby room is to be tiled with slate tiles as shown. The cost of tiling the room is $48/m2.

5m 2m

Find the area to be tiled.

How much will it cost to tile the room?

HOTEL LOBBY

The landscaping plan includes a garden positioned at the top, bordered by a circular arc and the property boundary. 5m

Find the area of the lawn in square metres.

Find the area of the garden in square metres. Round your answer to two decimal places.

c d

If roll-on lawn costs $12 per square metre, how much will it cost to cover the lawn and the garden?

Garden Paved Area Lawn

Find the paved area in square metres. Round your answer to two decimal places.

Paving costs $100 per square metre. Find the cost to pave the paved area.

12 m

House 10 m

The pattern for a simple t-shirt features a semicircular neck hole, found in the front piece only. 20 cm 25 cm

15 cm

70 cm

10 cm 50 cm Front

474

Back

Find the area of the back of the t-shirt.

Find the area of the front of the t-shirt, rounded to one decimal place.

If the front and back pieces are both cut from two separate rectangular pieces of fabric with dimensions 90 cm by 80 cm, how much fabric is wasted? Round your answer to the nearest square centimetre.

The fabric has to be purchased from a roll that is 80 cm wide and costs $8 per metre, how much will the fabric cost for 100 t-shirts?

Mathspace Victorian Curriculum – Year 9 mathspace.co

9.03 Nets of prisms After this lesson, you will be able to… • define and identify a ‘net’ as a two-dimensional representation of a prism. • visualise how a net can be folded to form a prism. • recognise and describe the nets of different types of prisms such as cubes, rectangular prisms and triangular prisms.

Nets of prisms Interactive exploration Discover this concept in action online

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A prism is a three-dimensional ﬁgure that has a constant (uniform) cross-section. The cross-section is parallel to the two identical ends.

A net of a prism is a representation of the faces arranged in such a way that it can be folded to form the three-dimensional prism. A prism can have many different nets. Nets can be useful to ﬁnd the total surface area of a prism by ﬁnding the areas of all its faces.

A cube and its net

A cube and another possible net

A rectangular prism and its net

A triangular prism and its net

Later, we will use nets as a way to ﬁnd the surface area of a 3D solid. 9.03 Nets of prisms mathspace.co

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Example 1 Choose the net that folds to give the shape shown:

Create a strategy Choose the net that: • has matching numbers of the different basic shapes. • has the same number of total faces as the solid. • can be folded up along its edges to make the solid.

Apply the idea The answer is option D.

Example 2 Consider the different views of this rectangular prism.

Top View

Front View

a What is the front view? A

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Side View

Create a strategy Visualise the 2D shape we would see when looking directly at the front of the rectangular prism.

Apply the idea We can see an elongated rectangle from the front view, so the answer is option A.

b What is the side view? A

Create a strategy Imagine the 2D shape visible when directly facing the side of the rectangular prism.

Apply the idea We can see a small rectangle from the side view, so the answer is option B.

c What is the top view? A

Create a strategy Consider the 2D shape displayed when viewing the rectangular prism directly from the top.

Apply the idea We can see a large rectangle from the top view, so the answer is option C.

Idea summary A prism is a three-dimensional ﬁgure that has a constant (uniform) cross-section. A net of a prism is a representation of the faces arranged in such a way that it can be folded to form the three-dimensional prism.

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9.03 Practice questions Explorer 1–3

Understanding 1

Adventurer 1–3

Trailblazer 1–3

Are the statements true or false? a

The net of a prism is a two-dimensional representation of the three-dimensional prism.

A prism can have only one possible net.

Nets can be used to determine the surface area of a prism.

A net of a rectangular prism will always have two square faces.

How many faces does a rectangular prism have?

In the net of this shape, how many faces would have a circle cut out of them?

Fluency 4

4, 5, 7ab, 8ac, 9

4, 5, 7bc, 8bc, 9

Name these solids: a

Which of the following nets fold into a cube? A

Which net matches this rectangular prism? Justify your answer. A

4 6 4

6 7 6

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6 4 6 6

6 7

7 7 7

4 4 4 4

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4–6, 7cd, 8cd, 9

Ex 1

Name the solid formed by each of these nets: a

Draw a net for each solid: a

Match each triangular prism with its net. a

i 26 cm

20 cm 16 cm

26 cm

18 cm

24 cm

20 cm

16 cm

20 cm 26 cm

26 cm

24 cm

iii

30 cm 26 cm

20 cm

24 cm 28 cm

20 cm

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iv 20 cm 30 cm

26 cm 24 cm 10 cm 20 cm

20 cm

26 cm 28 cm 24 cm

10 cm

20 cm

vi 20 cm

30 cm

26 cm

18 cm

24 cm

vii 20 cm

26 cm 24 cm 20 cm

28 cm

24 cm

viii

18 cm 30 cm

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20 cm

Reasoning

10, 11

10–12

Imagine you have 2 equilateral triangles and 3 congruent rectangles; draw three variations of a net diagram using these shapes.

Explain how you could ﬁnd the surface area of a rectangular prism using the net of a rectangular prism.

Suppose you are given a net with six congruent square faces. What three-dimensional shape can you form from it, and how can you prove it?

Problem-solving 13 13

13, 14

Rose wants to wrap a gift in a box using a glossy gift wrapper that costs $0.35 per square centimetre. How much will Rose pay to wrap her gift?

17.1 cm2 27 cm2

22.8 cm2

27 cm2

22.8 cm2

17.1 cm2

U-Pack Moving Company sells two sizes of moving boxes. U-Pack claims to keep their prices lower than competitors by selling their boxes ﬂat, and having customers assemble the boxes themselves. Box 1 sells for $0.02 per square centimetre:

Box 2 sells for $0.03 per square centimetre: 72.5 cm2 319 cm2

440 cm2

319 cm2

210 cm2

440 cm2

581.875 cm2

399 cm2

581.875 cm2

210 cm2 399 cm2

72.5 cm2

How much will it cost for a customer to buy Box 1?

How much will it cost for a customer to buy Box 2?

Which box is cheaper?

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9.04 Surface area of prisms After this lesson, you will be able to… • calculate the surface area of common prisms. • solve problems involving the surface area of prisms.

Surface area of prisms The surface area of a prism is the sum of the areas of all the faces. To ﬁnd the surface area of a prism, we can determine the basic shapes we need to add together. The net of a prism will show all of the faces as 2D shapes.

Another way to calculate the surface area for rectangular prisms and cubes is to use the formula. For any faces that are the same, we can ﬁnd the area of one face and multiply it by how many times that face appears. A rectangular prism has surface area: SA = 2 × top + 2 × front + 2 × side = 2lw + 2wh + 2lh = 2 (lw + lh + wh) For length, l, height, h, and width, w. A cube has surface area: SA = 6 × Area of front = 6s2 where s is the side length of the cube. The surface area of a triangular prism can be found by ﬁnding the sum of the areas of the two triangular faces and the three rectangular faces. The units for surface area will be in square units like mm2, cm2, and m2.

Interactive exploration Discover this concept in action online

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Example 1 4 cm

Find the surface area of the triangular prism shown.

3 cm 5 cm

19 cm

Create a strategy Add the areas of these ﬁve faces, where the top and bottom face are identical: +

Apply the idea Start with the formula for the area of a triangle, multiplied by 2

Substitute the values of b and h

Evaluate

Area of left rectangle = 19 × 3 2

= 57 cm

Area of front rectangle = 19 × 5 2

= 95 cm

Area of back rectangle = 19 × 4 2

= 76 cm

Multiply the length and width Evaluate Multiply the length and width Evaluate Multiply the length and width Evaluate

Surface area = 12 + 57 + 95 + 76 Add the areas = 240 cm2 Evaluate

Example 2 Find the surface area of the trapezoidal prism shown. Give your answer to the nearest two decimal places.

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Create a strategy Add the areas of the 6 faces together where the front and back face are identical trapeziums.

Use Pythagoras’ theorem to ﬁnd the missing dimension of the sloped side of the trapezoidal prism.

10 cm 5 cm

3 cm

a 13 cm

Apply the idea We can ﬁnd the area of each face using the following formulas: Start with the formula for the area of a trapezium formula, multiplied by 2

Substitute the dimensions

Evaluate

Left rectangle = 5 × 3 Area of a rectangle = length × width = 15 cm2 Evaluate Top rectangle = 10 × 3 Area of a rectangle = length × width = 30 cm2

Evaluate

Bottom rectangle = 13 × 3 Area of a rectangle = length × width = 39 cm2

Evaluate

We can ﬁnd the missing dimension of the sloped side using Pythagoras’ theorem:

Pythagoras’ theorem

Substitute a and b Evaluate

Take the square root of both sides

Evaluate Right rectangle = 5.830 952 × 3 Area of a rectangle = length × width = 17.49 cm2 Evaluate and round to two decimal places Surface area = 115 + 15 + 30 + 39 + 17.49 2

= 216.49 cm

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Add the areas Evaluate

Reﬂect and check If we were to unfold the prism into its net, we would get two trapezium faces and a large rectangle through the middle. 10 cm c cm

5 cm 13 cm

3 cm

c cm

5 cm

10 cm

The large rectangle will have a width equal to the height of the prism, and a length that is the perimeter of the trapezium (5 + 13 + 10 + 5.83 = 38.83 cm). 10 cm 5.83 cm

5 cm 3 cm

13 cm

5 cm 38.83 cm

Then the surface area is: SA = 2 × Atrapezium + Arectangle = 2 × 57.5 + 38.83 × 3 = 216.49 cm2

Example 3 A birthday gift is placed inside the box shown. What is the minimum amount of wrapping paper needed to wrap this gift? (Assume the box is in the shape of a rectangular prism).

12 cm 14 cm

5 cm

Create a strategy In a rectangular prism there are 6 rectangular faces. We have two faces that are 5 cm × 14 cm, two faces that are 14 cm × 12 cm and two faces that are 5 cm × 12 cm.

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Apply the idea SA = 2 (lw + lh + wh)

Sum the areas of the faces

= 2(14 × 5 + 14 × 12 + 5 × 12) Substitute the values = 596 cm2

Evaluate

Reflect and check If we wanted to do it without the formula, we could find the area of the top, front, and side, multiply each by 2 to get the bottom, other side, and back. Then finally add them all up. Atop = 12 × 5

Use the area of a rectangle to solve the top area 2

= 60 cm Afront = 14 × 5

Evaluate Use the area of a rectangle to solve the front area

= 70 cm

Aside = 12 × 14

Evaluate Use the area of a rectangle to solve the side area

= 168 cm

Evaluate

SA = 2 × Atop + 2 × Afront + 2 × Aside

Adding up all of the areas

= 2 × 60 + 2 × 70 + 2 × 168

Substitute

= 120 + 140 + 336

Perform the multiplication

= 596 cm

Evaluate

This matches what we found using the formula.

Idea summary The surface area of a prism is the sum of the areas of each face.

9.04 Practice questions Understanding 1

486

Explorer 1–3

Adventurer 1–3

How many faces does each prism have? a

Rectangular prism

Triangular prism

Cube

Hexagonal prism

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Trailblazer 1–3

Are these statements true or false? a

Surface area can be found by ﬁnding the area of the base and multiplying it by the height.

The net of a prism can be used to help calculate its surface area.

A prism is made up of two congruent, parallel, polygonal bases that are connected by rectangular faces.

No other face of a prism has the same area as the base.

Which net matches this rectangular prism? 6 9 10

A 10

6 6 6 6

9 6 9 6

9 6 9 9 10 9

9 10 9

10 10 10

4, 5ace, 6ac, 7–10, 12a, 13a

Fluency 4

9 6

4, 5ace, 6bc, 7–10, 12, 13

5bdf, 6cd, 7–13

The cube has a side length of 6 cm: a

Draw the net.

Find the surface area of the cube. 6 cm 6 cm

Find the surface area: a

3 cm

12 cm

d 8 cm

12 m

3m 5m

3 cm 6 cm

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f 9 mm 8 mm 10 mm

18 m

6 Ex 1

Find the surface area: a

15 cm

b 7 cm

3 cm

20 cm 9 cm

18 cm 9 cm

12 cm

4 cm 3 cm 5 cm

12 cm

2 cm

19 cm 21 cm

8 cm

Find the surface area rounded to two decimal places when necessary. a

12 cm

Ex 2

9 cm 5 cm

11 cm 7 cm

4 cm

5 cm

3 cm 12 cm

17 cm

The ﬁgure is a trapezoidal prism. a

Find the value of d rounded to one decimal place.

Find its surface area.

d cm 2 cm

12 cm

6 cm 15 cm

9 Ex 3

A birthday gift is placed inside the box shown: Find the minimum amount of wrapping paper needed to wrap this gift.

6 cm 9 cm

20 cm

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Sally is building a storage chest in the shape of a rectangular prism. The chest will be 93 cm long, 80 cm deep, and 18 cm high. Find the surface area of the chest.

The swimming pool has the dimensions shown. The sides and base of the pool are to be tiled. Calculate the area to be tiled, rounding your answer to two decimal places.

2.7 m

Find the value of x and the surface area of each prism. a

8 cm

13 cm

x 16 cm

6 cm

13 m 1.2 m

22 cm 24 cm

Find the surface area rounded to two decimal places when necessary. a

9 cm

5 cm 6 cm

10 m

12 cm

Reasoning

14, 15

14–16

14–17

If a cube and a rectangular prism have the same volume, will they also have the same surface area? Justify your answer with an example.

John and Ryan are asked to determine the surface area of a rectangular prism with dimensions 8 units by 5 units by 3 units. Ryan argues that the surface area is 120 square units, while John believes it’s 158 square units. Who is correct and why?

A triangular prism and a rectangular prism have the same base area and the same height. Vlad suggests that they will have the same surface area. Do you agree? Why or why not?

Write an expression for the surface area of this trapezoidal prism. Justify your formula.

Problem-solving 18, 20, 21 18

18, 20–23

d a

c l

19, 20, 22–25

A cube has surface area of 1032 cm2. a

Find the area of one of the square faces.

Find the length of one side of the cube, rounded to two decimal places.

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Find the side length of a cube that has a surface area of 726 cm2.

A rectangular box has a surface area of 312 m2. If width and length of the box are 8 m and 2 m respectively, ﬁnd the height of the box.

Find the surface area of the object shown.

5 cm 9 cm 2 cm

12 cm 18 cm

A box that has a rectangular base 2b centimetres in length and 3b centimetres in width. The height of the box is b centimetres. Calculate the surface area of the box in terms of b.

A rectangular prism has a length of 4x metres, width of 3x metres and height of 2x metres.

Determine the surface area in terms of x.

If the dimensions of this prism are increased by a factor of 2, calculate the increase in the surface area in terms of x.

The diagram shows a high rise building. a

Find the surface area of the external walls of the building.

If 2% of the surface area is taken up by metal frames and the rest is glass, how many square metres of glass are used on the walls? Round your answer to one decimal place.

19 m

18 m

45 m

15 m

Dramona is creating a rectangular prism-shaped box with a volume of 360 cm3. The box must have a square base, and the height must be twice the length of one side of the base. Determine the surface area of the box in square centimetres, rounding your answer to one decimal place.

9.05 Volume of prisms After this lesson, you will be able to… • apply the volume formula to find the volume of various prisms. • find the unknown dimensions of prisms using the volume formula. • distinguish between units of volume and units of capacity. • apply formulas to convert between different units of volume and capacity. • solve real-world problems involving the volume of prisms.

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Volume of prisms Volume is the amount of space a three-dimensional object takes up. A prism is a three-dimensional object with two identically shaped ends joined by rectangular faces. For all prisms: Volume = area of the base × height of the prism V = Abase × H The ﬁrst step in the calculation is to ﬁnd the area of the base, which may be a basic shape or a composite shape. Here are some helpful formulas for area that we have used previously: Square

s w

Rectangle

l h b

Triangle

Parallelogram

b a h

Trapezium

Here are some volumes that we have seen before: Vrectangular prism = length × width × height V=l×w×h Height Width

Length

Vcube = side3 V = s3

s s

Vtriangular prism = Area of base × prism height Height

V = Area of triangle × height

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Example 1 Find the volume of:

A = 35 cm2

8 cm

Create a strategy Use the volume formula: V = Abase x H

Apply the idea V = Abase x H = 35 × 8

Write the volume formula Substitute the base area and height

= 280 cm

Evaluate

Example 2 Find the volume of this hexagonal prism 10 cm 12 cm 6 cm 5 cm

Create a strategy Find the area of the hexagonal face, then multiply by the height of the prism.

10 cm

6 cm

5 cm

Apply the idea To ﬁnd the area of the base, we need to cut the hexagonal base into two triangles and a rectangle, then ﬁnd the sum of the areas of each part.

6 cm 2.5 cm 5 cm

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The height of the triangle will be For the triangular parts: Write the formula for the area of a triangle

Substitute b = 6 and h = 2.5

Evaluate

For the rectangle: A=l×w

Write the formula for the area of a rectangle

=6×5

Substitute l = 6 and w = 5 2

= 30 cm

Evaluate

The area of the base is the sum of the parts: 7.5 cm2 + 7.5 cm2 + 30 cm2 = 45 cm2 V = Abase × H = 45 × 12

Write the volume formula Substitute the area and height

= 540 cm

Evaluate

Reﬂect and check We could also have found the volume by breaking the entire prism into a rectangular prism and two triangular prisms and then adding up the volumes.

Example 3 A prism has a volume of 630 cm3. If it has a base area of 70 cm2, ﬁnd the height of the prism.

Create a strategy Use the formula for the volume of a prism and solve for the unknown height, H.

Apply the idea V = Abase × H

Write the formula

Substitute the values of the volume and the base area

Divide both sides by 70

Evaluate

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Idea summary The volume of any prism is given by:

V = Abase × H

Abase is the area of the base of the prism H is the height of the prism We may need to divide the base of a prism into basic shapes to ﬁnd its area.

Units of volume and capacity Volume is the amount of space a solid takes up. Capacity is the amount that something can hold. Volume is measured using units such as: • cubic millimetres (mm3) • cubic centimetres (cm3) • cubic metres (m3 ) • cubic kilometres (km3) The units of capacity include megalitres (ML), kilolitres (kL), litres (L) and millilitres (mL). When we consider liquids, the relationship between capacity and volume can be expressed as: 1 ML = 1000 kL 1 kL = 1000 L = 1 m3 1 L = 1000 mL = 1000 cm3 1 mL = 1 cm3

Example 4 This water bottle has a volume of 700 cm3.

a What is its capacity in millilitres?

Create a strategy Use the conversion for 1 cm3 = 1 mL.

Apply the idea

494

Volume = 700 cm3

Given

Capacity = 700 mL

Change the units from cm3 to mL

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b What is its capacity in litres?

Create a strategy Remember that 1 L = 1000 mL.

Apply the idea Capacity = 700 ÷ 1000 = 0.7 L

Divide by 1000 to convert from mL to L Evaluate

Idea summary Volume is measured using units such as cubic millimetres (mm3), cubic centimetres (cm3), cubic metres (m3), and cubic kilometres (km3), The units of capacity include megalitres (ML), kilolitres (kL), litres (L) and millilitres (mL). 1 ML = 1000 kL 1 kL = 1000 L = 1 m3 1 L = 1000 mL = 1000 cm3 1 mL = 1 cm3

9.05 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

Identify the base shape and the heights for these prisms: a

2 cm 8 cm 4 cm

21 cm

d 5 cm

11 cm 10 cm

13 cm

3 cm

3 cm 16 cm

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Complete the table: Volume 1 cm3 50 cm3 ⬚ cm3 1 m3 1 m3 1 m3

Capacity ⬚ mL ⬚ mL 750 mL ⬚ kL ⬚L ⬚mL

Are these statements true or false? a

The volume of a prism is calculated by multiplying the area of the base by the height of the prism.

A rectangular prism with the same height and base area as a triangular prism will have a different volume.

Calculating the volume of a prism requires you to draw the net of the prism.

The units of volume are always cubed units, regardless of the type of prism. 4, 5, 6a, 7, 9–12

Fluency 4 Ex 1

4, 5c–f, 6a–c, 7–10, 12–14

4, 5c–f, 6b–d, 8–10, 12, 14, 15

Find the volume: a

Area = 21 cm2 6 cm

Area = 12 m2

13 m

Find the volume: a

50 cm

14 cm

4 cm 6 cm

9 cm 5 cm

7 mm 6 cm

4 mm 20 mm

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f 10 cm

7m 5m

6 Ex 2

6 cm

40 mm

Find the volume: a

b 12 mm

8 mm

5m 4m

8 mm

16 mm

26 mm

12 mm

8 cm

6 cm 8 cm 5 cm

3 cm 1 cm

3 cm

4 cm 5 cm

A garden bed is 10 m in length, 4 m in width and 40 cm in height. Find the volume of soil in m3 that will be needed to ﬁll up the garden bed.

A box is 1 m long, 30 cm high and 40 cm wide. What is the volume of the box in cubic centimetres?

A prism has a volume of 360 cm3. If it has a base area of 60 cm3, ﬁnd the height of the prism.

Ex 3

The volume of the triangular prism is 117 m3.

Find the value of y.

7.6 m

4m 6.5 m

A storage box needs to have a volume of 153 090 cm , a height of 63 cm and a width of 81 cm. Find the length of the box.

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12 Ex 4

A bottle of orange juice has a volume of 450 cm3. a

What is its capacity in millilitres?

What is its capacity in litres?

A container has the shape of a rectangular prism, with dimensions 50 cm, 80 cm, and 20 cm. Find its volume in litres.

A medical refrigerator used to store containers has dimensions 65 cm × 52 cm × 26 cm. The containers have dimensions 50 mm × 40 mm × 20 mm.

Find the capacity of one container in millilitres.

Find the volume of the fridge in cubic millimetres.

How many containers can fit in the refrigerator?

The rectangular prism has a capacity of 1224 kL.

Find the length, b, of the rectangular prism. bm 17 m

Reasoning

16, 17

16, 18, 20

16, 18–20

A triangular prism and an octagonal prism have the same base area but different heights. Is it possible for them to have the same volume?

Stuart wants a locker which will give him the most space for his books at school. Locker A is 40 cm long, 30 cm wide and 160 cm tall. Locker B is 45 cm long, 30 cm wide and 145 cm tall. Which locker should Stuart choose? Justify your answer.

Explain what will happen to the volume of a rectangular prism if the length, width, and height are doubled.

If two prisms have the same volume, does it mean they have the same base area or height? Justify your answer with an example.

A certain prism has a larger base area than another but a shorter height. Can it have a smaller volume? Justify your answer with an example.

Problem-solving 22–24

21–24, 27

23–29

A length of gutter pipe has the shape of a rectangular prism and has a capacity of 1245 mL. The base is 10 cm wide, and the length is 1 m. What is the depth of the pipe?

An aquarium is in the shape of a rectangular prism with dimensions 120 cm in length, 60 cm in width, and 75 cm in height. If the aquarium needs to be ﬁlled to 90% of its capacity, determine the amount of water needed in litres.

A rectangular prism has a base area of 28 cm2 and a height of 7 cm. If the dimensions of the base are both doubled, ﬁnd the volume of the rectangular prism.

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Calculate the volume of a single sheet of paper, given that a pile of 500 sheets of paper measures 230 mm wide, 298 mm long and 30 mm tall.

A blue postage box has a base area of 65 cm2 and a height of 13 cm. A red postage box has the same height but triple the base dimensions of the blue box. a

Find the volume of the red postage box.

How many of the blue postage boxes can ﬁt into the red postage box?

Find the dimensions of two rectangular prisms, A and B. Use each digit from 1 to 9 exactly once across both prisms to form the numbers for their dimensions. The volume of Prism A must be exactly double the volume of Prism B.

The roof of a house is ﬂat and measures 16 m by 15 m in the shape of a rectangle.

Find the area of the roof.

If a water tank collects all the water from the roof, how many litres of water would it collect if there had been 20 mm of rain?

How many millimetres of rain need to fall on the roof in total, to ﬁll a water tank with a capacity of 24 000 L?

A rectangular box has the dimensions x cm by x cm by y cm. The sum of the dimensions is equal to 80 cm. a

Write down the formula for y in terms of x.

What is the volume of the box in terms of x, in expanded form?

A garage with dimensions shown has a cement ﬂoor. a

The cement ﬂoor is 0.4 m thick. What volume of cement is needed for the ﬂoor?

If a ventilation fan circulates 4 m3 of air per minute, how long would it take to circulate all the air in the garage?

2.5 m 5m

2.2 m

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INVESTIGATION

Volume versus area Investigate how the changes in the dimensions of prisms inﬂuences their volume and surface area.

Objectives To understand how different prisms can have the same volume but different surface areas based on variations in their base, height, and number of sides To determine how altering one dimension of a rectangular prism impacts the volume and surface area

Maximise volume and surface area Explore volume and surface area We will use applets to organise our data into tables and look for patterns that arise.

Interactive investigation Discover this concept in action online

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Investigate

500

Compare your table observations with a classmate. Did you begin with similar prism sizes? Were your scale factor results alike?

What happens to a prism’s volume if you multiply a dimension by a whole number greater than one?

When you multiply a dimension by a unit fraction, how does the prism’s volume change?

For surface area, what happens when you increase a dimension using a scale factor bigger than one? Did this surprise you?

If the scale factor is a unit fraction, how does the surface area react?

When adjusting a prism’s dimension using a scale factor, do volume and surface area changes mirror each other?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Volume versus area Using a spreadsheet applet, we’ll eﬃciently calculate and compare these properties for various prisms. Our goal is to understand how different prisms can have the same volume but different surface areas. Before we start, familiarise yourself with the spreadsheet tool, as it will be our primary means of recording and analysing data.

This sample spreadsheet offers a starting point for organising your data. There are some relationships established between cells. You can click on the Volume and Surface area for the rectangular prism to see the cell relationships. Change the dimensions and see how the spreadsheet updates. Initialise • Open your preferred spreadsheet application. You can make a copy of the Google spreadsheet example. • Label columns for “Base shape”, “Number of sides”, “Base area”, “Height”, “Volume”, and “Surface area”. Formula setup • In the appropriate cells, set up formulas to calculate the volume (cell for base area × cell for height) and surface area (this may involve multiplying more cells together) based on the shape and dimensions you input. • For more complex prisms, you may need to use conditions or additional formulas to account for the number of sides and specific shape properties. Experimentation • Start by inputting the dimensions of a simple prism, like a rectangular prism, into the spreadsheet and verifying the calculations against manual computations. • Now, change the dimensions of the prism to find another prism with the same volume but a different surface area. For example, increasing the height while reducing the base area, or altering the number of sides if using a polygonal base. • Document your findings in the spreadsheet, capturing the differing dimensions of prisms that yield the same volume.

Investigate 1.

What is the impact of changing one dimension (like height) on both the volume and surface area for any of the prisms you experimented with?

Describe a scenario where two prisms can have drastically different surface areas but the same volume. What implications might this have in real-world applications, such as packaging or storage?

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Discussion 1.

In the ﬁrst activity, we changed one dimension of the prism using a scale factor. Predict what might happen if you were to change two or all three dimensions with the same scale factor. How would the volume and surface area be affected?

Consider other 3D shapes beyond prisms (like pyramids or cylinders). How might the relationship between volume and surface area play out for these shapes?

Discuss with a classmate: How did the spreadsheet tool enhance or limit your investigation? What other digital tools or strategies might further aid in such mathematical explorations?

Did you know?

Measurement is essential in cooking to ensure consistency and flavor. Precise measurements of ingredients, like flour, sugar, and spices, can make the difference between a perfectly baked cake and a recipe disaster. Professional chefs and home cooks alike rely on accurate measurements for reliable results in every dish.

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9.06 Accuracy of measurement After this lesson, you will be able to… • understand what is measurement error and its implications. • identify factors that contribute to measurement errors. • calculate absolute error and percentage error. • determine the upper and lower bounds for a given measurement. • use upper and lower bounds to calculate areas or volumes.

Absolute error We use different devices for measurement, including a ruler for length, a weighing scale for weight, a thermometer for temperature, and a clock for time. The accuracy of these devices is important in providing reliable results. Accuracy refers to how close a measured value is to the true or actual value. The quality and design of the measurement device can influence the accuracy but no measuring device is perfectly accurate; there will always be some degree of uncertainty in every measurement. This uncertainty can be due to factors such as the limitations of the device itself, environmental conditions, or even human error in reading the measurements. The absolute error of a measurement is half of the smallest unit on the measuring device. The smallest unit is called the precision of the device. Absolute error =

1 × precision 2

For example, the ruler below has a precision of 0.5 cm, because the smallest scale markings are 0.5 cm apart. The absolute error of any measurement made with the ruler will be 0.25 cm (i.e. half of 0.5 cm). 0

10 11 1 2 1 3 1 4 1 5 1 6 1 7 1 8

If we have taken a measurement, we can find the absolute error by finding the difference between the measured value and the true value. These lines ∣value∣ around a value mean we take the positive value only, ∣2∣ = 2 and ∣−2∣ = 2. Absolute error = ∣Measured value − True value∣ To compare the sizes of errors across different measurements and situations, it is more useful to use the percentage error, rather than the absolute error. The percentage error of a measurement is the absolute error expressed as a percentage of the measurement. The absolute error and the measurement must be in the same units before calculating the percentage error. Percentage error =

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Example 1 A dishwasher weighs exactly 33.1 kg, but was measured by Luke to be 32 kg. a What is the absolute error?

Create a strategy Find the distance between the true value and the measured value. We only need the difference between the values; we do not need to know whether the measured weight is above or below the true weight.

Apply the idea The measured value is 32 kg and the true value is 33.1 kg. Absolute error = 33.1 – 32 = 1.1 kg

Substitute the values Evaluate

Reﬂect and check If we wanted to use the formula, we would get: Absolute error = ∣Measured value − True value∣

Write the formula

= ∣32 − 33.1∣

Substitute the values

= ∣ − 1.1∣

Subtract

= 1.1 kg

Evaluate

b What is the percentage error, rounded to one decimal place?

Create a strategy Use the formula for percentage error.

Apply the idea Absolute error × 100% True value 1.1 = × 100% 33.1 = 3.3%

Percentage error =

Example 2 A set of scales weighs a parcel as 165 g. What is the absolute error of the measuring device?

Create a strategy Identify the unit of precision. 504

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Write the formula Substitute the values Evaluate using your calculator

Apply the idea The scales are using units of grams, which means the precision is 1 g. 1 × precision 2 1 1g Absolute error = × precision 2 = 0.5 g

Absolute error =

Write the formula Substitute precision = 1 g Evaluate

Idea summary The precision value is the smallest unit on the measuring device. • Absolute error of a measuring device 1 Absolute error = × precision 2 • Absolute error if you know the true value Absolute error = |Measured value − True value| The absolute error is the positive difference between the measured value and the true value. When calculating the percentage error, use the formula: Percentage error =

Absolute error × 100% True value

Upper and lower bounds The limits of accuracy give an interval where the true value will lie. The lower value of the interval is called the lower bound, and the higher value is the upper bound. lower bound

measured value absolute error

upper bound absolute error

the true value being measured lies somewhere within this interval

For example, if a student’s height is measured as 160 cm to the nearest centimetre, the unit of precision is 1 cm, so the absolute error is 0.5 cm. This makes the lower bound 159.5 cm and the upper bound 160.5 cm 180 160 140 120 100 80 60 40 20 0

1 cm

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If measurements with some error are used to calculate an area or volume, the lower bound will give the minimum area or volume. Similarly, the upper bound will give the maximum area or volume.

Example 3 Between which limits does the cost of a CD lie if it is known to be $50 when rounded to the nearest $5?

Create a strategy Calculate the absolute error and then add/subtract this from the estimated cost.

Apply the idea The price of the CD has been measured to the nearest $5.

Halve the unit of precision

Evaluate Upper bound = 50 + 2.5 = $52.50 Lower bound = 50 − 2.5 = $47.50

Add the absolute error to the measured value Evaluate Subtract the absolute error from the measured value Evaluate

The cost of the CD lies between $47.50 and $52.50.

Example 4 State the limits of accuracy for a distance measured to be 13.45 km.

Create a strategy Calculate the absolute error and then add/subtract this from the measured distance.

Apply the idea The distance has been measured to the nearest 0.01 kilometre. Halve the unit of precision

Evaluate

Upper bound = 13.45 + 0.005 = 13.455 km Lower bound = 13.45 − 0.005 = 13.445 km

Add the absolute error to the measured value Evaluate Subtract the absolute error from the measured value Evaluate

So the limits of accuracy for the distance are 13.445 km and 13.455 km.

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Example 5 The dimensions in the figure have been measured to the nearest centimetre.

10 cm 13 cm

a Calculate its volume.

13 cm

Create a strategy Use the formula: Volume = length × width × height

Apply the idea Length = 13 cm, width = 13 cm, and height = 10 cm. Volume = length × width × height

Write the formula

= 13 × 13 × 10

Substitute the values

= 1690 cm

Evaluate

b What is the smallest possible volume? Give your answer to three decimal places.

Create a strategy Find the lower bounds of the measurements, then use these to calculate the volume.

Apply the idea The lengths were measured to the nearest centimetre.

Halve the unit of precision

Evaluate Subtract the absolute error from each measured value to ﬁnd the lower bounds: Length = 12.5 cm, width = 12.5 cm, and height = 9.5 cm. Minimum volume = 12.5 × 12.5 × 9.5 Use the lower bounds to calculate the minimum volume = 1484.375 cm3

Evaluate using your calculator

c What is the maximum possible volume?

Create a strategy Find the upper bounds of the measurements, then use these to calculate the volume.

Apply the idea Add the absolute error found in (b) to each measured value to ﬁnd the upper bounds: Length = 13.5 cm, width = 13.5 cm, and height = 10.5 cm. Maximum volume = 13.5 × 13.5 × 10.5 = 1913.625 cm3

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d What is the maximum possible absolute error in the calculation?

Create a strategy Subtract the measured volume from the maximum possible volume.

Apply the idea Maximum possible error = 1913.625 – 1690 3

= 223.625 cm

Subtract the volumes Evaluate using your calculator

e What is the maximum possible percentage error in the calculation? Give your answer rounded to one decimal place.

Create a strategy Divide the maximum possible absolute error by the calculated volume and multiply by 100%.

Apply the idea We found the maximum possible error in part (d) and the measured volume in part (a), so calculate the percentage error: Substitute the values

Convert as a percentage

Reﬂect and check Notice that the percentage error increases as we go from lengths to volume. For 13 cm, 1D side length: Absolute error × 100% True value 0.5 = × 100% 13 = 3.85%

Percentage errorOne side =

For 13 cm × 13 cm, 2D square base: Absolute error × 100% True value Maximum area − True area = × 100% True area 13.5 × 13.5 − 13 × 13 = × 100% 13 × 13 = 7.84%

Percentage errorBase area =

For 13 cm × 13 cm × 10 cm, 3D volume: Percentage errorvolume = 13.2%

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Idea summary The limits of accuracy for a measurement are the upper and lower bounds. • Lower bound = measurement − absolute error • Upper bound = measurement + absolute error

9.06 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Measure each object to the nearest millimetre: a

b 0 10 20 30 40 50 60 70 mm

0 10 20 30 40 50 60 70 mm

d 0 10 20 30 40 50 60 70 mm

0 10 20 30 40 50 60 70 mm

Match each measuring tool with the unit of measure. a

Kitchen scales

Ex 1

Metres (m)

Ruler

Millilitres (mL)

Liquid measuring cup

iii

Grams (g)

Tape measure

Centimetres (cm)

Is each of the statements true or false? a

The true value lies within the lower bound and upper bound.

To ﬁnd the upper and lower bounds, start with the measured value then add and subtract the absolute error, respectively.

To calculate percentage error we divide the absolute error by the measured value, then multiply by 100%.

Fluency 4

Trailblazer 1–3

4, 5, 6a–d, 7–9, 11–14

6c–f, 7–8, 10, 12, 13, 15

6e–h, 8, 10, 14–17

A microwave weighs exactly 23.5 kg, but was measured to be 24.1 kg. a

Determine the absolute error of this measurement.

Find the percentage error, rounded to one decimal place.

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5 Ex 2

Find the absolute error: a

A measurement of 30 kg, which was measured to the nearest kilogram.

A measurement of 400 g, which was measured to the nearest gram.

The amount $700, which was calculated to the nearest $10.

Estimate the absolute error for these measurements: a

18.9 mL

269 mm

9.9 g

18 s

98.6 °F

15.12 kg

17.8 m

117.38 °C

90.03 cm

An object is measured at point P using a tape measure: P

40 cm

Find the smallest unit marked on the tape.

What measurement should be recorded for P?

What is the absolute error of measurement P?

100

110

Could these be the actual measurements of P? i

Ex 3, 4

89.02 cm

90.64 cm

iii

89 cm

The lower bound.

For each of the measurements, ﬁnd: i

The upper bound.

A distance measured to be 6.4 m

A height measured to be 5 km

A bag of sugar that weighs 14 kg to the nearest 10 grams

Puncak Jaya, Indonesia’s greatest mountain, standing 4884 m high rounded to the nearest metre

For the weight scale: a

Find the smallest unit labelled on the scale.

Find the absolute error.

If a certain object is measured at 70 kg, ﬁnd the lower and upper bounds of this measurement.

80 90

20 10

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0 kg

100 110

For the speedometer: a

What speed is indicated on the speedometer?

Find the absolute error of this speed reading.

Find the maximum speed limit needed to ensure the driver is not exceeding 60 km/h.

60 40

100

120 140 km/h 160

20 0

180 200

A chef measures the amount of olive oil to be used in a cup, marked in millilitres. Find the absolute error of any measurement she makes.

The capacity of a bottle is measured as 1.25 litres, correct to the nearest 10 millilitres. Calculate the percentage error for this measurement.

The length of a piece of rope is measured to be 19.99 m, using a ruler. Find the largest possible length of this rope.

The height of a tower is measured as 12.1 m. Find the shortest possible height of the tower.

What are the maximum and minimum possible numbers?

A town’s population is estimated to be 170 people, to the nearest 10 people.

A stack of paper is estimated to have 3000 sheets to the nearest 100 sheets.

The dimensions of the rectangle have been measured to the nearest 0.1 of a metre. a

Find the maximum possible area of the rectangle.

Find the minimum possible area of the rectangle.

6.1 m 15.3 m

17 Ex 5

The dimensions in the ﬁgure have been measured to the nearest centimetre. a

Calculate its volume.

Find the smallest possible volume. Give your answer to three decimal places.

Find the maximum possible volume.

Find the maximum possible absolute error in the calculation.

Find the maximum possible percentage error in the calculation. Give your answer, rounded to one decimal place.

Reasoning 18

11 cm 9 cm

12 cm

18, 19

The number of people at a concert is approximately 5050. Explain why we cannot ﬁnd the absolute error of this approximation.

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Aine is painting a circle for a class assignment. The circle has a radius of 2 m. a

Calculate the area of the circle using the exact value of π, rounded to two decimal places.

One of her friends argues that using 3.14 for π will make the calculations easier. Calculate the area of the circle using 3.14, rounded to two decimal places.

Compare the areas calculated with the exact value of π, and the approximation. What is the difference between the two calculations, rounded to two decimal places?

Discuss the potential implications of using rounded or approximated values in these realworld situations.

Problem-solving 20, 21 20

20–24

c = 4 and d = 8 have both been rounded to the nearest whole number. a

Calculate the lower bound of the value of c × d.

Calculate the upper bound of the value of

To make paper easier to ﬁle, two holes of diameter 6 ± 0.5 mm are made using a hole punch. The centres of the two holes are 80 ± 0.5 mm apart in the diagram: A

20–22

80 ± 0.5 mm

Find the smallest unit of measurement of the holes.

Find the closest possible distance between points A and B.

A ﬁeld has dimensions 15.4 m × 17.6 m, to the nearest 10 cm. a

Find the upper and lower bounds of the area of the ﬁeld.

Find the upper and lower bounds of the perimeter of the ﬁeld.

The dimensions of a room are given as 3.6 m by 8.6 m, to the nearest 10 cm. Find the area of carpet that should be purchased to make sure it covers the whole room.

A school ﬁeld has a 100 m track that is measured to the nearest metre. The record for the fastest 100 m sprint is 13.9 seconds rounded to the nearest 0.1 second.

512

Find the fastest speed, in metres per second, the record holder could have run. Round your answer to two decimal places.

Find the slowest speed, in metres per second, the record holder could have run. Round your answer to two decimal places.

Mathspace Victorian Curriculum – Year 9 mathspace.co

9.07 Surface area of cylinders After this lesson, you will be able to… • calculate the surface area of cylinders. • solve problems involving the surface area of cylinders.

Surface area of a cylinder A cylinder has three faces: two identical circular bases and a curved surface that joins the two bases together. The surface area of a cylinder is the sum of the areas of these three faces.

The area of a circle is A = π r2. Now we need to ﬁnd the area of the curved face that connects the two circles.

Interactive exploration Discover this concept in action online

mathspace.co

When unrolled the curved surface is a rectangle. One side length is equal to the height of the cylinder and the other side length is the circumference of the base circle. The area of the curved part of a cylinder is 2π rh, where r is the radius and h is the height. 2π r

2π r h

To ﬁnd the surface area of the whole cylinder, we need to add the area of the top and bottom circles to the area of the curved part. Surface area of a cylinder = 2 × (Area of circular base) + area of curved face = 2 × (π × r2) + 2π r × h

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Example 1 Consider the following cylinder. 7m 4m 4m 7m

Find the area of the curved part of the cylinder, rounded to two decimal places.

Create a strategy Use the curved part area formula of the cylinder: Curved part area = 2π rh

Apply the idea Curved part area = 2π rh

Write the formula

= 2π × 4 × 7 2

≈ 175.93 m

Substitute the values of r and h Evaluate and round

Example 2 Consider the cylinder shown in the diagram.

5 cm

6 cm

a Find the surface area of the cylinder in square centimetres. Round your answer to one decimal place.

Create a strategy Find the radius then use the formula for the surface area of a cylinder.

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Apply the idea d r= d 2 r=d 2 r=6 d r=6 2 2 =6 2 = 3cm 6 = = 3cm 2 2 = 2π r 2 + 2π rh Surface area = =2 3cm Surface area = 2 π r +22π rh 3cm π r×2 3+ 2+π2rhπ × 3 × 5 Surface area = =2 2π = 2π r×2 3+222+π22rhπ × 3 × 5 Surface area = = 150.8 2π × 3cm 2 +22π × 3 × 5 = 150.8 2π × 3cm + 2π × 3 × 5 = 150.8 cm22 = 150.8 cm

Write the formula of the radius in terms of diameter (d) Substitute the value of d Evaluate Write the surface area formula of the cylinder Substitute the values of r and h Evaluate and round

Reﬂect and check If we had used an approximation for π, like 3.14, then our answer would be slightly different, SA = 150.7 cm2.

b Use your answer from part (a) to ﬁnd the surface area of the cylinder in square millimetres.

Create a strategy Use the fact that 1 cm2 = 100 mm2.

Apply the idea Surface area = 150.8 × 100 mm2 = 15 080 mm2

Convert cm2 to m2 Evaluate

Reﬂect and check If we used the exact values and had not rounded in part (a), then our answer would be 4800π ≈ 15 079.6 mm2. In this case, they would be the same when rounded to the nearest square millimetre.

Example 3 The area of the circular face on a cylinder is 8281π m2. The total surface area of the cylinder is 25 662π m2. a Find the radius of the cylinder, r.

Create a strategy Use the formula for the area of the circle: A = π r2, and equate the area in the formula with the given area.

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Apply the idea π r2 = A

Write the formula Substitute the value of A Divide both sides by π Take the square root of both sides Evaluate

Reﬂect and check We only take the positive square root as r is a length, so must be positive.

b Find the height h of the cylinder.

Create a strategy Use the formula for the surface area of the cylinder: SAcylinder = 2π r2 + 2π rh

Apply the idea Surface area = 2π r2 + 2π rh

Write the formula Substitute the values of r and the surface area Simplify each term Subtract 16 562π from both sides Evaluate the subtraction Divide both sides by 182π Evaluate

The cylinder has a radius of 91 m and a height of 50 m.

Idea summary

Surface area of a cylinder = 2π r2 + 2π rh r is the radius h is the height of the cylinder

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9.07 Practice questions Explorer 1–3

Understanding 1

Adventurer 1–3

Trailblazer 1–3

In terms of π and x, ﬁnd the: a

Circumference of the circle

Area of the circle

Is each statement true or false? a

The surface area of a cylinder includes the area of both bases and the curved surface.

If you double the radius of a cylinder, the surface area will also double.

A cylinder that is taller always has a larger surface area than a shorter one.

For this cylinder:

4 cm

Does this net match the cylinder?

To ﬁnd the surface area of the cylinder, is the radius 8 cm and the height 4 cm?

8 cm

4 cm

8 cm

Fluency 4 Ex 1

4, 5, 6ab, 7, 8a–c, 9

4, 5, 6bc, 7, 8c–e, 9

4–7, 8d–f, 9–11

For each pair of cylinders and their nets, ﬁnd the curved surface area of the cylinder. Round your answers to two decimal places. a

b 3m

9m 5m

7m 3m

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Find the area of the curved part of a cylinder with a radius of 9.2 m and a height of 15.1 m. Round your answer to one decimal place.

For each cylinder: i

Find the curved surface area of the cylinder, rounded to two decimal places.

Find the total surface area of the cylinder, rounded to two decimal places.

2 cm

3m 6 cm

4m 8m 9m

Find the exact surface area of a cylinder with diameter 6 cm and height 21 cm by leaving your answer in terms of π.

Find the surface area of each cylinder, rounded to two decimal places.

Ex 2

5 cm 7m

13 cm

23 cm

5 cm 87 cm 10 cm

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800 mm

200 mm

53 cm

55 cm

9 Ex 3

The area of the circular face on a cylinder is 225π m2. The total surface area of the cylinder is 3210π m2. a

Find the radius of the cylinder.

Find the height of the cylinder.

This cylinder has a surface area of 16 173 m2:

26 m

Find the height of the cylinder. Round your answer to the nearest whole number. hm

This cylinder has a surface area of 54 425 cm2:

71 cm

Find the height of the cylinder. Round your answer to the nearest whole number. h cm

Reasoning

12, 13

12–14

Matea suggests that if the height of a cylinder is doubled then the surface area of the cylinder will be doubled. Is he correct? Justify your response with an example.

Consider a cylinder with a ﬁxed surface area. If you increase the radius of this cylinder, what will happen to its height? Explain your reasoning.

If you have a toilet paper roll (a cylinder) and you want to increase the total amount of paper on the roll, will this affect the surface area of the roll and paper combined? Justify your answer.

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Problem-solving 15–20

15–22

15, 18, 20–25

A cylindrical can of radius 7 cm and height 10 cm is open at one end. Find the external surface area of the can. Round your answer to two decimal places.

Paul is using a toilet paper roll for crafts. He has measured the toilet paper roll to have a diameter 4 cm and a length 10 cm.

4 cm

Find the external surface area of the toilet paper roll. Round your answer to two decimal places. 17

10 cm

A paint roller is cylindrical in shape. It has a diameter of 6.2 cm and a width of 38.1 cm. Find the area painted by the roller when it makes one revolution. Round your answer to two decimal places.

The diagram shows a water trough in the shape of a half cylinder. Find the exact surface area of the outside of this water trough, leaving your answer in terms of π.

3.15 m

1.21 m

Amy and Vincent each have a cylinder. Amy’s cylinder has a diameter of 8 cm and a height of 9 cm. Vincent’s cylinder has a diameter of 9 cm and a height of 8 cm. a

Find the surface area of Amy’s cylinder, rounded to two decimal places.

Find the surface area of Vincent’s cylinder, rounded to two decimal places.

Which cylinder has a larger surface area and by how much?

If a spherical ball with a radius of 4.9 m ﬁts exactly inside a cylinder, ﬁnd the surface area of the cylinder rounded to one decimal place.

The two identical spherical balls with radii of 1.8 m ﬁt exactly inside a cylinder. Find the surface area of the closed cylinder rounded to one decimal place.

Find the surface area of the ﬁgure, rounded to two decimal places.

3 cm 5 cm 5 cm

12 cm

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Find the surface area of the brickwork for this silo rounded to two decimal places. Assume that there is a brick roof and no ﬂoor.

3.16 m

5.42 m

A beekeeper has a unique beehive made up of two complete cylinders of beeswax. Write down an expression for the total surface area of the beehive. Use h to represent the height of the smaller cylinder. Leave your answer in terms of π.

If the total surface area is 336π cm2, determine the radius of the base of the beehive, rounded to two decimal places, given that h = 8 cm.

At a luxury hotel’s lobby, there’s an ornate chocolate fountain made from a cylindrical block of solid chocolate. This block has been cut in half vertically to allow for a cascade of ﬂowing chocolate. If the original chocolate cylinder had a height of 2.3 m and a diameter of 150 mm, how much chocolate surface is exposed after the cut? Answer in m2 and round your answer to two decimal places.

9.08

Volume of cylinders

After this lesson, you will be able to… • calculate the volume of a cylinder using its base area and height. • calculate the volume of fractions of cylinders. • calculate the volume of a cylinder with a hollow section, such as a pipe. • solve problems involving the volume of cylinders.

Volume of cylinders Interactive exploration Discover this concept in action online

mathspace.co

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A cylinder is very similar to a prism with a circular base. The volume of a cylinder can be found in a similar way: Volume = Area of base × Height Height

Remember the area of a circle is equal to π r2. So, the volume of the cylinder can be written as: Volume = Area of base × Height of prism

Base

= π r2 × h

Radius

= π r2 h where r is the radius.

Example 1 Find the volume of a cylinder rounded to one decimal place if its radius is 5 cm and its height is 13 cm.

Create a strategy Use the formula of the volume of a cylinder.

Apply the idea We have been given values for r = 5 and h = 13 into the formula. V = π r2 h

Use the formula 2

= (π × 5 ) × 13

Substitute the values

= π × 25 × 13

Evaluate the squares

= 1021.0 cm

Evaluate

Reﬂect and check As an exact value, this would be 325π cm3.

Idea summary The volume of the cylinder is:

V = π r2 h r is the radius of the cylinder h is the height of the cylinder

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Volumes formed from part of a cylinder For prisms and cylinders, we have seen that:

V = Abase × h V Abase h

volume of the solid area of the end face height of the solid

When we are asked to find the volume of half or a quarter of a cylinder, we multiply the full volume by the fraction we are trying to find. To find the volume of a tube, we can find the area of the end face (base) by subtracting the smaller circle from the larger one. We can then use the area of the base to find the volume of the solid.

Example 2 Calculate the volume of the solid. Assume that the solid is a quarter of a cylinder. Round your answer to one decimal place.

14 cm 5 cm

Create a strategy Multiply the volume of a cylinder formula by a quarter.

Apply the idea We are given r = 5 and h = 14. Multiply the formula by

Substitute the values

Evaluate

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Example 3 A 13 cm concrete cylindrical pipe has an outer radius of 6 cm and an inner radius of 4 cm as shown. Find the volume of concrete required to make the pipe, rounded to two decimal places.

13 cm 4 cm 6 cm

Create a strategy Use the formula of the volume of a cylinder.

Apply the idea To ﬁnd the base area, subtract the area of the inner circle from the outer circle. Abase = Aouter circle − Ainner circle 2

Find the area of the pipe face

=π×6 −π×4

Area of circle is A = π r2

= 36π − 16π

Evaluate the squares

= 20π cm

Evaluate as an exact value

V = Abase × h

Use the formula

= 20π × 13

Substitute the values 3

= 816.81 cm

Evaluate 3

The volume of concrete required is 816.81 cm .

Reﬂect and check We can identify the volume in one set of workings: V = Abase × h Use the formula = (π × (6)2 − π × (4)2) × 13 cm3 3

= (π × 36 − π × 16) × 13 cm

Substitute the values Evaluate the squares

= 816.81 cm Evaluate This can help with accuracy, because if we had rounded the area of the base to 20π = 62.83 cm2, then our answer would be V = 62.83 × 13 = 816.79 cm3. Being off by 0.02 sometimes does not make a difference, but if we make that rounding error for 1 000 000 pipes, then it can add up to be signiﬁcant.

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Idea summary When asked to calculate the volume of part of a cylinder, multiply the total volume by the portion of the fraction required. To find the volume of a tube, we can find the area of the end face (base) by subtracting the smaller circle from the larger one. We can then use the area of the base to find the volume of the solid.

9.08 Practice questions Understanding 1

Explorer 1–3, 5

Adventurer 1–3, 4a, 5

Trailblazer 1–5

What is the shape of the base of the solid?

6 cm

10 cm

What fraction of the cylinder is shown? a

19 cm 14 cm 8 cm 5 cm

Calculate the area of each circle, rounded to two decimal places. a

8 cm

14 cm

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Find the shaded area as an exact value. a

32 mm

12 mm 10 mm

20 mm

Is each statement true or false? a

The volume of a cylinder is found by multiplying the area of the base by the height.

If two cylinders have the same height but different radii, they will have the same volume.

If two cylinders have the same volume, they must have the same height and radius.

If the radius of a cylinder is halved, the volume of the cylinder will be reduced by half. 6ab, 7ab, 8a–c, 9ab, 10ab, 11ab

Fluency 6 Ex 1

6bc, 7bc, 8c–e, 9bc, 10bc, 11bc

6cd, 7cd, 8d–f, 9cd, 10bd, 11bd, 12

Find the volume of each of the cylinders. Round your answers to one decimal place. a

A cylinder with a radius of 3 cm and height of 16 cm.

A cylinder with a radius of 9 m and height of 22 m.

A cylinder with a diameter of 17 cm and height of 25 cm.

A cylinder with a diameter of 2.8 cm and height of 31 mm.

Find the volume of each of the cylinders as an exact value. a

A cylinder with a radius of 5 mm and height of 14 mm.

A cylinder with a radius of 8 cm and height of 14 cm.

A cylinder with a diameter of 38 km and height of 10 km.

A cylinder with a diameter of 32 m and height of 120 cm.

Find the volume of each of the cylinders. Round your answers to one decimal place. a

5 cm

10 cm

15 cm 35 cm

d 20 cm 4m

8m 526

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55 cm

34 cm

f 9.8 cm 80 mm 26 cm

Find the volume of each of the cylinders as an exact value. a

8 mm

17 cm

22 mm 20 cm

37 km

26 m 21 km 17 m

10 Ex 2

Find the volume of each solid, rounding your answers to one decimal place. a

8 mm 21 mm 14 cm

5 cm

3 cm

12 m

18 mm

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11 Ex 3

Find the volume of each of the concrete cylindrical pipes. Round your answers to two decimal places. a

b 7 cm 5 cm 11 cm

15 m 5m 8m

d 5 cm 3 cm 1.4 m

13 m 3m 7m

Find the volume of the ﬁgure shown, as an exact value.

3 mm

Reasoning

13, 15

13–15

12 mm

13–15

There are two types of cylindrical juice cans available for Lyn to purchase at her local store. One has a diameter of 16 cm and a height of 18 cm, and the other has a twice the diameter and height of the smaller can. Lyn believes the volume will double, is she correct? Justify your answer.

A cylindrical cup has the dimensions as shown:

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Calculate the volume of the cup in cm3, rounded to two decimal places.

Convert your answer from part (a) to the nearest mm3.

Calculate the exact volume of the cup in cm3.

Convert your answer from part (c) to mm3, then round to the nearest mm3.

Explain why part (d) is more accurate than part (b)? Does this matter?

Mathspace Victorian Curriculum – Year 9 mathspace.co

6 cm

10 cm

Bob has a choice between two types of cylindrical soup cans at his local store. The ﬁrst type has a diameter of 12 cm and a height of 14 cm, and the second type has a diameter of 14 cm and a height of 12 cm. Assuming the price is the same for both cans, which one should Bob buy to get the maximum amount of soup? Justify your answer using the formula for the volume of a cylinder.

Problem-solving 16–18

16–19

18, 20, 21

A cylindrical lip gloss container has a volume of 35 cm3 and the radius of 2 cm. Using the rule for volume of a cylinder, ﬁnd the height of the cylindrical lip gloss container, rounding your answer to one decimal place.

A cylindrical spa has a radius of 1.8 m and the water is 160 cm deep. Find the volume of water in the spa in litres. Round your answer to two decimal places.

A cylindrical candle has a radius of 46 mm and a height of 0.13 m. Find the exact volume of the candle in cubic centimetres.

A dumbbell is made of three cylindrical sections. The outer sections both have a diameter of 12 cm, and the inside section has a diameter of 4 cm.

12 cm

If each section is 10 cm long, calculate the total volume of the dumbbell. Round your answer to two decimal places. 20

4 cm

10 cm 10 cm 10 cm

A solid is formed by extruding a sector as shown in the diagram. The sector has a radius of 6 cm and is extruded to a height of 12 cm. Find the volume of the solid, leaving your answer in terms of π. A

60° 6 cm

12 cm

A hollow cylindrical pipe has the dimensions shown: a

Calculate the volume of the pipe, rounded to two decimal places.

Calculate the cost of the pipe if 1 cm3 of metal weighs 5.7 g, and costs $0.50/ g.

15 cm

2.4 cm

1.8 cm

Round your answer to the nearest cent.

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9 Chapter review 1

Find the area of these shapes: a

28 mm 5 mm

25 mm

15 mm

12 mm

Find the exact area of the circle.

Find the area to two decimal places. 7m

Find the radius of a circle that has an area of 361π mm2

Jake is setting up a rectangular playground for children.

530

If the playground is 22.5 m wide and 8 m long, calculate the total area of the playground.

Each piece of play equipment requires an area that is the equivalent of 7.5 square metres. How many pieces of playground equipment can Jake ﬁt in the playground?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Find the total area: a

7 mm

5 mm 21 mm 10 mm 15 mm

9 mm

14 mm

23 mm

Find the shaded area: a

9 cm

b 4 cm

16 cm

8 cm

18 cm 6 cm

Find the shaded area, rounded to two decimal places: a

9 cm 24 cm 18 cm

A pool has a rectangular shape measuring 9 m in length and 6 m in width. Around the outside of the pool, there is a walkway made of square stones with side lengths of 60 cm. How many square stones are needed? Justify your answer.

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Name the solid formed by each net: a

Draw a net for each solid: a

Match each rectangular prism with its net. a

6 cm 3 cm 8 cm

8 cm 6 cm 10 cm

4 cm 3 cm

8 cm

6 cm

6 cm 3 cm

iii

3 cm

6 cm

10 cm

4 cm

6 cm 8 cm

Find the surface area rounded to two decimal places when necessary. a

10 cm

21 cm

3 cm 9 cm 8 cm

532

4 cm

Mathspace Victorian Curriculum – Year 9 mathspace.co

14 cm

12 cm

8 cm

5 cm 6m

For the triangular prism shown: a

Find the value of x.

Find the surface area.

17 m

28 m 30 m

A storage container is shaped as a rectangular prism and has a length of 3x metres, width of x metres and height of 4x metres. If the dimensions of this container are doubled, determine the increase in the surface area in terms of x. Justify your answer.

Find the volume: a

5 cm

b 4m

12 cm 8 cm

8m 7m

8 mm

4 mm 11 mm

3 mm

14 mm 10 mm 10 mm

20 mm

30 mm

14 mm

A pentagonal prism has a volume of 175 cm3. If it has a base area of 25 cm2, ﬁnd the height of the prism. Area = 25 cm2 h cm

A rectangular box has dimensions 45 mm × 35 mm × 25 mm. Find the capacity of one box in millilitres.

A water storage tank is in the shape of a rectangular prism with dimensions 130 cm in length, 65 cm in width, and 80 cm in height. If the tank needs to be ﬁlled to 85% of its capacity, determine the amount of water needed in litres.

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Estimate the absolute error for these measurements: a

23.5 mL

21 s

14.75 kg

112.5°C

The dimensions in the ﬁgure have been measured to the nearest centimetre. a

Calculate its volume.

Find the smallest possible volume. Give your answer to three decimal places.

8 cm 5 cm

7 cm

Find the maximum possible volume.

Find the maximum possible absolute error in the calculation.

Find the maximum possible percentage error in the calculation. Give your answer, rounded to one decimal place.

An athlete practices his long jumps at a beach. He jumps a distance of 7.⬚ m measured to the nearest tenth of a metre. a

If the maximum possible percentage error is 0.685%, ﬁnd the measured distance.

Find the maximum and minimum possible distances of the jump.

Calculate the surface area of each cylinder, rounded to two decimal places. a

b 15 cm

8 mm

40 cm

12 mm

The area of one circular face of a cylinder is 144π m2. The total surface area of the cylinder is 1296π m2. a

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Find the radius of the cylinder.

Find the height of the cylinder.

A cake has two tiers. The top layer has a radius of 7 cm, and the bottom layer has a radius of 10 cm. Each layer is 6 cm high. a

Find the amount of light green icing needed to cover the top tier of the cake, rounded to two decimal places.

Find the amount of dark green icing needed to cover the bottom layer of the cake, rounded to two decimal places.

Mathspace Victorian Curriculum – Year 9 mathspace.co

7 cm 6 cm 6 cm 10 cm

Calculate the volume of each cylinder. Round your answers to one decimal place. a

7 mm

20 m

18 m

15 mm

Calculate the volume of each solid rounded to one decimal place. a

25 mm

10 mm 14 cm 4 cm

Find the volume of the concrete cylindrical pipe. Round your answer to two decimal places.

10 m

A cylindrical can has a volume of 120.1 cm3 and a height of 8 cm. Find the radius of the cylindrical can rounded to two decimal places.

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Big ideas • Probability helps us make predictions about future events by analyzing past or modeled data, and the rules can change depending on the type of experiment. • Tools like Venn diagrams, arrays, and tree diagrams help organise and calculate probabilities, making it easier to understand and analyse data.

Curriculum Outcomes VC2M9P01 list all outcomes for two-step chance experiments both with and without replacement, using lists, tree diagrams, tables or arrays; assign probabilities to outcomes and events VC2M9P02 calculate relative frequencies from given or collected data to estimate probabilities of events involving ‘and’, inclusive ‘or’ and exclusive ‘or’ VC2M9P03 design and conduct repeated chance experiments and simulations using digital tools to estimate probabilities that cannot be determined exactly

10 Probability Chapter outline 10.01 10.02 10.03 10.04 10.05

Experimental probability Venn diagrams and two-way tables Arrays and simple tree diagrams With and without replacement Multi-stage probability Investigation: Simulation of compound events Chapter 10 review

538 548 564 575 587 603 606

Astronomers use statistics to calculate the probability of finding other planets like Earth.

10.01

Experimental probability

After this lesson, you will be able to… • differentiate theoretical and experimental probability. • calculate the experimental probability of an event. • calculate the expected number of occurrences for specific outcomes based on experimental probabilities. • interpret data from a frequency table to calculate experimental probability. • solve problems involving experimental probability.

Relative frequency Experimental probability, also called relative frequency, is based on experiments, observations and trials. Imagine throwing 200 thumbtacks and trying to ﬁnd the probability of a thumbtack falling on its back or its pointed end.

Pointed end

Back

How it falls Back Pointed end

Frequency 73 127

Here are the results recorded in a frequency table.

We can use this experimental data to predict a future event. The probability of the event occurring is:

We can add the relative frequency in a column to the frequency table.

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How it falls

Frequency

Back

Pointed end

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Relative frequency

Sometimes “experiments” involve looking at historical data. The following table shows the winner of the Eurovision Song Contest from 1999 to 2018: Year 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008

Winning country Sweden Denmark Estonia Latvia Turkey Ukraine Greece Finland Serbia Russia

Year 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018

Winning country Norway Germany Azerbaijan Sweden Denmark Austria Sweden Ukraine Portugal Israel

Using the data of past winners, we could calculate the probability that Sweden will win the next Eurovision Song Contest. Based off the data, the probability that Sweden wins the Eurovision Song Contest is:

We can use this information to predict how many of the next 60 contests Sweden could win. This is calculated by multiplying the relative frequency by the number of trials. In this case, Sweden could expect to win:

We might expect Sweden to win 9 contests in the next 50 years. Note: There are often many other factors that might affect Sweden winning in the next 50 years, and sometimes past performance cannot be used as a predictor for future performance. In these cases, we need to consider other factors in the interpretation of results.

Interactive exploration Discover this concept in action online

mathspace.co

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Example 1 Ryan ﬂips a coin 20 times. a How many times would he expect a head to appear?

Create a strategy Find the probability of tossing a head, then multiply it by the number of trials.

Apply the idea The theoretical probability of tossing a head is . Write the formula

Multiply by the number of trials

Evaluate Ryan can expect that a head will appear 10 times.

b Ryan then tossed the coin 20 times and recorded that heads came up 11 times. Write the relative frequency of getting a head as a fraction.

Apply the idea Write the formula

Substitute the values

c Comment on the expected result compared to the actual outcome.

Apply the idea Ryan expected 10 heads, and resulted in 11. With such a small number of trials, the variance is negligible.

Example 2 To prepare for the week ahead, a restaurant keeps a record of the number of each main meal ordered throughout the previous week. Find the probability of randomly selecting chicken. Meal Chicken Beef Lamb Vegetarian

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Frequency 21 55 44 31

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Create a strategy Find the total amount of meals ordered the previous week. Calculate the probability. Convert to a percentage.

Apply the idea Total number of meals = 21 + 55 + 44 + 31 Add the frequencies from the frequency table = 151

Evaluate Write the formula

Substitute the given values

Multiply by 100%

Evaluate and round

Example 3 This graph shows the outcomes from a coin toss experiment.

Frequency

Coin Toss Experiment

Heads Tails Outcome

a How many times was the coin ﬂipped?

Create a strategy The total frequency will be the number of heads + the number of tails.

Apply the idea Based on the graph, the frequency of heads tossed is 26 and the frequency of tails tossed is 14. Total number of flips = 26 + 14

Sum the outcomes for all the possible events

= 40 Evaluate

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b What was the relative frequency of heads in this experiment?

Create a strategy Use the formula for finding the relative frequency of an event =

Apply the idea Based on the graph, the frequency of heads tossed is 26 and the total number of trials based on part (a) is 40. Divide 26 by 40 Simplify the fraction

Idea summary • To calculate the relative frequency (experimental probability): Relative frequency of event =

Number of favourable outcomes Total number of trials

• To calculate the expected number of trials with a speciﬁc outcome: Expected no. of favourable trials = Relative frequency × Total number of trials • Experimental probability does not need to equal theoretical probability. Experimental probability should get closer to the theoretical probability value as the number of trials increases.

10.01 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

If the relative frequency of an event is 1, the event: A

Happened for most of the trials

Didn’t happen

Happened on each trial

In what situations should we consider using experimental probability rather than theoretical probability?

True or False: As the number of trials in an experiment increases, the experimental probability gets closer to the theoretical probability.

Fluency 4

4–7, 9, 10, 12–15, 17, 19, 20, 22, 23, 26

4–7, 9–14, 16, 18–20, 22, 24, 26, 27

A bag contains 24 red marbles and 26 blue marbles. What is the theoretical probability of drawing a red marble?

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4, 5, 7–11, 13, 14, 17–19, 21, 24, 25, 27, 28

Ex 1

A bag contains 28 red marbles, 29 blue marbles, and 27 black marbles. a

What is the theoretical probability of drawing a blue marble?

A single trial is drawing a marble from the bag, writing down the colour, and putting it back. If this trial is repeated 400 times, how many blue marbles should you expect?

Mohamad watched cars drive past his house for ﬁve minutes. In this time, 12 cars went by, and 10 of those cars were white. What was the relative frequency of a white car driving past?

Kenji rolled a 6-sided die 30 times.

How many times would he expect a 3 to appear?

After rolling the die 30 times, he recorded that 3 had appeared 17 times. Write the relative frequency of rolling a 3.

Rosey has a bag with 2 red balls, 2 blue balls, and 2 green balls in it. She took a ball out of the bag and returned it 36 times, recording the colour each time. a

How many times would she expect to get a red ball?

After she ﬁnished, she noticed that she had drawn a red ball 15 times. Write the relative frequency of getting a red ball as a fraction.

A retail store served 989 customers in October, and there were 21 complaints during that month. What is the relative frequency that a customer complains? Round your answer to the nearest whole percent.

In the past year, an insurance company found that out of 2523 claims made, 1492 of them were from drivers under the age of 25.

Find the relative frequency that a claim is ﬁled by someone under the age of 25. Round your answer to the nearest whole percent.

Find the relative frequency that a claim is ﬁled by someone aged 25 or older. Round your answer to the nearest whole percent.

Five schools compete in a basketball competition. The results from the last season are given in the table: Match Lakehurst v St Trinian’s Summer Heights v Marquess Lakehurst v Ackley Bridge St Trinian’s v Marquess Summer Heights v Ackley Bridge

Winner

Match St Trinian’s v Ackley Bridge Marquess v Lakehurst Lakehurst v Summer Heights Marquess v Ackley Bridge St Trinian’s v Summer Heights

Lakehurst Marquess Ackley Bridge Marquess Ackley Bridge

Winner St Trinian’s Lakehurst Summer Heights Ackley Bridge St Trinian’s

What is the relative frequency that Ackley Bridge wins a match? 12

At the main traﬃc light in town, when traveling from north to south, the green light is illuminated for 27 seconds, followed by the yellow light for 3 seconds, and then the red light for 10 seconds. This cycle then repeats. If a car approaches the traﬃc light, what is the probability that the light will be: a

Green?

Yellow?

Red? 10.01 Experimental probability mathspace.co

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13 Ex 2

A food festival keeps track of the number of each type of dessert sold during the weekend: a

How many desserts were sold in total?

Calculate the relative frequency that a customer will purchase a chocolate dessert. Round your answer to the nearest whole percent.

Find the relative frequency of rolling a number greater than or equal to 3.

Find the relative frequency of not rolling a prime number.

Find the relative frequency of rolling an even number.

Find the relative frequency of rolling a 1, 2, 3, 4, 5 or 6.

Boxes of matchsticks are examined and the number of matchsticks in each box is recorded in the table:

Have 89 matchsticks?

Have more than 90 matchsticks?

Have less than 90 matchsticks?

This frequency table shows the number of people that came to donate blood and their respective weights: a

If this group is considered to be representative of the population, what is the probability that someone in the population weighs 50 to 54 kg?

What is the probability that someone weighed between 60 and 69 kg inclusive?

A die is rolled 60 times and the results are recorded in the following table: Using this frequency table, ﬁnd the probability of rolling a:

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Frequency 108 72 60 0

Outcome 1 2 3 4 5 6

Frequency 17 14 12 13 12 16

Number of matchsticks 87 88 89 90 91 92 93

Number of Boxes 0 0 7 5 6 5 3

Weight in kg (x) 40–44 45–49 50–54 55–59 60–64 65–69 70–74 Total

Frequency ( f ) 4 1 1 3 10 8 2 29

Number 1 2 3 4 5 6

Frequency 11 12 12 12 11 2

The table shows the results of rolling a die multiple times:

If the number of matchsticks in another box were counted, what is the relative frequency that it will:

Dessert Chocolate Vanilla Strawberry Mango

3 or higher

3 or lower

Mathspace Victorian Curriculum – Year 9 mathspace.co

How many times was the experiment repeated?

Find the probability of tossing: i

3 tails

at least 2 heads

iii

at least 1 tail

only 1 head

Outcome HHH HHT HTH HTT THH THT TTH TTT

This bar graph shows the outcomes of a six-sided die rolling experiment:

Frequency 13 13 14 11 13 14 11 13

Rolling a 6-sided die experiment 14

How many times was the die rolled?

What was the relative frequency of rolling a 4 in this experiment?

10 8 6 4 2 0

The size of several earthquakes was measured over a period of time, and the results are presented in the graph: a b

Find the probability that an earthquake measures 5, 6, or 7. Find the probability that an earthquake measures less than 5.

This histogram shows the number of people who waited in line for a rollercoaster at a theme park, and the duration of their wait time: a

Find the probability that someone waited in line for 40 minutes or less.

Find the probability that someone had to wait for at least 50 minutes to be served.

3 4 5 Outcome

6 5 4 3 2 1 0

7 Number of Earthquakes

Number of people

Ex 3

The table shows the outcomes of tossing three coins multiple times:

Frequency

100 90 80 70 60 50 40 30 20 10 0

1 2 3 4 5 6 7 8 9 10 Size of Earthquakes

35 40 45 50 55 60 65 Time (mins)

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This histogram shows the heights of people that were surveyed at a particular rollercoaster ride in a theme park: a

Estimate the probability that someone randomly chosen was between 130 and 150 cm tall.

Find the probability that someone randomly chosen was at most 130 cm tall.

30 25 Frequency

20 15 10 5 0

90 110 130 150 170 190 Height (cm)

The relative frequency that a commuter uses public transport is 30%. Out of 200 commuters, how many would you expect to use public transport?

If the probability of an event is , how many times would you expect it to occur in 76 trials?

A medical student is predicting how many people in their town will have certain genes. The population of their town is 1 600 000.

The probability of a person having the gene for red hair is 3%. How many people in the town are expected to have this gene?

The probability of a person having the gene for tetrachromacy is 0.03%. How many people in the town are expected to have this gene?

A factory produces tablets. In March, it produced 7000 tablets, of which 140 were found to be faulty. a

What is the probability that a tablet produced by the factory is faulty?

The factory plans to produce 8000 tablets in April. How many should they expect to be faulty?

A transportation company recorded the results of the types of vehicles used for deliveries during the past month. The table shows the number of deliveries made using different types of vehicles: Vehicle Type Number of Deliveries a How many deliveries were made in total during the month? Bicycle 120 Car 80 b Calculate the probability of a delivery being made using a bicycle. Motorcycle 100 c Which type of vehicle was used for the Truck 100 majority of deliveries? d

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If an additional 600 deliveries are expected next month, how many deliveries do you expect will be handled using motorcycles?

Mathspace Victorian Curriculum – Year 9 mathspace.co

This table shows the number of books borrowed from a local library during a month: a

What is the probability of a book being borrowed on the 5th of the month? Round your answer to the nearest whole percent.

What was the highest probability of a book being borrowed on any day during the month? Round your answer to the nearest whole percent.

What is the probability of a book being borrowed at least once during the entire month? Round your answer to the nearest whole percent. 29

Borrowed 36 32 35 30 33 32

29, 30

Beth is testing a coin to see if it is fair. She has ﬂipped the coin 100 times and recorded 52 tails. a

What is the probability of ﬂipping tails with this coin? Write your answer in decimal form.

If Beth wants to communicate the fairness of the coin to the manufacturer based on her experiment, what should she tell them?

Beth tested the coin again. This time she ﬂipped the coin 1000 times, and recorded 504 tails. What is the probability of ﬂipping tails based on this second experiment? Write your answer in decimal form.

Considering the results of the two experiments, can Beth conclude whether the coin is fair? Explain your answer.

Maria is tossing a coin. She continues tossing the coin until she gets a Tail. Her ﬁrst set of tosses resulted in Heads, Heads, Tails, so she stopped after three tosses. She then repeated the experiment 19 more times and recorded her results in the following table:

Number of Tosses before a Tail appears 1 2 3 4 5

Frequency 7 3 4 3 3

Based on Maria’s experiment, what is the probability that it takes 5 tosses of the coin before a Tail appears?

Theoretically, what is the probability that it takes 5 tosses of a coin before a Tail appears?

Was the experimental probability Maria found the same as the theoretical probability? If not, why might there be a difference?

Problem-solving 31, 32 31

1st 5th 10th 15th 20th 25th

Number of Books 50 45 48 46 49 47

Reasoning 29

Day

Homer used a spinner to choose “YES” or “NO”. The table shows three different days, how many times he spun the spinner, and how many times it landed on “YES”:

31, 32 Day 1 2 3

31–33 Number of spins Number of “Yes” 6 3 30 9 84 20

Create a possible spinner that Homer used in his experiment.

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The column graph shows the four countries that university students applied to for exchange in the last month:

Worker Applications by Country

Frequency

30 20 10 0

Germany

Philippines Country

USA

What is the relative frequency of the country with the fewest applications?

If the monthly applications are the same throughout the year, how many people in total would apply for the UK over the next six months?

If the monthly applications remain constant, how many months would it take for the Philippines to receive more than 500 applications?

Consider the table showing the status of a domino game being played by four players:

Each player draws as many dominos as they like from a bag. A domino has two numbers from 1 to 6. A high value domino occurs if the sum of its two numbers is greater than or equal to 6 while a low value domino occurs otherwise.

Player A B C D

Number of Number of high dominoes drawn valued dominoes 10 9 7 1 8 7 6 4

A player wins by randomly selecting a low value domino from their drawn sample. At this point in the game, which player has the lowest chance of winning?

10.02 Venn diagrams and two-way tables After this lesson, you will be able to… • translate data between a Venn diagram and a two-way table. • organise data using Venn diagrams and two-way tables. • interpret data presented in Venn diagrams and two-way tables.

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Venn diagrams A Venn diagram is a helpful tool when organising data into categories, especially when categories overlap. A Venn diagram, used in probability calculations, tells us how many favourable outcomes there are in each category. The rectangle is called the sample space. The sum of everything inside the rectangle represents all the events, and the sum of the probabilities of each section within the rectangle will be 1. In this case, 5 + 6 + 3 + 15 = 29 is the total. Football

The sum of all the numbers in the left circle tells us how many students play football. We can see that 5 + 6 = 11 students play football.

Baseball

P (Football) =

15 Football

Everything in the outside box does not ﬁt into either category.

Baseball

There are 15 students who do not play football or baseball.

P (Not Football nor Baseball) = 15

Football

The very middle section highlighted here represents students that are in both categories. It represents the 6 students who play football and baseball.

Baseball

P (Football and Baseball) = 15

Football

The part of the circle highlighted here are all the students who play football but do not play baseball. We could refer to these as “the students who only play football”.

Baseball

Using notation: P (F ∩ B) =

P (Football and not Baseball) = 15

In Venn diagrams, special notations help represent various situations. The entire rectangle represents the sample space, denoted by the letter ‘S’.

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The entire area inside both circles is shaded, representing the union of events A and B, which is written as A ∪ B.

In two overlapping circles, the area where the overlap is shaded indicates the intersection of events A and B, which is written as A ∩ B.

The area inside the rectangle but outside the circle that is shaded within the sample space represents the complement of event A. It can be represented by A′, A or Ac.

Example 1 A group of students were asked about their siblings. The two categories show if they have at least one brother, and if they have at least one sister.

Have a brother

Have a sister

a What is the probability that a student selected from the group will have at least one sibling?

Create a strategy Use the probability formula:

Apply the idea Number of students with at least 1 sibling = 6 + 5 + 9 Sum the numbers inside the circles = 20 Evaluate Total number of students = 20 + 8 Sum the numbers inside and outside the circles = 28 Evaluate

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Write the formula

Substitute the values

Simplify The probability that a selected student will have at least one sibling is .

b What is the probability that a student chosen from the group will have at least one brother?

Create a strategy Use the probability formula:

Apply the idea Number of students with at least 1 brother = 5 + 9 Sum the numbers inside the ‘Have a brother’ circle = 14

Evaluate

Based on part (a), the total number of students is 28. Write the formula

Substitute the values

Simplify The probability that a selected student will have at least one brother is . c What is the probability that a student chosen from the group will not have a sister?

Create a strategy Use the probability formula:

The section from the Venn diagram that relates to “Does not have a sister” is everything outside of the “Have a sister” circle.

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Apply the idea Number of students who have no sister = 9 + 8 = 17

Sum the numbers outside the ‘Have a sister’ circle Evaluate

Based on part (a), the total number of students is 28. Write the formula Substitute the values The probability that a selected student will not have a sister is

Idea summary Venn diagrams are a type of diagram that uses circles to group and organise things. The overlap or intersection, ∩, will usually use the word “and”. Combined areas or unions, ∪, will usually use the word “or”. The whole area contained in the rectangle is called the sample space.

Two-way tables Two-way tables represent data that is classiﬁed by two criteria.

Enjoys drawing Doesn’t enjoy drawing

Right 4 2

Left 9 15

The table has two criteria: a student’s main writing hand, and whether or not a student enjoys drawing.

To read a two-way table, look at the column and row that a cell is in. For example, there are 9 students who are left-handed and enjoy drawing, and 2 students who are right-handed that don’t enjoy drawing. Tables will often include totals of each column, row, and the total sum. This is found by adding up every cell in that row or column. Enjoys drawing Doesn’t enjoy drawing Total

Right 4 2 6

Left 9 15 24

Total 13 17 30

The categories of the rows and columns should be chosen such that each person or object can only go in one of the cells.

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Example 2 This table describes train departures from a train station during the months of May and June. May June

Departed on time 113 108

Departed late 31 33

a Complete the table by adding totals for the columns and rows.

Apply the idea May June Total

Departed on time 113 108 221

Departed late 31 33 64

Total 144 141 285

b How many trains departed during May and June?

Create a strategy Add all the values in the original table, or use the total from part (a).

Apply the idea Number of trains = 113 + 108 + 31 + 33 = 285

Add Evaluate

c What percentage of the trains in June were delayed? Write your answer as a percentage to one decimal place.

Create a strategy Find the number of delayed trains in June as a percentage of the total trains departing in June.

Apply the idea According to the table in part (a), the total number of June trains is 141. Find the 33 delayed June trains as a percentage of the total number of June trains: Percentage delayed =

Write the fraction

= 0.234 04 …

Evaluate the division

ª 23.4%

Convert to a percentage

Reﬂect and check This percentage also tells us that if a train in June was randomly selected, the probability that it would have been delayed is 23.4%.

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d What fraction of the total number of trains across the two months departed on time in May?

Create a strategy Find the number of on time trains in May as a fraction of the total trains.

Apply the idea From part (a), we found that there are 285 total trains. We can see from the table that 113 departed on time in May. Write as a fraction

Reﬂect and check This fraction also tells us that the probability that a train selected at random departed on time in May is

Idea summary Two-way tables represent data that is classiﬁed by two criteria.

Two-way tables and Venn diagrams A two-way table presents similar information to a Venn diagram. We can convert between a two-way table and a Venn diagram by matching up their parts. Lefthanded

Righthanded

Entered

Didn’t enter

The column marked “Left-handed” is a category used in the Venn diagram and the row marked “Entered” is the other category used.

Entered

Left-handed

This image shows how a typical two-way table and Venn diagram are related.

The number that is in both categories, 4, will go in the overlap of the two circles. The remaining value, 9, in the selected row, represents the “Entered and right-handed” students and will go into the “Entered” circle but not in the overlap because they are not “Left-handed”. Any cells that are in neither the highlighted row nor the highlighted column will go into the surrounding box. 554

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Example 3 A student makes a Venn diagram of students who are late to school, and students who catch the bus to school.

Late to school

Caught bus

Using the Venn diagram, complete this table: Late

Not late

Total

Caught bus Didn’t catch bus Total

Create a strategy We can use the Venn diagram to construct the table by looking at the corresponding regions.

Apply the idea The categories are Late/Not late and Caught bus/Didn’t catch bus.

Caught bus Didn’t catch bus Total

Late 13

Not late

Total

The top left cell is for the number of people who are late to school and caught the bus, that is, 13 people. The 9 that is written outside of the circles refers to 9 students who did not do either of these things, so it goes in the ‘not late’ and ‘didn’t catch a bus’ cell.

Total

3 is the number of students that caught the bus and are not late, so it goes in the top right cell. 7 is the number of students that did not catch the bus and are late, so it goes in the bottom left cell.

Late 13

Not late 3

Late 13

Not late 3

Total 16

The total for the ‘Caught bus’ row is 13 + 3 = 16, which matches the total inside the ‘Caught bus’ circle.

The total for the ‘Late’ to school column is 20, and this matches the total inside the ‘Late’ to school circle in the Venn diagram. We can complete the other totals by summing rows and columns.

Reﬂect and check We can check if the table is correct using the totals columns. The totals of the horizontal row (20 + 12) should be the same as the total for the vertical totals column (16 + 16).

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Idea summary We can convert between a two-way table and a Venn diagram by matching up corresponding parts.

10.02 Practice questions Understanding 1

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

Match each Venn diagram with what the shaded area represents. i

Sample space

A′, the complement of event A

iii

A ∩ B, the intersection of events A and B

A ∪ B, the union of events A and B

b S

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A gardener creates this diagram of her favourite plants:

Flowering plants

How many of the plants are ﬂowering plants?

How many of the plants are both ﬂowering and succulent?

Sunflower

How many of the plants are either ﬂowering or succulent, but not both?

Daisy

Succulent plants

Orchid

For the given two-way table, is each statement true or false?

Tulip

A B

Christmas cactus

Aloe Vera

Crown of Thorns

Bamboo

Ivy

21 11 32

23 45 68

Total 44 56 100

Events A and B are complementary.

100 is the total number of outcomes.

We add all nine numbers in the table to get the total number of outcomes.

The number of favourable outcomes for event A is 21.

If 23 was a typo, and meant to be 32, that is the only number that we would need to change to correct the error.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Total

38 students were asked whether they prefer online or in-person classes, and whether they like group projects. Data is recorded in this two-way table: Prefer online classes Prefer in-person classes

Ex 1

Dislike group projects 5 13

How many students prefer online classes?

How many students prefer online classes or like group projects, or both?

How many students prefer at most one of the two options (either online classes or liking group projects, but not both)?

Fluency 5

Like group projects 8 12

5–11, 13, 15, 16

5–10, 12, 13, 15, 16

A small magazine asked people from different states to send in a vote on whether they supported Daylight Saving Time. The results are shown in this ﬁgure: a

What proportion of people voted “YES”?

What proportion of “NO” votes are from NSW?

Considering just the voters from NSW and ACT, what proportion of the votes are NSW “YES” votes?

Clara creates a Venn diagram about the musical instruments she could assign to a student in the music class. She is told that the instrument will be chosen at random. a

What is the probability that the chosen instrument will be a stringed instrument that isn’t traditional?

What is the probability that the chosen instrument won’t be traditional?

What is the probability that the chosen instrument will be traditional?

Coach Aaron creates a Venn diagram about potential team captains based on their leadership skills and goal shooting ability:

5–7, 9, 10, 14, 15, 17 NSW YES NO

ACT YES NO

QLD YES NO

NSW YES NO

ACT YES NO

QLD YES NO

NSW YES NO

ACT YES NO

QLD YES NO

NSW YES NO

QLD YES NO

Traditional

Stringed

Leadership skills

Good shooter

How many players are there on Coach Aaron’s team, excluding Aaron?

What is the probability that a randomly selected team captain has leadership skills or is a good shooter?

Coach Aaron thinks a player is suitable to be a captain if they have both leadership skills and are a good shooter. How many suitable players are there on the team?

Aaron is told that the team captain will be chosen randomly. What is the probability that the chosen captain will be suitable?

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A group of students were asked about the dogs or cats they have as pets and if they walk them regularly or not. The results are in the two-way table. Fill in the missing values.

10 Ex 2

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Dogs Cats Total

240 cyclists were surveyed on whether they would support the use of smart glasses during competitive races. The results are detailed below: a

Fill in the missing value in the table.

What is the probability that a cyclist chosen at random is an amateur that does not support using smart glasses?

45 2

Does not walk regularly 20

Support Professional Amateur

Total

Do not support 48 40

The table below illustrates the take-offs of ﬂights from an airport for the months of May and June: Month May June Total

Walks regularly

Took off on time 135 140

Delayed 45 60

Total

Complete the table.

How many ﬂights took off during May and June?

Find the percentage of ﬂights in June that were delayed. Round your answer to one decimal place.

Find the percentage of the total number of ﬂights over the two months that took off on time in May. Round your answer to one decimal place.

Find the exact probability that a ﬂight selected at random in June would have taken off on time.

Find the exact probability that a ﬂight selected randomly over the 2 months was delayed.

A beverage preference survey asked individuals to select their preferred drink: tea or coffee. Alongside this, participants were also asked about their peanut butter preference. The table below summarises the responses:

Tea Coffee Total

Likes smooth peanut butter 45 40

Likes crunchy peanut butter 60 35

Total

Complete the table.

How many individuals participated in the survey?

If one person is chosen at random, what is the probability that they prefer tea and like smooth peanut butter?

If one person is chosen at random, what is the probability that they prefer coffee?

Mathspace Victorian Curriculum – Year 9 mathspace.co

15 Ex 3

In a study, some individuals were asked whether they enjoy listening to classical music. A partially completed two-way table of the results is shown: a

Complete the table.

Of those in the study, one is chosen at random. Find the probability that they said they don’t enjoy classical music.

Don’t enjoy

Enjoy Teenagers Seniors Total

Total

18 12

40 60

A group of individuals were asked whether they liked sushi or pizza. The result is summarised in the table: a

Complete the table.

How many people like both foods?

If one person is chosen at random, what is the probability that they like neither food?

If one person is chosen at random, what is the probability that they like only one of the foods?

Likes sushi Doesn’t like sushi

Likes pizza

Doesn’t like pizza

Total

In a study, some people were asked whether they were vegetarian or not. 27 responders said they were vegetarian, of which 8 were children. 18 children said they were not vegetarian, and 21 adults said they are not vegetarians. a

Complete the table.

Calculate the proportion of responders who are vegetarian.

Calculate the proportion of adults who are vegetarian.

What is the probability that a person chosen at random is an adult?

What is the probability that a person chosen at random is a child who is not a vegetarian?

Not Vegetarian vegetarian

Total

Children Adults Total

A librarian notes that 30 books are checked out in a day. The books are categorised based on whether they are ﬁction and whether they have been borrowed. a

Using the Venn diagram, complete the table: Borrowed

Available

Fiction

Total

Fiction Nonﬁction Total

What is the probability that a book is ﬁction and available?

What is the percentage of books that are ﬁction and have been borrowed?

Borrowed

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Students in Luke’s class were asked if they owned a laptop and if they owned a tablet. The following two-way table shows the results of the survey: Owns a tablet Doesn’t own a tablet Total

Owns a laptop Doesn’t own a laptop 5 4 10 9 15 13

Total 9 19 28

Construct a Venn diagram that represents the information provided in the two-way table.

What is the probability that a student owns both a laptop and a tablet?

150 random individuals were surveyed regarding their preference for reading material format and their employment type. The individuals were categorised as either freelancers or fulltime employees, and as preferring physical or digital books. a

Construct a Venn diagram that represents the information provided in the two-way table. What is the probability that an individual prefers digital books and is a full-time employee?

What is the proportion of individuals who are freelancers?

Is it true to say that individuals who are freelancers generally prefer physical books?

Reasoning

Prefers physical books Prefers digital books Total

18, 19

Freelancer

Full-time employee

Total

150

18–20

The results from a recent survey showed that 56 people speak Spanish or French. Of these, 36 speak Spanish, and 18 speak both Spanish and French. The number of people who speak French is asked. Trent’s answer is 20 and Joanna’s answer is 38. Who is right? Why?

In a class, 5 people play both football and tennis, 13 people in total play tennis, and 11 in total play football.

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How many people only play football?

If a random student is chosen from the group, what is the probability that the student only plays tennis?

Goran believes that 14 people only play 1 sport. Yujuan believes that 19 people only play 1 sport. Who is correct? Justify your answer.

Four more people enter the class and are asked to contribute their answers to the survey. They all said that they don’t play either. How does this affect your results from part (b)?

At a school, 142 students are asked to join either the soccer club or the basketball club (or both). 75 students join the soccer club and 78 students join the basketball club. a

Samuel calculates that 11 students joined both clubs. Explain how he got his answer.

Can we determine what is the most popular club from this survey? Why or why not?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Problem-solving 21, 23, 25, 26, 30 21

22, 24, 26, 27, 30

How many of the movies are horror ﬁlms?

How many of the movies ﬁt into only one genre?

How many of the movies are an action films combined with at least one other genre?

If she randomly chooses a movie, what is the probability that it is a movie that is classiﬁed as:

Action

Comedy

Rosey is deciding which movie to watch from a streaming website. A Venn diagram of her options sorts movies into three categories based on genre:

5 3

6 13 Horror

action and horror?

comedy only?

iii

comedy?

comedy or action?

action and comedy?

not horror?

Linda is planning to redesign her garden. She catalogues her plants based on certain criteria. A Venn diagram of her garden sorts the plants into three categories: Flowering plants, plants that require direct sunlight, and plants taller than 1 metre.

22, 24, 26–30

How many plants are there in Linda’s garden?

Linda wants to choose a ﬂowering plant. What is the probability she will select a ﬂowering plant?

Linda is looking for a plant that ﬂowers but doesn’t grow taller than 1 metre. What is the probability that she will select a suitable plant?

Linda wants a ﬂowering plant that grows taller than 1 metre. What is the probability that she will select a suitable plant?

Flowering

Direct sunlight

4 7

2 13

Taller than 1 metre

A student is analysing the favourite art activities among her classmates. 22 people enjoy sketching, 19 enjoy pottery, and 7 enjoy both sketching and pottery. a

Construct a Venn diagram that represents the given information.

How many people only enjoy pottery?

What is the probability that a student randomly selected from the group: i

only enjoys sketching?

enjoys sketching or pottery but not both?

iii

enjoys sketching and pottery?

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In a group of 190 primary and senior students, 88 are primary students. The students fell into three categories of travel to school: by bus, car, or walking. 109 students get to school by bus, of which 52 are primary students. 50 students get to school by car, of which 20 are primary students. a

b 25

a senior student?

a senior student that catches the bus to school?

iii

either a senior student or a student that catches a bus but not both?

What is another way to describe the groups of students discussed in part (a)?

In a study, a number of people were asked whether they were musicians or not. 29 responders said they were a musician, of which 8 were children. 19 children said they were not musicians, and 12 adults said they were not musicians. a

What is the probability that a student chosen at random is:

What is the probability that a student chosen at random is: i

a musician?

an adult musician?

iii

a musician, but not an adult?

What is another way to describe the same groups from part (a)?

A manager is assessing the participation of employees in two different training programs: leadership development and project management. Out of a team of 40 members, 32 attended the leadership development program and 15 attended the project management training. a

Given that everyone in the team attended at least one program, ﬁnd out how many employees attended both programs.

Construct a Venn diagram that represents the given information.

What is the probability that an employee attended only the leadership development program?

A jeweller has sorted 25 gems based on two categories: • Based on the colour – emerald • Based on the cut - rectangular

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Put the information into a two-way table.

If a gem is selected at random, what is the probability that it is: i

An emerald or rectangular?

Not emerald or rectangular?

iii

Both emerald and rectangular or neither?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Ruby Emerald Amethyst

The employees of Koala Airlines were discussing where they should hold their end of year party. Of all of the employees: • 53 would go to a restaurant. • 67 would go to a bowling alley. • 76 would go to a theme park. • 16 selected both the restaurant and bowling alley. • 29 selected both the bowling alley and theme park. • 17 selected both the restaurant and theme park. • 4 would not go to a restaurant or a bowling alley or to the theme park. • 5 would go to all three places.

Construct a Venn diagram that represents the given information.

How many people in total work at Koala Airlines?

If the restaurant isn’t available, what proportion of the employees still have another option?

What is the probability that an employee chosen at random had chosen: i

the bowling alley or the theme park?

the bowling alley?

iii

not the bowling alley?

A gardener observed that some plants were affected by pests. 8 plants showed signs of aphid infestation, 14 plants showed evidence of mite damage, and 12 plants had slug damage. Only one plant had signs of all three pests. There were two plants with aphid and mite damage but no slug damage; there were three plants with aphid and slug damage but no mite damage. There were 25 affected plants in total in the garden. a

Draw a Venn diagram to illustrate the information.

What is the probability that if they chose one affected plant at random, it would show mite and slug damage but not aphid damage?

It costs $1.20 per plant to treat for aphids and $2.50 per plant to treat for mites. The gardener cannot use the individual treatments together; instead, if a plant has both aphids and mites, they have to use the all purpose spray that costs $4.10 per plant. What is it going to cost to treat the aphids or mites?

The treatments were 60% successful. Create a new Venn diagram for the plants that are still affected, and work out the probability that a plant at random has aphids but not mites.

During a food quality check, a new device is introduced to determine if fruits have any pesticides. To validate its precision, a batch of 200 fruits was taken. Some are known to have pesticides and others are organic (have no pesticides). Given this two-way table:

Fruit has pesticides Fruit is organic Total

Detected Pesticides 40

Detected as Organic

Total 60

200

Complete the table.

For what percentage of fruits did the device detect pesticides?

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10.03 Arrays and simple tree diagrams After this lesson, you will be able to… • interpret arrays to identify all possible outcomes and calculate probabilities. • use tree diagrams to track multi-stage experiments. • construct and interpret tree diagrams to identify outcomes and calculate probabilities. • solve problems involving arrays and tree diagrams.

Arrays A table, also known as an array, is useful for showing all possible outcomes of two-stage events in the rows and columns.

H T

1 H1 T1

2 H2 T2

3 H3 T3

4 H4 T4

5 H5 T5

6 H6 T6

For example, if we tossed 1 coin and 1 die, we can show the outcomes for the coin along the ﬁrst column on the left (H, T) and the outcomes for the die across the top row (1, 2, 3, 4, 5, 6).

Each cell in the array is an outcome of rolling a die and tossing a coin. There are 12 possible outcomes in the sample space. We can use the probability formula to determine probabilities from the array.

Example 1 The table shows the possible results for the event of rolling 2 six-sided dice and summing the total. 1 2 3 4 5 6

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1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

Mathspace Victorian Curriculum – Year 9 mathspace.co

What is the probability that a player rolls two dice that sum to 8?

Create a strategy

Apply the idea From the table, we can see that there are 6 × 6 = 36 possible outcomes in total, 5 of which are an 8.

9 10

Substitute the values

Reﬂect and check Let’s check, the ﬁve ways that we can get an 8 are (5, 3), (3, 5), (6, 2), (2, 6) and (4, 4).

Example 2 This spinner is spun and a six-sided die is rolled. The product of their respective results is recorded.

a Create an array that shows the sample space for the experiment.

Create a strategy Place the die numbers on the left and spinner numbers on top, then ﬁll each cell with their product.

Apply the idea The array for the product of the spun spinner and rolled six-sided die is given by:

Die

1 2 3 4 5 6

2 2 4 6 8 10 12

Spinner 3 3 6 9 12 15 18

5 5 10 15 20 25 30

8 8 16 24 32 40 48

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b What is the probability of an odd product?

Create a strategy Use the array from part (a) then the probability formula.

Apply the idea Using the array, we can see that there are 24 total outcomes and 6 are odd products. This means that the probability of an odd product is: Write the formula

Substitute the odd and total outcomes

Simplify

Reﬂect and check The only way to get an odd product is to multiply two odd numbers. There are three odd numbers on the die, 1, 3, 5, and two odd numbers on the spinner, 3, 5, so there are 3 × 2 = 6 possible odd products. This matches what we found with the array.

c What is the probability of rolling a 5 on the die and scoring an even product?

Create a strategy Use the array from part (a) to ﬁnd the number of times a 5 is rolled and an even product is scored. Then use the probability formula.

Apply the idea Using the array in part (a), we can see that there are 24 total outcomes and 2 outcomes resulted from rolling a 5 and scoring an even product. This means that the probability of rolling a 5 on the die and scoring an even product is given by: Write the formula Substitute the 2 favourable outcomes and 24 total outcomes

Simplify

d What is the probability of spinning a 3 on the spinner or scoring a product which is a multiple of 4?

Create a strategy Use the array from part (a) to find the number of times a 3 is spun or a product which is a multiple of 4 is scored. Then use the probability formula.

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Apply the idea Based on the constructed array from part (a), there are 24 total outcomes and 16 outcomes that are either a spinner value of 3, or a product that is a multiple of 4. 5

3 24 6

12 20

32 8

5 40 10 15 25 6 48 12 18 30

This means the probability of spinning a 3 on the spinner or scoring a product which is a multiple of 4 is given by:

Write the formula Substitute the 16 favourable outcomes and 24 total outcomes Simplify

Idea summary An array is useful for showing all possible outcomes of two events. The rows display one event and the columns display the second event.

Tree diagrams A tree diagram is useful in multi-stage experiments. Probability Branch

0.1

H Outcome

0.9

The important components of the tree diagram are: • Branch • Probability (only shown when the events have unequal probabilities) • Outcome

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T T

T H T

T H

Here is a three-stage example where all the outcomes are equally likely. There are 8 outcomes and each has a probability of : • TTT • TTH • THT • THH • HTT • HTH • HHT • HHH

In tree diagrams where all branches are equally likely, we can find the probability in two ways: 1. By counting the number of favourable branches and dividing by the total number of branches. 2. Using the complementary events and doing P ( A) = 1 − P ( A′) For some scenarios, the options for the second split of branches will be different from the first step because the option from the first step is no longer available.

Example 3 Three fair coins are tossed and the outcomes are displayed in the tree diagram shown.

T T

T H T

T H

a List the sample space of the event.

Create a strategy List all of the possible outcomes from the tree diagram.

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Apply the idea The array for all the possible outcomes of three fair coins tossed is given by: H H H H T T T T

1st Toss H H H H T T T T

2nd Toss 3rd Toss H H H T T H T T H H H T T H T T

b Find the probability of obtaining at least one head.

Create a strategy Express the number of times one or more heads occurs as a fraction of the total possible outcomes.

Apply the idea Based on the given tree diagram, one or more heads occurs on 7 occasions. This means that the probability of obtaining at least one head is

Reﬂect and check Another way to do this would be to reword this as “not all tails”. This would mean: Use complementary events

Only one branch has all tails, TTT

Simplify

c What is the probability of obtaining TTH in this sequence?

Create a strategy

Apply the idea

Look for how many times this outcome (TTH) occurs in the sample space.

Based on the given tree diagram, there is one occurrence of TTH out of 8 possible outcomes. This means that the probability of obtaining TTH in the sequence is

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d What is the probability of obtaining two heads and a tail in any order?

Create a strategy

Apply the idea

Look for how many times we can get 2 heads and 1 tail.

Based on the given tree diagram, there are three possible ways to get 2 heads and one tail out of 8 possible outcomes. This means that the probability of obtaining THH in any order is:

We could have the tail first THH, the tail second HTH or the tail third HHT.

Idea summary A tree diagram is useful in tracking multi-stage experiments. Each branch represents one event of the multi-stage events.

10.03 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Are these statements true or false? a

In a two-dimensional array, every row must contain the same number of elements.

We can use a two-dimensional array for a three-stage experiment.

We can use a tree diagram for a three-stage experiment.

Each cell in an array is equally likely.

Lukas uses a tree diagram to represent an experiment where a fair six-sided die is rolled three times. Does each pair of events have an equal probability? a

Rolling an even and an odd.

Rolling a ﬁve and not rolling a ﬁve.

Rolling a prime and not rolling a prime.

A player is rolling two dice and calculating their sum. They draw an array of all the possible dice rolls with the sum of two dice. List the sample space for the sum of two dice.

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Trailblazer 1–3

Mathspace Victorian Curriculum – Year 9 mathspace.co

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

Fluency Ex 1

4, 6–8, 10–12

4–9, 11, 12, 13a

A player rolls 2 six-sided dice and the sum is recorded in an array.

1 2 3 4 5 6

What is the probability that sum is

less than 5

more than 6

7 or higher

an even number

What is the probability that the sum is 6 or less?

What is the probability that the sum is more than 6?

What is the sum of the probabilities from (a) and (b)?

Explain the sum in part (c).

Blue (B) B, D B, S

White (W ) W, D W, S

Red (R) R, D R, S

How many different combinations are possible?

Find the probability that he is wearing:

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

Yellow (Y ) Y, D Y, S

A blue shirt and a striped tie

A yellow shirt and a dotted tie

iii

A red shirt

A red or white shirt

A striped tie

A white shirt or a striped tie

A coin is tossed and a die is rolled at the same time. Complete the array to show the possible outcomes: Heads Tails

2 3 4 5 6 7 8

George has 4 different coloured shirts: blue (B), white (W ), red (R) and yellow (Y ). He also has 2 different patterns of ties: dots (D) and stripes (S).

Ex 2

1 2 3 4 5 6 7

Using the table from question 4, showing the results of the sum of 2 six-sided dice:

Dots (D) Stripes (S)

4, 5, 8, 9, 11, 13

1 H, 1

3 H, 3 T, 2

4 H, 4 T, 3

6 H, 6 T, 5

How many possible outcomes are there?

Determine the probability of rolling a 1.

Determine the probability of rolling an even number.

Determine the probability of tossing heads and rolling an even number.

Determine the probability of tossing tails and rolling an odd number.

This spinner is spun and a six-sided die is rolled. The sum of their results is recorded. a

Construct an array to represent all possible outcomes.

Find the probability of an even sum.

Find the probability of rolling a 3 on the die and scoring an odd sum.

Find the probability of spinning a 4 on the spinner or scoring a sum of 7.

Find the probability of the sum being greater than 8.

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These two spinners are spun and the result of each spin is recorded:

Complete the given array.

How many possible outcomes are there?

Find the probability that the spinner lands on a consonant and an even number.

Find the probability that the spinner lands on a vowel or a prime number.

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1 2 3

1, A 2, C

Complete the given array.

Ties 11

Xavier is choosing an outﬁt for the day and has 3 shirts (cyan, pink and white) and 4 ties (black, grey, red and yellow) to select from. a

Ex 3

Black (B) Grey (G) Red (R) Yellow (Y )

Cyan (C) C, B C, G ⬚, ⬚ C, Y

Shirts Pink (P ) ⬚, B P, G P, R ⬚, ⬚

White (W ) W, B W, ⬚ W, R W, Y

How many different outﬁts are possible?

What is the probability that he wears a black tie?

What is the probability that he wears a white shirt?

What is the probability that he wears a pink shirt or a yellow tie, but not both?

Each day Chris chooses one accessory, a hat or a tie. There is an even chance that Chris wears a tie or hat on any given day. The tree diagram represents the possibilities of Chris’ choice on three consecutive days.

H H

H T H

What is the total number of outcomes?

List the outcomes.

Find the probability that Chris:

Wears a hat on all three days.

Wears a hat on the ﬁrst day, a hat on the second day and a tie on the third day.

iii

Wears a hat once and a tie twice in any order over the three days.

Doesn’t wear a hat on the ﬁrst day

Wears a tie at least once

Mathspace Victorian Curriculum – Year 9 mathspace.co

T T

In a town in Australia, only ravens (R) and sparrows (S) are found. Samuel waits to see two birds. He is equally likely to see each type of bird. a

Find the probability that at least one of the birds he sees is a sparrow.

Find the probability that the last bird he sees is a raven.

Find the probability that he only sees 1 sparrow.

R S

Construct a tree diagram showing: a

All possible outcomes of red or black cards that a person could have if they draw three cards from a deck.

All the ways the names of three candidates; Alvin, Sally and Peter, can be listed on a ballot paper.

Reasoning

14, 15

14–16

Students in a Mathematics class were asked to use either a tree diagram or an array to list the sample space, when rolling 2 dice. Which method would be more eﬃcient? Justify your answer.

A fair coin is tossed 8 times and the result was heads every time. What is the probability that the next toss will also be heads?

Frankie has a ten-sided die and Myah has a six-sided die. Do they both have the same chance of rolling a 6 and then a 2? Explain.

Problem-solving 17, 18 17

17–19, 21

20–22

A fair die is rolled twice. Find the probability of rolling: a

A 3 and a 6 in any order

A 3 and then a 6

A double

An odd and an even number

Two spinners labelled 1 to 4 are spun at the same time and the results are added: Spinner 1

Spinner 2

Find the probability of: a

An even result

A result greater than 6

A result less than 5 and even

A result less than 5 or even

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A two-digit number is formed by spinning the two spinners, with Spinner 1 being the ﬁrst digit and Spinner 2 being the second digit in the number: Spinner 1

Spinner 2

Find the probability of the number formed being:

Equal to 25

Even

Prime

Divisible by 5

Divisible by 5 and odd

Containing at least one 2

Two dice are rolled, and the combination of numbers rolled on the dice is recorded. a

find the probabilities for the two numbers rolled: i

P(1 and 4)

P(1 then 4)

iii

P(difference = 4)

P(product = 12)

P(difference £ 2)

P(difference = 3)

vii P(product = 20) b

viii P(difference £ 1)

If the numbers appearing on the uppermost faces are added, determine whether these statements are true or false: i

A sum greater than 7 and a sum less than 7 are equally likely.

A sum greater than 7 is more likely than a sum less than 7.

iii

A sum of 5 or 9 is more likely than a sum of 4 or 10.

An even sum is more likely than an odd sum.

These two spinners are spun and the product of their spins is recorded. Find the probability that: a

The product is less than 15.

The product is odd.

A 7 was spun, given that the product is less than 20.

The product is less than 40.

The product is odd, given that a 4 was spun.

5 2

Melinda loves board games and has 4-sided, 6-sided, 10-sided, 12-sided, and 20-sided dice. Two different dice are rolled and their sum is recorded.

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If the probability of a sum greater than 30 is

If the probability of a sum of 3 is

If the probability of an even sum is , which two dice were rolled?

Mathspace Victorian Curriculum – Year 9 mathspace.co

, which two dice were rolled?

10.04 With and without replacement After this lesson, you will be able to… • differentiate between selecting items with and without replacement. • construct tree diagrams to represent multi-stage experiments. • calculate probabilities for scenarios where items are either replaced or not replaced. • evaluate real-world examples to apply the concept of replacement in probability.

With and without replacement Selecting items with replacement means putting each selected item back before you select another item. When we select with replacement, the probabilities at each stage stay the same. Sample space B

For example, when randomly selecting a pen from a pencil case with two blue pens and one red pen, the probability of getting a blue pen on the ﬁrst selection is . If we get a blue pen ﬁrst and replace it before we make the second selection, the probability of getting a blue pen on the second selection is still , as shown in the tree diagram. The probability of selecting two blue pens is .

Selecting items without replacement means that we don’t put each selected item back before selecting another. When we select without replacement, the probabilities at each stage will change. Sample space B

In the same pencil case, if we get a blue pen ﬁrst and don’t replace it, then for our next selection there is still one red pen, but only one blue pen remaining. This means that the probability of getting a blue pen on the next selection is . The probability of selecting two blue pens is now 2 1 2 1 × = = . 3 2 6 3

When constructing tree diagrams for experiments without replacement, we must think carefully about the number of each item left after each selection, and change our probabilities accordingly.

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Exploration To decide which chores Jacques needs to do, he pulls out pieces of paper from a hat. The options are sweeping (S), mopping (M), and vacuuming (V). No chore is done twice on the same day. Day 1 chore

Day 2 chore

Day 1 chore S

V M

V V

Day 2 chore S M V S M V S M V

Tree diagram 1: two chores on the same day Tree diagram 2: two chores on different days 1.

Which tree diagram shows the experiment with replacement?

What is the probability he will need to sweep and mop if they are done on the same day?

What is the probability he will need to sweep and mop if they are done on different days?

Is there a difference in the probabilities when selecting with or without replacement?

Example 1 Esther was given four animal crackers as a snack. She has a donkey (D), an elephant (E), a goat (G), and a hippo (H). She eats one cracker and then another. a Is this an example of with or without replacement?

Create a strategy With replacement means the ﬁrst choice is still available for the second choice. Without replacement means the ﬁrst choice is eliminated in the second selection.

Apply the idea Since the ﬁrst cracker is eaten, it is not available in the second round. This means it is without replacement.

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b List the sample space for the two crackers she eats if the order she eats them matters to her.

Create a strategy Construct a tree diagram to easily list all the outcomes, making sure that none are missed.

Apply the idea First cracker

Second cracker Sample space E G H

DE DG DH

D G H

ED EG EH

D E H

GD GE GH

D E G

HD HE HG

So the sample space is {DE, DG, DH, ED, EG, EH, GD, GE, GH, HD, HE, HG}.

Reﬂect and check Since for the ﬁrst animal cracker she has 4 choices and then after eating one she has 3 choices for the next one, it makes sense that there are 4 × 3 = 12 outcomes in the sample space.

Example 2 Construct a tree diagram showing all the ways a captain and a vice-captain can be selected from Matt, Rebecca and Helen.

Create a strategy Any of the three people can be chosen as captain. Once a captain has been selected, any of the two remaining people can be vice-captain.

Apply the idea There are three names to be selected from, so this means that there are 6 possible outcomes.

Captain

Vice-captain

Matt

Rebecca Helen

Rebecca

Matt Helen Matt

Helen

Rebecca

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Example 3 Four cards numbered 1 to 4 are placed face down on a table. Two cards are drawn with replacement. a Construct a table array of this situation.

Create a strategy Since this is with replacement, our options will be the same for both draws.

Apply the idea For the ﬁrst selection, we have 1, 2, 3, 4 as possible options. Since each card is replaced, the options will stay the same for the second selection.

1 2 3 4

1 2 3 (1, 1) (1, 2) (1, 3) (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (4, 3)

4 (1, 4) (2, 4) (3, 4) (4, 4)

1 2 3 4

1 2 3 (1, 1) (1, 2) (1, 3) (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (4, 3)

4 (1, 4) (2, 4) (3, 4) (4, 4)

b Calculate the probability that the two cards are both even.

Create a strategy Use the array table from part (a).

Apply the idea From the array table, we can see that there are 16 outcomes in the sample space. Four outcomes are our desired outcome with both numbers being even (2, 2), (2, 4), (4, 2), (4, 4). Therefore, the probability that the two cards are both even is .

Reﬂect and check A good check is that the size of the sample space is the number of possibilities in the first step times the number of possibilities in the second step. Sample space size = 4 × 4 = 16 which matches what we saw in the array table.

c Construct an array table that lists the sample space if the experiment is done without replacement.

Create a strategy Since this is without replacement, the first card is eliminated in the second selection.

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Apply the idea

When the experiment is done without replacement, the options for the second selection will change. The table array will show only valid outcomes where the first and second draws are different: In this array, the diagonal elements are excluded because a card cannot be drawn twice when there is no replacement.

1 2 3 4

2 (1, 2)

3 4 (1, 3) (1, 4) (2, 1) (2, 3) (2, 4) (3, 1) (3, 2) (3, 4) (4, 1) (4, 2) (4, 3)

The sample space without replacement consists of the remaining 12 pairs.

Example 4 Qian packed three shirts in his suitcase for a holiday. Two are pink and one is white. He selects two at random to bring on a day trip. a Construct a tree diagram of this situation.

Create a strategy Since he is bringing two distinct shirts, this is with out replacement.

First shirt

Second shirt

Sample space

Apply the idea For his first choice, he will have three shirts to choose from. For his second choice, he will only have two shirts to choose from. The options for the second shirt will change based on the first choice.

b Find the probability that Qian selects at least one pink shirt.

Create a strategy Use the tree diagram from part (a).

Apply the idea

Sample space

We can highlight the desired outcomes on the tree diagram. P

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Start with the probability formula

Substitute

Simplify

Reﬂect and check He must bring at least one pink shirt as he only has one shirt that isn’t pink. A probability of 100% means we are certain this will happen.

c Find the probability that Qian selects at most one pink shirt.

Create a strategy Use the tree diagram from part (a).

Apply the idea

Sample space

We can highlight the favourable outcomes. P

P(at most 1 pink shirt)

Start with the probability formula

Substitute

Simplify

Reﬂect and check This would be the same as the probability of him choosing one pink shirt and one white shirt.

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Example 5 Noive selects two letters from the word ONTO without replacement to see if a new word is formed. a Is the outcome NN possible?

Create a strategy Look at the number of letter Ns which remain after one N has been selected.

Apply the idea No, the outcome NN is not possible. There is only one N and letters are being selected without replacement, so once one N is selected first, there are none available for the second selection.

Reﬂect and check It is possible to have a double letter as an outcome, but there must be a double in the original word, for example OO. b Find the probability that the word ON is formed, in that order.

Create a strategy Create a tree diagram showing each of the possible two letter outcomes. Then using the tree diagram, we can divide the number of outcomes with the word ON by the total number of outcomes.

Apply the idea

Sample space

For the ﬁrst letter selection, it can be any of the letters of ONTO.

For the second letter, we will have to choose from the remaining three letters as this is without replacement.

Start with the probability formula

Substitute

Simplify

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c Find the probability that the word ON or NO is formed.

Create a strategy Use the tree diagram from part (b).

Apply the idea

Sample space

From the tree diagram, there are four outcomes which spell ON or NO.

Start with the probability formula Substitute Simplify

Reﬂect and check Another strategy is to ﬁnd the probability of ON and the probability of NO and add them together. There is no overlap between these outcomes, so we don’t need to subtract it. Sum the probabilities of the outcomes Substitute Add the fractions Simplify

Idea summary Selecting items with replacement means putting each selected item back before you select another item. Selecting items without replacement means that we don’t put each selected item back before selecting another. When constructing tree diagrams for experiments without replacement, we must only list the options that are still available after the ﬁrst selection.

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10.04 Practice questions Understanding 1

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

Identify whether each experiment is with replacement or without replacement. a

A card is randomly selected from a normal deck of cards, and then returned to the deck. The deck is shuﬄed and another card is selected.

Each of four students is allowed to randomly pick an item from the teacher’s prize bag (and keep it).

A butterﬂy is captured and its gender is recorded. It is released back into the garden before another butterﬂy is captured.

Without looking, a child selects a chocolate from a box of milk and dark chocolates. They eat it, then select another.

For the same experiment done with and then without replacement, will the sample space change or stay the same?

A bag has 3 pink marbles and 2 blue marbles. Two marbles are drawn without replacement. Is each statement true or false?

The sample space will have 5 items in it.

The sample space will have 20 items in it.

The sample space is {PP, PB, BP, BB} and each outcome is equally likely.

It is more likely to draw two pink marbles than two blue marbles.

Three cards labelled 2, 3, and 4 are placed face down on a table. Two of the cards are selected randomly to form a two-digit number. The outcomes are displayed in the tree diagram. List the sample space.

4 2

Fluency 5

6–9, 11, 14–17

With replacement

Without replacement

A container holds four counters coloured red, blue, green and yellow. Draw a tree diagram representing all possible outcomes when two draws are done, and the ﬁrst counter is: a

Ex 1

5–9, 11, 12, 14, 15, 17

A bag has 2 blue marbles, 1 white marble, and 1 green marble. Draw a tree diagram for drawing two marbles. a

5–10, 12–14, 17

Replaced before the second draw

Not replaced before the next draw

Aditya has three lollies. A blue one, a green one, and a yellow one. They take one, eat it, and then eat one more. a

Is this experiment with or without replacement?

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Ex 2

A project team consists of four members: Sarah, David, Olivia, and Michael. The team needs to designate a leader and a coordinator. Construct a tree diagram to demonstrate all the possible ways a leader and a coordinator can be chosen.

Three cards labelled 7, 8 and 9 are placed face down on a table. Two of the cards are selected randomly without replacement to form a two-digit number.

Ex 3

Construct a tree diagram of this situation.

List the sample space for the possible two-digit numbers formed.

What is the probability that both cards are odd?

A bucket contains 3 green buttons and 1 black button. Two buttons are selected from the bucket. The ﬁrst button is replaced before the second button is selected. a

Construct a tree diagram of this situation.

Find the probability of selecting two black buttons.

A bag has ﬁve counters labelled 1 to 5. Two counters are drawn and the larger number is recorded as the outcome. a

Complete this array that lists the sample space if the experiment is done with replacement. 1 1

1 2 3 4 5 b

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3 3 3 3

4 4 4 4 4

5 5 5 5 5 5

1 3 4

Complete this array that lists the sample space if the experiment is done without replacement. 1 2 3 4 5

2 2 2

2 2

3 3 3

4 4 4 4

5 5 5 5 5

If the experiment is done without replacement, ﬁnd the probability of the larger number being: i

Even and prime

Odd

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iii

Even or prime

Ex 4

A container holds three cards coloured red, blue and green. a

Draw a tree diagram representing all possible outcomes when two draws occur, and the card is not replaced before the next draw.

Find the probability of drawing the blue card ﬁrst.

Find the probability of drawing a blue card in either the ﬁrst or second draw.

While playing a card game, Xander has a 9, a 6, a 5, and a 3. Two cards are drawn without replacement. a

Construct a tree diagram for the situation.

Calculate the probability that:

At least one of the cards is a multiple of 3

Construct a tree diagram of this situation.

Find the probability that Elliott-Rose selects: i

At least one cotton dress

Two cotton or two linen dresses

iii

At most one cotton dress

No cotton dresses

Two dresses of the same material and at least one linen dress

A band has three guitarists and one bassist. Two different musicians are chosen for solos at random. Find the probability of the two soloists being: Guitarists

With replacement

Bassists

Without replacement

Two letters are randomly selected in order and without replacement from the word MONEY. a

List the sample space.

Y is considered a vowel here. Find the probability of: i

The word MY

The word ME or MY

iii

The letters having at least one vowel

Starting with an M or an N

Starting with an M and ending with a vowel

Starting with an M or ending with a vowel

Reasoning 18

A number game uses a basket with 6 balls, all labelled with numbers from 1 to 6. Two balls are drawn at random. Find the probability that the ball labelled 3 is picked once if the balls are drawn: a

Ex 5

Neither of the cards are odd

Elliott-Rose has five dresses. Three are cotton and two are linen. She selects two for a weekend away.

a 16

18, 20

18–20

Explain how selection with or without replacement affects the number of outcomes and branches in a tree diagram.

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In a bag, there are 3 red marbles and 2 yellow marbles. Two marbles are drawn. This tree diagram shows the situation. a

Are the marbles drawn with or without replacement? Explain how you know.

Why do you think it is written as R1, R2, and R3 instead of writing them all as just R? Explain.

R1 R2 R3 Y1 Y2 R1 R2 R3 Y1 Y2 R1 R2 R3 Y1 Y2 R1 R2 R3 Y1 Y2 R1 R2 R3 Y1 Y2

Two letters are randomly selected in order and without replacement from the word YUMMY. a

Is the outcome YY possible? Explain.

Give an example of an outcome that is more likely than UM. Justify your answer.

Explain how the probability of getting MY would change if this was done with replacement.

Problem-solving 21, 23 21

586

22–25

In a bakery, there are 6 wrapped pastries. Three are chocolate, two are custard, and one is strawberry. A customer buys two pastries at random. What is the probability that the customer gets: a

21–24

Two chocolate pastries

At least one chocolate pastry

A bag has counters with different numbers on them. Two counters are drawn without replacement. The smaller number is recorded as the outcome. a

Describe an event that would have a probability of 0.

Describe an event that would have a probability of 1.

Describe an event that would have a probability of less than 0.5.

4 6 7

At the start of the week, Milana has three bananas and two apples in her fruit bowl. Each morning before school, she selects a piece of fruit at random to take and eat at lunchtime. a

Is this experiment with replacement or without replacement?

Determine the probability that: i

She eats apples on Monday and Tuesday.

She eats an apple on Tuesday.

iii

There are no bananas left on Thursday morning.

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A hand has ﬁve cards numbered from 3 to 7. Two cards are drawn and their product is the outcome. a

If the experiment is with replacement, show how an array could be used to list the sample space.

If the experiment is without replacement, show how an array could be used to list the sample space.

If the hand had 20 cards, how many ﬁlled cells would there be in the array: i

With replacement

Without replacement

There are four numbered cards. Two of the cards are randomly chosen and the possibilities for the sum of their numbers is: {15, 10, 8, 11, 9, 4}

Find the missing number on the fourth card.

If two cards are chosen at random, ﬁnd the probability that the sum of their numbers is: i

Even

10.05

At least 10

Multi-stage probability

After this lesson, you will be able to… • understand and apply the concept of tree diagrams with probabilities for experiments with and without replacement. • calculate probabilities for multi-stage experiments using tree diagrams. • construct tree diagrams for various multi-stage probability problems.

Tree diagrams with probabilities We have seen that the probabilities of outcomes are different depending on if the experiment was done with or without replacement. The more outcomes, the harder it can be to list the outcomes in a simple tree diagram.

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Exploration One tree diagram does not have probabilities on the branches, while another does. Sample space C

C C L L

CC CC CL CL

C C L L

CC CC CL CL

C C L L

CC CC CL CL

C C C L

LC LC LC LL

C C C L

LC LC LC LL

2 4

Without probabilities on the branches

With probabilities on the branches

C 3 5

2 5

2 4

3 4

1 4

Could these tree diagrams represent the same situation? Explain.

For the tree diagram without probabilities on the branches, how could you ﬁnd the probability of CC?

Knowing the probability of CC from question 2, how could you ﬁnd the probability of CC from the other tree diagram?

What are the beneﬁts and drawbacks of each of the diagrams?

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When the outcomes are not equally likely, the probability will be written on the branches. In this example, there are probabilities on the branches.

W 60% 30%

If you add the probabilities in each trial, the sum of the branches equals to 1 (or 100%). This also helps us to check that all the outcomes are listed.

10% D

Here is an example of a multi-stage experiment, where the probabilities are written on the branches. D

0.7 0.7

0.3

0.7

0.3

0.7

0.3

0.7

0.3

0.7

0.3 N

This tree diagram shows that the probability that a child gets home during the day is P (D) = 0.7 and the probability that a child gets home during the night is P (N ) = 0.3. This is looked at over three days.

By multiplying across the branches of the tree diagram, from left to right, we get the probability of that outcome. So, P(DDD) = 0.7 × 0.7 × 0.7 = 0.343. Next, we get that the probability of P(DDN) = 0.7 × 0.7 × 0.3 = 0.147. When we add the probabilities of all the possible outcomes, it should equal to 1 (or 100%). When constructing tree diagrams for experiments without replacement, we must think carefully about the number of each item left after each selection, and change our probabilities accordingly.

Example 1 A standard deck of 52 cards has four queens. Ilham shuffles the deck then selects two cards, with replacement, noting whether or not they got a queen for each selection. a Construct a tree diagram of this situation with the correct probability on each branch.

Create a strategy Since each card is replaced, the probabilities will stay the same for the second selection. The total number of outcomes will be 52. The number of favourable outcomes will be the number of queens in the deck.

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Apply the idea For the ﬁrst and second selections, there are 52 cards in total, with 4 queens and 48 cards that are not queens. Let Q represent the event of drawing a queen and N represent the event of not drawing a queen.

4 52

Q 48 52 N

4 52

48 52 N

Q 48 52 N

b Calculate the probability that Ilham selects only one queen.

Create a strategy Use the tree diagram from part (a).

Apply the idea Let Q represent the outcome of drawing a queen. Let N represent the outcome of not drawing a queen. Ilham selects only one queen in the outcomes QN and NQ. Add the probabilities of each outcome Multiply along the branches of the tree diagram to calculate probabilities

Evaluate the multiplications

Evaluate the sum

Simplify

Reﬂect and check Notice that N is the same as Q′ since P (Q) + P (N ) = 1.

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Example 2 Noive selects two letters from the word PEPPER, without replacement. a Is the outcome RR possible?

Create a strategy

Apply the idea

Look at the number of Rs left after one R has been selected.

No, the outcome RR is not possible. There is only one R, and letters are being selected without replacement, so once one R is selected first, there are none available for the second selection.

b Find the probability that the letters P and R are selected, in that order.

Create a strategy Create a tree diagram with the correct probabilities on each branch, by considering the number of each item (and the total number of items) at each stage of the experiment.

Apply the idea For the first selection, there are 6 letters in total, with 3 of the letter P, 2 of the letter E and 1 of the letter R. For the second selection, there are 5 letters remaining: • If a P was selected first, there are 2 of the letter P, 2 of the letter E and 1 of the letter R. • If an E was selected first, there are 3 of the letter P, 1 of the letter E and 1 of the letter R. • If an R was selected first, there are 3 of the letter P, 2 of the letter E and none of the letter R.

2 5

P 2 5

1 5

3 6 2 6

3 5

1 5 3 5

1 6 R

E R P

1 5

E R P

2 5

Multiply along the branches of the tree diagram

Evaluate

Simplify

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c Find the probability that the letters P and R are selected, in any order.

Create a strategy Use the tree diagram from part (b).

Apply the idea Probability = P(PR) + P(RP ) From part (b) we know that P(PR) =

Sum the probabilities of the outcomes .

Multiply along the branches of the tree diagram

Evaluate

Simplify

Substitute the probabilities

Evaluate

Simplify

Reﬂect and check In general, order does not matter when we are making selections, so P ( AB) = P (BA).

Example 3 Luke plays three tennis matches. In each match, he has a 60% chance of winning: first match

second match

60%

40%

60%

40%

60%

40%

60%

40%

W W 60%

60%

40%

60%

40% L

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a Find the probability that he will win all his matches.

Create a strategy Multiply the probabilities along the branches of the tree that lead to the event that he will win all three matches.

Apply the idea Based on the tree diagram, the probability that he will win his ﬁrst match is 0.6. The probability that he will win his second match is 0.6. The probability that he will win his third match is 0.6. 60%

40%

60%

40%

60%

40%

60%

40%

W 60% W 60%

40%

60%

40% L

The probability that Luke will win all his matches is: Probability = 0.6 × 0.6 × 0.6

Multiply all the probabilities

= 0.216

Evaluate the multiplication

b Find the probability that he will lose all his matches.

Create a strategy Multiply the probabilities along the branches of the tree that leads to the event that he will lose all three matches.

Apply the idea Based on the tree diagram, the probability that he will lose his ﬁrst match is 0.4. The probability that he will lose his second match is 0.4. The probability that he will lose his third match is 0.4. 60%

40%

60%

40%

60%

40%

60%

40%

W 60% W 60%

40%

60%

40% L

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The probability that Luke will lose all his matches is: Probability = 0.4 × 0.4 × 0.4

Multiply all the probabilities

= 0.064

Evaluate the multiplication

Reﬂect and check This is the least likely outcome as it is the product of all of the least likely outcomes at each stage.

c Find the probability that he will win more matches than he loses.

Create a strategy Use the fact that Luke needs to win more matches than he loses, so he must win either 2 or 3 matches.

Apply the idea Based on the tree diagram, we need to get the sum of the probabilities of the branches that result in 2 or three wins. i.e. WWW, WWL, WLW, LWW. 60%

40%

60%

40%

60%

40%

60%

40%

W 60% W 60%

40%

60%

40% L

The probability that Luke will win more matches than he loses is: Probability = P(WWW) + P(WWL) + P(WLW) + P(LWW) = (0.6 × 0.6 × 0.6) + (0.6 × 0.6 × 0.4) + (0.6 × 0.4 × 0.6) + (0.4 × 0.6 × 0.6) = 0.648

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Idea summary A tree diagram is useful in tracking multi-stage experiments. To find the probability we can: 1. Multiply along the branches to calculate the probability of individual outcomes. 2. Add down the list of outcomes to calculate the probability of multiple options. 3. The final percentages should add to 100%, or the final fractions should add to 1. When constructing tree diagrams for experiments without replacement, we must think carefully about the number of each item left after each selection, and change our probabilities accordingly.

Multiply probabilities We have seen that we can multiply along the branches of a tree diagram to ﬁnd the probability of an outcome in a multi-stage probability experiment. Instead of drawing the whole tree diagram, we can use what the probability will be for each step, based on the previous step, and then multiply them.

Exploration These two tree diagrams show drawing two cards from a standard and well shuﬄed deck of cards. The colour of the card is noted. 26 52

R R

R 26 52

26 52

25 51

26 52

26 51

26 52

26 51

25 51

26 52 B

B 26 52

With replacement

Without replacement

Without drawing another tree diagram, how could you ﬁnd the probability of drawing three red cards?

For the experiment with replacement, which outcomes have the same probabilities? Why do you think that is?

For the experiment without replacement, which outcomes have the same probabilities? Why do you think that is?

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We can find the probability of multi-stage events by multiplying the probabilities at each stage. The probabilities at each stage might stay the same or change.

Example 4 Nino randomly selects three cards from a standard deck of 52 cards. Find the probability that he selects a jack, then a queen of the same suit, then a king of the same suit, if his selections are made: a With replacement

Create a strategy Consider the number of each item at each stage in the experiment.

Apply the idea There are 4 jacks in the deck, so the probability that Nino selects a jack ﬁrst is

No matter which suit of jack is selected on the ﬁrst pick, there is only 1 queen of the same suit, so the probability he selected this is

. The probability he selected a king of the same suit is also

Substitute

Evaluate

b Without replacement

Create a strategy Consider the number of each item at each stage in the experiment.

Apply the idea As in part (a), the probability that Nino selects a jack ﬁrst is

If that card isn’t replaced, there are now only 51 cards remaining, so the probability that he selects the queen of the same suit next is

After that selection, there are only 50 cards remaining, so the probability that he then selects the king of the same suit is

Substitute

Evaluate

Reﬂect and check You may notice that our answer here is larger than the answer to (a). This makes sense, because when each selected card is not replaced, the probability of picking any other card increases slightly.

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Example 5 A bag contains ten pieces of paper with the numbers 1 to 10. Three pieces of paper are drawn without replacement. Find the probability of: a All numbers being even.

Create a strategy Since this is without replacement, we need to ﬁnd the probability of an even number at each stage. Then we can multiply them all together.

Apply the idea Initially, there were 10 numbers and 5 of them are even: 2, 4, 6, 8, 10. For the ﬁrst draw, the probability of drawing an even would be

The second draw will only have 4 even numbers and a total of 9 numbers left. The probability of drawing an even on the second draw is . The third draw will only have 3 even numbers and a total of 8 numbers left. The probability of drawing an even on the third draw is .

Substitute

Multiply

Simplify

The probability of drawing three even numbers is

Reflect and check The actual number we select is not important, we only need to know if it is even or odd. Another way to look at this would be: We have five even and five odd numbers to start: EEEEEOOOOO, so P (Even first)

b At least one number being even.

Create a strategy There are many ways for this to happen, so adding up all of those possibilities will take a long time. Instead, we can recognise that this is the same as there being not all odd.

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Apply the idea The probability of all odds, will be the same as the probability of all evens. We have ﬁve even and ﬁve odd numbers to start: EEEEEOOOOO, so P (Odd first) Complementary events Substitute Multiply Subtract The probability of selecting at least one even number is

Reﬂect and check This outcome is highly likely.

Idea summary We can multiply the probabilities of each step of a multi-stage experiment without drawing a tree diagram. The probabilities for the second and third stages must consider what happened in the previous stages.

10.05 Practice questions Understanding 1

598

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

Are these statements true or false? a

If there are no probabilities shown on the branches of a tree diagram, then each branch is equally likely.

To ﬁnd the probability of an outcome on a tree diagram, we can add along the branch to ﬁnd its probability.

The sum of the probabilities of all outcomes is 100% or 1.

We must draw a tree diagram to determine multi-stage probabilities.

Each outcome in a sample space is always equally likely.

Alexis uses a tree diagram to represent an experiment where a fair six-sided die is rolled three times. A “win” is rolling a 6 and a “loss” is not rolling a 6. a

What probability should go on the “win” branch?

What probability should go on the “loss” branch?

Mathspace Victorian Curriculum – Year 9 mathspace.co

A fair coin is tossed and then the spinner is spun: a

Construct a tree diagram representing the situation.

Will each outcome be equally likely?

R Y

Y G

A bag has 3 green marbles, 5 pink marbles, and 2 yellow marbles. Two marbles are drawn without replacement. a

How many marbles are left after the ﬁrst draw?

How many pink marbles are left if a green was drawn ﬁrst?

How many pink marbles are left if a pink was drawn ﬁrst?

Fluency Ex 1

Ex 2

5, 6, 8–10, 12, 14–16

5, 6, 8, 9, 11–13, 15, 16

5, 7–9, 11–13, 15, 17

A bucket contains 5 green buttons and 7 black buttons. Two buttons are selected one after another from the bucket. The ﬁrst button is replaced before the second button is selected. a

Construct a tree diagram of this situation with the correct probability on each branch.

Find the probability of selecting two black buttons.

A remote community does not always have access to fresh fruit. For two days in a row, they have fresh berries. On both days, the probability being able to get raspberries (R) is 0.6 and being able to get strawberries (S) is 0.4. a

Find the probability that someone was able to get strawberries at least once.

Find the probability that the fruit available on the last day were raspberries.

Find the probability that strawberries were only available for one day.

0.6 R 0.6

0.4

S R

0.6

0.4 S

0.4

Students are given a choice of activity for free time. Of the 7 students in the room, four chose a board game (G) and three chose a book (B). Two students are selected to deliver a message to the oﬃce. a

Construct a tree diagram of this situation with the correct probability on each branch.

Find the probability that both of the students selected were playing a game.

Find the probability at least one book reader is selected.

Two marbles are randomly drawn without replacement from a bag containing 1 black, 3 green and 5 pink marbles. a

Construct a tree diagram to show the sample space.

Find the probability of drawing: i

A black marble and a pink marble, in that order.

A green marble and a black marble, in that order.

iii

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No green marbles.

2 black marbles.

A pink marble and a green marble, in that order.

vii A pink and a green marble, in any order. Ex 3

An archer has three arrows that each have a probability of arrows are shot at a target: a

Construct a tree diagram with the probability on each branch.

Find the probability that all three arrows will hit the target.

Find the probability that at least one arrow will miss the target.

Find the probability that at least one arrow will hit the target.

From a standard pack of cards, one card is randomly drawn and then put back into the pack. A second card is then drawn. Calculate the probability that: a

Ex 4

Both cards are red

Both cards are of different colours

600

At least one of the cards is a diamond

Both cards are the same colour

A number game uses a basket with 6 balls, all labelled with numbers from 1 to 6. Two balls are drawn at random. Find the probability that the ball labelled 3 is picked once if the balls are drawn: With replacement

Without replacement

A standard deck of cards is used and 3 cards are drawn out. Find the probability, in fraction form, that: a

All 3 cards are diamonds if the cards are drawn with replacement.

All 3 cards are diamonds if the cards are drawn without replacement.

Three letters are randomly selected from the word MATHEMATICAL. Find the probability that the letters M, A, and T are selected, in that order, if the selection is made: a

Ex 5

Neither of the cards are diamonds

Amelia randomly selects two cards, with replacement, from a normal deck of cards. Find the probability that:

a 13

of hitting a target. If all three

With replacement

Without replacement

There are 4 green counters and 8 purple counters in a bag. Two counters are drawn without replacement. a

Find the probability of choosing a green counter, then choosing a purple counter.

Find the probability of choosing a purple counter ﬁrst or choosing two greens.

Three marbles are randomly drawn with replacement from a bag containing 6 red, 4 yellow, 3 white, 2 black and 4 green marbles. Find the probability of drawing: a

Three white marbles

No green marbles

At least one red marble

At least one white marble

Beth randomly selects three cards, with replacement, from a normal deck of cards. Find the probability that: a

The cards are queen of diamonds, king of spades, and king of diamonds, in that order.

The cards are all black.

The ﬁrst card is a 9, the second card is a heart and the third card is red.

The cards are all hearts.

None of the cards is a 9.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Reasoning 18

18, 20

18–21

In the black or red game, you guess whether the next card is black or red. The deck of cards is re-shuﬄed every round. In this particular game, the previous 11 cards drawn were black. a

What is the probability that the next card is also black?

Would this be the case if the cards were not re-shuﬄed and just put into a discard pile? Explain your answer.

Explain how selection without replacement affects the probabilities in a tree diagram. In your answer, refer to the numerator and the denominator of the fractions for the probabilities.

4 gardeners and 3 bakers are at a craft show. Two winners are randomly selected for the large and small prize baskets. The large prize is selected ﬁrst. We want to ﬁnd the probability of a gardener winning the large basket. see that it is . Brook says that we can calculate P (GB) + P (GG), since those are the two favourable outcomes.

3 7

Who is correct? Explain.

1 2

2 3

1 3

Five six-sided dice are rolled. Explain how to ﬁnd the probability that the results are all less than 5.

Problem-solving 22, 24, 25

G 4 7

Atanasio says that we can just look at the ﬁrst branch and

1 2

23, 25, 26

23, 26–29

In a pie shop, there are 15 pies. Five are vegetarian, six are beef, and four are chicken. A customer buys two pies at random. a

What is the probability that the customer gets two vegetarian pies?

What is the probability that the customer gets at least one beef pie?

Based on historical data, it rains 19 days out of 31 in July in Bangkok, Thailand. a

Find the chance it will rain in Bangkok on the 7th July 2027.

Find the probability, to the nearest percent, that: i

It will rain two days in a row.

From a Monday to a Wednesday in July, it only rained on the Monday.

iii

From a Monday to a Wednesday in July, it rained on only one of the days.

In a school, 25% of students ride skateboards and 20% of students have dark hair. One student is selected at random. Find the probability that the student: a

Has dark hair and rides a skateboard

Has light hair and does not ride a skateboard

Has dark hair and does not ride a skateboard

Has light hair and rides a skateboard

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A coin is tossed, then the spinner is spun and either lands on A, B or C: a

Construct a probability tree diagram that correctly shows all possible outcomes and probabilities.

Find the probability of: i

Landing on tails and the spinner landing on C

Landing on tails, or the spinner landing on C, or both

B C

In a basketball free throw contest, a player has two attempts to make a free throw. If they succeed on the ﬁrst attempt, they do not take the second. A player shoots with probabilities: • First attempt: 0.6 • Second attempt: 0.8

Construct a tree diagram showing the probability of the ﬁrst two attempts either being a success or fail.

Find the probability that the player needs to make a second attempt.

Find the probability that the player misses on both attempts.

A sailor has four meal packets to choose from on her last day but only wants to eat three meals. Two of the packets are oatmeal (O) and two of them are toasted muesli (T ). She will eat three of them throughout the day, chosen at random for breakfast, lunch and dinner. The probability tree shows the options for the day. a

Fill in the probabilities matching the edges of the probability tree: A

O A

T O

? T

D E

? T

Find the probability that she has the same type of meal for breakfast and lunch.

The probability of Mary hitting the bullseye in a game is 0.97. If Mary plays 13 games in a tournament, what is the probability that Mary misses the bullseye at least once in the whole tournament? Round your answer to the nearest percent.

A gardener has a rectangular ﬂower bed divided into two sections, A and B. In section A, the probability of a rose growing is , while the probability of a daisy growing is . In section B, the probability of a rose growing is , while the probability of a daisy growing is . The gardener randomly picks a ﬂower from anywhere in the garden. a

Let x be the probability of drawing a ﬂower from section A. Draw a tree diagram to represent the situation.

If the probability that a daisy is picked is determine the areas of sections A and B.

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, and the total area of the garden is 100 m2,

INVESTIGATION

Simulation of compound events Investigate the relationship between the probability of simple events and related compound events using simulations.

Objectives To describe the relationship between simple events and related compound events. To compare theoretical probability to experimental probability. To use digital tools to conduct experiments and analyse results.

Simple and compound events A simple event is an event with a single outcome. For example: • Flipping a fair coin and getting tails. • Rolling a standard die and getting an odd number. • Drawing a card from a standard deck and getting a queen. A compound event is an event that consists of two or more simple events. For example: • Tossing a coin twice and getting heads. • Drawing two cards and getting a diamond. • Tossing a coin and getting heads, and rolling a die and getting a 6. Heads

Tails

Probability Tossing a fair coin is an example of a simple event. Complete the table by writing the probability of each outcome. 1

Probability Rolling a standard die is an example of a simple event. Complete the table by writing the probability of each outcome. Tossing a fair coin and rolling a standard die is an example of a compound event • Create a list or array of the possible outcomes. • Then, create a table that shows the probability of each outcome. What is the relationship between the total number of outcomes for the simple events and the total number of outcomes for the compound event?

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Investigate Use your tables to answer these questions: 1.

When rolling a standard die, what is the probability of rolling an even number?

When tossing a coin and rolling a die, what is the probability of tossing a tails and rolling an even number?

What is the probability of tossing a tails or rolling an even number?

If you were to toss a coin and roll a die 30 times, how many times would you expect to toss a head and roll an even number?

In your own words, explain the difference between a simple event and a compound event.

Simulation of compound events Now that we have looked at the theoretical probabilities of simple and compound events, we will use a simulation tool to explore the experimental probabilities of compound events.

Interactive exploration Discover this concept in action online

mathspace.co

Investigate

604

Based on your simulations, what is the experimental probability of tossing a head and rolling an even number on the die? How does it compare to the theoretical probability?

Do the outcomes of the experiment match what you expected? Why or why not?

If you were to simulate 100 trials instead of 30, how do you think the experimental probabilities would compare to the theoretical probabilities? Would they be closer or farther away? Explain your reasoning.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Guess the game! In this activity, we will consider three different games: • Game A: Rolling one die. If you roll a 1 or a 2, the outcome is a 0. If you roll a 3 or a 4, the outcome is a 1. If you roll a 5 or a 6, the outcome is a 2. • Game B: Rolling one die. If you roll a 1, 2, or 3, the outcome is a 0. If you roll a 4 or 5, the outcome is a 1. If you roll a 6, the outcome is a 2. • Game C: Tossing two coins. The outcomes 0, 1, and 2 represent the number of heads. Construct three tables, one for each of the three games, showing the theoretical probabilities of each outcome.

Interactive exploration Discover this concept in action online

mathspace.co

Investigate 1.

Explain how you determined which game was played based on the simulation results.

How did the number of times a game was played change the results? Did changing the number of games make it easier or harder to decide which game was played?

Discussion 1.

How do the probabilities of simple events help us understand the probabilities of compound events?

How did the results of the probability simulations compare to your initial expectations about the likelihood of different compound events? How might increasing the number of trials affect the accuracy of your conclusions?

How does the concept of compound probability extend our understanding of probability in real-world situations, like sports, game design, statistics, or decisionmaking? Give an example.

Design and conduct your own repeated chance experiments and simulations. Investigate the results of an event that involves replacement and compare it to the results of an event that does not involve replacement. How can you use digital tools to simulate outcomes?

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10 Chapter review 1

Jarvis rolled a 6-sided die 60 times. a

How many times would he expect a 4 to appear?

After rolling the die 60 times, he recorded that 4 had appeared 20 times. What is the experimental probability of rolling a 4?

Do you think this is a fair die? Explain.

A French bakery tracks the number of each type of macaron sold in one week: a

How many macarons were sold in total?

Calculate the experimental probability of a randomly selected customer purchasing a:

Raspberry macaron

Macaron type Chocolate Lemon (dairy-free) Raspberry Salted caramel Vanilla (dairy-free) 3

606

Frequency 204 51 72 191 82

Anju has a blank 12-sided die. On each side, she writes a 1, 2, 3, or 4. She gives the die to four friends who each roll it several times. The results are presented in this table: Briana Clinton Desi Elijah

Dairy-free macaron

1 5 9 22 25

2 6 4 19 27

3 1 2 3 5

4 2 1 4 6

How might Anju have labelled her die? Explain.

The die has regular pentagonal faces, each with an area of 208.2 mm2. Using your answer from part (a), what surface area of the die has numbers that are perfect squares?

A manager at a marketing agency needs to send one employee from the team to a conference in Paris, France. She creates a Venn diagram showing different skills that her team has. A suitable team member to send will be able to speak French and code with React JSX.

208.2 mm2

Speaks French

Codes with React JSX

How many team members are there?

What is the probability that a randomly selected team member can speak French or code with React JSX?

How many suitable team members are there?

The manager is told that a team member was selected by her supervisor at random. What is the probability that the selected team member is suitable?

Mathspace Victorian Curriculum – Year 9 mathspace.co

A group of teachers were asked if they have a pet and if they have ever gone out of state. A partially completed two-way table of the results is shown. a

Find the missing values.

If a random teacher is randomly selected, ﬁnd the probability that:

6 50

The teacher does not have a pet and has never been out of state.

iii

The teacher has a pet or has been out of state.

The teacher has a pet if we know they have been out of state.

This Venn diagram and two-way table are both used to represent the same sample space. Complete the Venn diagram and the two-way table. Likes yeast spread

Likes yeast spread

Does not like yeast spread

Australian resident Not Australian resident Total

16 68

1 2 3 4

What is the probability that the difference is: i

less than 2

iii

Total 49

A player rolls 2 four-sided dice and the total positive difference is recorded in an array. a

Total 54

The teacher has a pet.

Pet No pet Total

Australian resident

Has been Has not out of been out state of state

more than 1

Find the sum of parts ii and iii. Explain that sum.

1 0 1 2 3

2 1 0 1 2

3 2 1 0 1

4 3 2 1 0

One fair 4-sided and one fair 6-sided die are rolled at the same time. Their product is recorded. a

Construct an array to represent all possible outcomes.

Find the probability of an even product.

Find the probability of rolling a 3 on the 4-sided die and an even product.

Find the probability of rolling a multiple of 3 or an even number.

Two dice are rolled, and the two numbers are recorded. Is each statement true or false? Justify your answer.

1 2 3 4 5 6

1 1, 1 2, 1 3, 1 4, 1 5, 1 6, 1

2 1, 2 2, 2 3, 2 4, 2 5, 2 6, 2

If the two numbers are added, a sum less than or equal to 5 and a sum greater than 5 are equally likely.

If the two numbers are multiplied, an even product is more likely than an odd product.

If this was drawn as a tree diagram, it would have 36 branches.

3 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3

4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4

5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5

Chapter 10 review mathspace.co

6 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6

607

Demetria teaches dance and gives students a sticker after class. She has one star, one unicorn, and one tractor left. Three students take a sticker, one after the other. a

Is this experiment with or without replacement?

List the sample space of the possible ways the stickers were handed out.

Four cards labelled 3, 5, 6, and 8 are placed face down on a table. Two of the cards are selected randomly without replacement to form a two-digit number. a

Construct a tree diagram of this situation.

List the sample space for the possible two-digit numbers formed.

What is the probability of the resulting number being: i

Even and a multiple of 5

Even or a multiple of 5

A calming corner has 6 cuddly toys and 4 fidget spinners. One toy is selected, used, and replaced. Then another toy is selected and used. a

Construct a tree diagram of this situation with the correct probability on each branch.

Find the probability of selecting one of each type of toy.

A laundry basket has 3 black, 2 white and 6 red shirts. There is only room for two more shirts in the washing. Two shirts are selected at random. a

Construct a probability tree diagram to show the sample space.

Find the probability of drawing: i

2 red shirts.

2 white shirts.

iii

Two shirts of the same colour.

Two different coloured shirts.

A red shirt, then a white shirt in that order.

Exactly one red shirt.

Alice is hatchet throwing. Each of her three hatchets has a probability of of hitting the bullseye. If all three hatchets are thrown at a target:

Construct a tree diagram with the probability on each branch.

Find the probability that all three hatchets will hit the bullseye.

Find the probability that at least one hatchet will miss the bullseye.

Find the probability that at least two hatchets will hit the bullseye.

Hasan randomly selects three cards from a normal deck of cards. a

608

If he draws the cards with replacement, ﬁnd the probability that: i

The cards are the ace of diamonds, the two of hearts, and the three of spades in that order.

The cards are all red.

iii

The cards are all diamonds.

None of the cards are queens.

An ace, a king, and a queen, all of the same suit, in that order.

Mathspace Victorian Curriculum – Year 9 mathspace.co

If he draws the cards without replacement, ﬁnd the probability that: i

The cards are the three of hearts, the four of diamonds, and the ﬁve of spades, in that order.

The cards are all black.

iii

At least one card is black.

The cards are all black or diamonds.

Five six-sided dice are rolled. Explain how to ﬁnd the probability that less than two 4s were rolled. Give your answer as an exact value.

When Lila goes out to dinner, she needs to choose a restaurant and a meal, chicken or duck. For restaurants, she is half as likely to choose Arjun’s as Benny’s. If she goes to Arjun’s, she is four times more likely to order duck than chicken. If she goes to Benny’s, the probability she orders chicken is x. a

Draw a tree diagram to represent the situation

At both restaurants, the chicken costs $20 and the duck costs $30. The probability of her spending $20 is

. Find the probability that Lila chooses duck at Benny’s.

What is the expected cost of Lila’s meal to the nearest cent?

Did you know?

The Birthday Paradox: In a group of just 23 people, there’s a 50% chance that two people share the same birthday! This is known as the “Birthday Paradox.” Though it seems unlikely, probability shows that the chance of two matching birthdays rises quickly as the group gets larger. This is because as more people are added, there are many more pairs of birthdays to compare, making it more likely that a match will occur. It’s a great example of how our intuition about probability can sometimes be very different from reality! Chapter 10 review mathspace.co

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Big ideas • Understanding and analysing data, along with knowing the right measures of centre and spread help us make better conclusions and informed decisions. • Different tools like stem-and-leaf plots, frequency tables, and histograms help us organise and better understand data.

Curriculum Outcomes VC2M9ST01 analyse reports of surveys in digital media and elsewhere for information on how data was obtained around everyday questions and issues involving at least one numerical and at least one categorical variable, to estimate population means and medians VC2M9ST02 analyse how different sampling methods, and different samples using the same method, can affect the results of surveys and how choice of representation can be used to support a particular point of view VC2M9ST03 represent the distribution of multiple data sets for numerical variables using comparative representations such as back-to-back stem-and-leaf plots and histograms; describe data, using terms including ‘skewed’, ‘symmetric’ and ‘bi-modal’; compare data distributions using mean, median and range to describe and interpret numerical data sets with consideration of centre, spread and shape, and the effect of outliers on these measures VC2M9ST04 choose appropriate forms of display or visualisation for a given type of data; justify selections and interpret displays for a given context VC2M9ST05 plan and conduct statistical investigations involving the collection and analysis of different kinds of data; report findings and discuss the strength of evidence to support any conclusions

11 Statistics Chapter outline 11.01 Types of data 11.02 Data displays Investigation: Opinion Polls 11.03 Back-to-back stem and leaf 11.04 Grouped histograms 11.05 Shape of data Investigation: Statistical reports in the media 11.06 Compare data sets Investigation: Plan and conduct statistical investigations Chapter 11 review

612 619 634 636 644 658 668 670 685 689

Statistics help scientists study animal populations. They tag animals to monitor their movements and patterns to determine the health of the whole population.

11.01

Types of data

After this lesson, you will be able to… • distinguish between the types of data. • understand census and sampling in data collection. • identify the importance of representative data and the risks of bias. • apply strategies to avoid bias in data collection and interpretation.

Types of data Univariate data is the word commonly used in statistics to refer to data that consists of observations on a single characteristic. Data can be classified into two types, categorical data and numerical data. Numerical data is data that can be counted, ordered and measured. Categorical data is non-numerical, it describes the qualities or characteristics of a data set. Data that can be broken into groups or categories is called categorical data or qualitative data. Categorical data can be divided into two categories: • Nominal data involves categories that have no natural order. For example, if you asked someone what their favourite subject is, they might say “Maths” or “English”. • Ordinal data involves categories that have a natural order. For example, if you were tracking the position of people coming in after a race you would have “first”, “second”, etc. Things that have a natural order. Data that can be counted or measured is called numerical data or quantitative data. Numerical data can be divided into two categories: • Discrete data involves distinct or separate values. The data cannot take “in between” values. For example, if you were looking at the numbers of people that had entered a building over time, the responses could be 5 or 6, but we would not say 5.5, as we can’t have half a person. If you have to count to collect each piece of data, the data is discrete. • Continuous data is measured, so the values are not distinct. Within the range of possible values a piece of data could be any decimal or fractional number. If we measure the length of a person’s foot, we might get any decimal value, limited only by the precision of our ruler. The flowchart shows us the different types of data and sub-categories: Data

Numerical Made of numbers

Continuous Measurable

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Discrete Countable

Mathspace Victorian Curriculum – Year 9 mathspace.co

Categorical Made of categories

Nominal Doesn’t have any ordering

Ordinal Elements are ordered

Example 1 Which one of the following data types is discrete? A Your height

The time it takes to swim 200 metres

C Daily temperature

The number of pets in your family

Create a strategy

Apply the idea

Choose the option which can be counted (involves distinct or separate values).

The correct answer is option D: The number of pets in your family.

Example 2 Classify each data as ordinal, nominal, continuous or discrete. a A class survey about where they went on their most recent holiday

Create a strategy

Apply the idea

Determine whether the answers would be words Names of places are words. So it’s categorical, or numbers. If they would answer with a word, and as there is no natural order, it is nominal. the data is categorical. If they would respond with a number, the data is numerical. b Weights of kittens

Create a strategy

Apply the idea

Determine whether the answers would be words The weight of a kitten can be measured. So or numbers. If they would answer with a word, it is numerical. Weight is a measurement that the data is categorical. If they would respond can have any number of decimal places, so it is with a number, the data is numerical. continuous.

Idea summary The ﬂowchart summarises the different types of data and sub-categories: Data

Numerical Made of numbers

Continuous Measurable

Discrete Countable

Categorical Made of categories

Nominal Doesn’t have any ordering

Ordinal Elements are ordered

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Data collection A census is when data is collected from a whole population. A sample is when data is collected from a part of a population. For example, if we wanted to know whether or not students at the school liked their uniform, we could survey every student in the school (a census). However, since this would be difficult and time consuming, we may choose to survey a smaller group of students (a sample). When we take a sample it is important that it is representative of the population. This means that the sample should reflect the characteristics of the entire group. For example, if we only survey Year 10 boys, this wouldn’t represent all students, since the school has different year levels and both boys and girls. If a survey gives unreliable results, it’s often because the data is biased. To avoid bias in a survey, here are some things to keep in mind: • Who is in the sample? Make sure it’s a good representation of the whole population. • How many people are surveyed? The more people you survey, the more accurate the results will be. After thinking about who you ask and how many you ask, the next thing to consider is how you ask the questions. • A leading question suggests a particular answer or influences how someone responds. For example, asking “Do you watch the most popular sport, soccer?” may push someone to answer “yes” because the question makes it seem like soccer is the correct choice. A better question would be, “Do you watch soccer?” • A loaded question is a question that contains an assumption or forces someone to answer in a way that might not be fair. For example, asking “How often do you skip practice?” assumes the person skips practice regularly, which may not be true and is emotive as skipping practice sounds bad. A better approach would be, “Do you attend practice regularly?” By avoiding leading and loaded questions, we can get more reliable and unbiased results.

Example 3 Yvonne, Lachlan and Tricia are investigating the favourite sports of students at their school. In each case, consider the sample and the survey question to determine whether the results are likely to be biased or not. a Yvonne asks students on the school soccer team, “What’s your favourite sport?”

Create a strategy Consider the following: • Who is being surveyed? • How many people are being surveyed? • Whether the question being asked actually addresses the question at hand. • Whether the question is leading.

Apply the idea The members of the soccer team will be more likely to answer soccer since they play it (the sample is not representative of the population). So the results are likely to be biased.

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b Lachlan randomly selected students from the school and asked, “What’s your favourite sport?”

Apply the idea Lachlan randomly selected students so the sample should be representative of the population. The question is not leading. The results are not likely to be biased.

c Tricia randomly selected students from the school. To each student she said, “The local AFL team is donating money to our school this term. What’s your favourite sport?”

Apply the idea The question is leading – because it mentions the donation from the local AFL team, students may feel pressured to choose AFL. So the results are likely to be biased.

Idea summary A census is when we collect data from a whole population. A sample is when we collect data from part of a population. Bias can affect the results of the data collection, or appear in the presentation of the data after collection. Consider: • • • •

Who is being surveyed? How many people are being surveyed? Whether the question being asked actually addresses the question at hand. Whether the question is leading.

11.01 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

True or false: a

Nominal data is a type of numerical data.

Numerical data can be discrete or continuous.

Qualitative data involves words or categories.

Quantitative data is another word for categorical data.

What is the difference between a sample and a census?

What are 3 ways we can reduce bias when collecting data?

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Fluency 4

Ex 1

Ex 2

616

4–13

Classify each data set as numerical or categorical. a

Amount owing on layby

Types of vegetables

Brands of tablets

Types of folders

Colours of folders

Brands of phones

Maximum snowfall

Daily UV index

Types of dogs

Favourite colours

Cost of a computer game

Types of fruits

For each of these categorical data types, identify whether they are nominal or ordinal: a

Ice cream flavour

Smartphone application

c e

Clothing size

Blood type

Customer satisfaction rating

Pain level

Membership rank

Cuisine preference

For each of these numerical data types, identify whether they are discrete or continuous: a

Number of classrooms at a school

Daily humidity

The ages of a group of people

The time taken to run 200 metres

The number of people taller than you

The number of coins in a coin collection

The amount of snowfall in a city

For each survey question, classify the data as ordinal, nominal, discrete or continuous. a

How far do you travel to school each day?

How many siblings do you have?

What modes of transport do you use to get to school?

What kinds of pets do you own?

How many pets do you own?

How tall are you?

How much does your schoolbag weigh?

Are these survey questions biased or fair? a

Do you want soup for lunch or a sandwich?

What do you do on a Sunday morning?

Do you think the government should be allowed to cut down some of the oldest trees in the area to construct a metro railway line in the city?

Do you eat at least the recommended number of servings of fruits and vegetables to ensure a healthy and long life?

Identify the type of bias that could be present in these biased survey questions: a

Do you prefer the natural beauty of hardwood floors in your home?

Do you feel that the TV news is an inaccurate portrayal of life’s problems?

Don’t you think this newspaper is biased?

Do you prefer the look and feel of thick lush carpet in your living room?

Mathspace Victorian Curriculum – Year 9 mathspace.co

Ex 3

Determine whether each question is likely to give biased results or not. If you think results will be biased, state why. a

Damon asked four of his friends, “What is your favourite subject at school?”

Lulu is interviewing people outside of a train station. She asks, “Should the government spend more money on public transportation?”

Lys is interviewing people about how shark nets are affecting shark attack numbers. He says, “Shark nets cost $1 000 000 per year and barely help. Do you support shark nets?”

After the government decided to increase the minimum retirement age, a survey is to select a group of people to ask their opinions on the changes. Should each of these groups be included in the survey?

Politicians and policy makers

Members in the community that are directly impacted by the changes

Members in the community not impacted by the changes

A radio station conducts a poll asking its listeners to call in to say if they are for or against restrictions on scalpers selling tickets for gigs at a higher price. State whether these are reasons why this is not an appropriate way to conduct a poll:

A large variety of people are likely to call.

A person can call more than once, so they could be counted more than once.

People with stronger views are more likely to call than those who don’t have a strong view.

A question is inappropriate for a survey given it fulﬁlls any of the following criteria: • Question is unclear • Emotional language • Question is too personal Using the criteria list, determine why each of the questions are inappropriate: a

Many people have worked incredibly hard and even died making this bridge. Do you like the bridge?

How many large electronic devices are in your home?

How many people have died in your family?

Reasoning 14

14–16

14–17

Explain why each survey question would give unreliable results? Rewrite each question to provide more reliable results. a

The Prime Minister believes that taxes are too high, do you think taxes are too high?

Do you have a Snapchat or Twitter account?

Do your siblings ever drive you to school?

Do you often waste time on social media?

What brand of shoe polish do you use every day?

Explain why it might be important to collect both categorical and numerical data in a survey about dietary habits.

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A city council wants to understand the commuting habits of its residents. What types of data should they collect and why?

A study claims that 75% of students prefer online learning. However, the survey was conducted by interviewing only 30 students from a single school. Comment on the reliability of this study and how the study might be improved.

Problem-solving 18–20 18

18, 20, 21

18, 20–22

Patricia surveys her class about their favourite music. Her results are shown in the table: a

According to the survey, which genre is most popular among her class?

This was her survey question:

Genre No. students

Country 15

Pop 2

“The coolest kids like country music, and nobody likes pop. Do you like country music or pop music?” Do you trust the results of Patricia’s survey? Explain your answer. 19

You are part of a school council and you are tasked with ﬁnding out the most popular afterschool activities among students. a

Describe how and where you would carry out this survey, including which groups of people you would focus on.

What type of data would you be collecting?

How would you ensure that your data collection methods are unbiased and accurate?

You are a researcher studying the relationship between hours of sleep and academic performance among high school students. a

What type of data would you need to collect to investigate this?

How would you go about collecting this data?

A clothing company wants to understand their customer’s shopping habits better to improve their sales. a

What kind of data should they collect?

How could this data help improve their business strategy?

A company is conducting a survey to determine the average income of people in a city. They randomly select 50 employees from a local technology company and 50 employees from a small retail store to participate in the survey. The company ﬁnds that the average income of employees from the technology company, T, is $60 000, while the average income of employees from the retail store, R, is $30 000. However, the technology company mainly employs professionals with higher salaries, while the retail store employs a mix of part-time and full-time workers.

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Write an equation for the overall average income, A, of all 100 employees surveyed, in terms of T and R. Simplify if possible.

Explain why using the overall average income, A, in this case might lead to a biased representation of the city’s average income.

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11.02

Data displays

After this lesson, you will be able to… • understand the concept of univariate data and its representation through frequency tables, stem-and-leaf plots and dot plots. • apply knowledge of measures of centre and spread to univariate data displays. • calculate the mode, mean, median and range from frequency tables, stemand-leaf plots and dot plots.

Frequency tables Frequency tables display univariate data. We can ﬁnd the mode, mean, median and range from a frequency table. The frequency table shows the number of goals scored in a football game, and the number of games that had that score (the frequency). Number of goals scored (x) 0 1 2 3 4 5 Total

Frequency ( f ) 4 11 5 1 3 1 25

The range is a measure of the spread of scores. We find the range by subtracting the lowest score from the highest. In the table, the range is the highest – lowest = 5 – 0 = 5 The mode is the most frequently occurring score in the data. We can find the mode by identifying the score with the highest frequency in the table. In the table, the mode is the most frequent score which is 1 goal, with a frequency of 11. The mean is the average score of the data set. We find the mean by summing all the scores and dividing by the total number of scores. Let’s add a column to the frequency table for fx, the frequency ( f ) multiplied by the score (x) for each row. Number of goals scored (x) 0 1 2 3 4 5 Total

Frequency ( f ) 4 11 5 1 3 1 25

fx 0×4=0 1 × 11 = 11 2 × 5 = 10 3×1=3 4 × 3 = 12 5×1=5 41

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The total for the fx column, gives us the total number of goals scored in the 25 games played. To ﬁnd the mean, we divide the total number of goals scored (total of the fx column) by the amount of games played (total of the frequency column).

In this case, the mean is

= 1.64.

The median is the middle score when all the scores are arranged in order. To help with the median we can add a column called the cumulative frequency. This column is the sum of the frequencies of the score and each of the scores above it. The first row is the frequency of the first row. The second row is 11 more than the one before it. The third row is 5 more than the one before that, and so on for the full table. Number of goals scored (x) 0 1 2 3 4 5 Total

Frequency ( f ) 4 11 5 1 3 1 25

Cumulative frequency 4 4 + 11 = 15 15 + 5 = 20 20 + 1 = 21 21 + 3 = 24 24 + 1 = 25

The last row of the cumulative frequency should be the same as the total frequency. If we look at the score of 2 goals row, we’ll see that the cumulative frequency is 20. That means 20 games scored 2 goals or less. In a list of ordered scores, the median is found by selecting the middle score. In a frequency table, the position of the median can be found by adding 1 to the total frequency and then halving it.

In this example, the total frequency is 25, so the position of the median is median is the 13th score.

, that is, the

We can see that scores of 0 took up the first 4 positions, then scores of 1 took up the 5th to 15th positions. As the 13th position is found in this range, the median is 1 goal. If the median is between 2 positions, we would need to find the mean of the 2 scores. For example, if the median position is 12.5 then we need to calculate the mean of the 12th and 13th scores. We call the mean, median, range, mode measures of centre and spread.

Example 1 For the following data set: 27, 25, 24, 24, 24, 24, 24, 25, 23, 24, 26, 23, 27, 23, 24, 27, 23, 23, 27, 25, 24, 24, 27, 25, 23

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a Complete the frequency distribution table: Frequency ( f )

Score (x) 23 24 25 26 27 Total

Create a strategy Calculate frequency by counting score occurrences, multiply score (x) and frequency ( f ) for the fx column, and evaluate column totals for ﬁnal results.

Apply the idea Here’s the complete table: Frequency ( f ) 6 9 4 1 5 25

Score (x) 23 24 25 26 27 Total

fx 23 × 6 = 138 24 × 9 = 216 25 × 4 = 100 26 × 1 = 26 27 × 5 = 135 615

b Calculate the mean, rounded to one decimal place.

Create a strategy Use the formula

Apply the idea Substitute the values in part (a)

Evaluate

c Find the range.

Create a strategy Subtracting the lowest score from the highest score.

Apply the idea The highest score is 27 and the lowest score is 23. Range = 27 – 23 =4

Subtract the scores Evaluate

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d Find the mode.

Create a strategy

Apply the idea

Look at the table in part (a) and identify the score with the highest frequency.

24 has a frequency of 9, which is the highest frequency of all the scores. So the mode is 24.

e Find the median.

Create a strategy To ﬁnd the median we need to add a cumulative frequency column, then use the formula for median position.

Apply the idea Score (x) 23 24 25 26 27 Total

Frequency ( f ) 6 9 4 1 5 25

Cumulative frequency 6 6 + 9 = 15 15 + 4 = 19 19 + 1 = 20 20 + 5 = 25

Write the formula

Substitute the total frequency

Evaluate

The median will be 13th score. Looking at the cumulative frequency column of the table, the 13th score is 24 so the median is 24.

Construct a column graph of the data.

Create a strategy Use the values we found in part (a).

Apply the idea

Frequency ( f )

Based on the frequency distribution table, the column graph should look like:

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10 9 8 7 6 5 4 3 2 1 0

23 24 25 26 27 Score (x)

Idea summary When ﬁnding measures of centre and spread from a frequency table: • The range is found by subtracting the lowest score from the highest score. • The mode is the score with the highest frequency. • The cumulative frequency is the sum of the frequencies of the score and each of the scores above it. Adding a cumulative frequency column to a frequency table helps us ﬁnd the median. • The position of the median score can be found with the formula:

• Adding an fx ( frequency times score) column to a frequency table is helpful for ﬁnding the mean. The mean can be found with the formula:

Stem-and-leaf plots A stem-and-leaf plot is a univariate data display, where scores are split using place value. The ﬁrst digit or digits in each value become the ‘stem’ and the last digit becomes the ‘leaf’. • The stems are arranged in ascending order, to form a column, with the lowest value at the top • The leaf values are arranged in ascending order from the stem out, in rows, next to their corresponding stem (each stem can have more than one leaf) • There are no commas or other symbols between the leaves, only even spacings

Score of 10

Score of 56

Stem

Leaf

Key: 2 | 1 = 21

8 9 All of these are in the 40s

When ﬁnding measures of centre and spread from a stem-and-leaf plot: • The range for the stem-and-leaf plot is the smallest score subtracted from the largest score. In the above example, it is 58 − 10 = 48 • The mode is the most common score, not to be confused with the most common leaf. In this case, there are two modes, 35 and 41. • The median is the middle score. In the stem and leaf above, the median is 36. • The mean is the sum of all the scores divided by the number of scores. The number of scores can be found by counting the number of leaves (don’t count the stems). In the above, the mean is

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Example 2 A cyclist measured his heart rate immediately after ﬁnishing each event in which he competed. The results are recorded in a stem and leaf plot. Stem

Leaf

4 4 4 6 8 8

0 0 4 6 8

0 2 3 3

2 3 4 7 7 8

Key: 16|1 = 161 a What was the difference between his slowest and fastest heart rate?

Create a strategy Subtract the smallest leaf on the lowest stem from the largest leaf on the highest stem.

Apply the idea The slowest recorded heart rate is 161 and the fastest recorded heart rate is 198. Difference = 198 – 161 = 37 beats per minute

Subtract the heart rates Evaluate

Reﬂect and check As a reminder, always remember to add the appropriate unit to your ﬁnal answer.

b For each extra heartbeat per minute, it takes 7 seconds longer for him to recover. How much longer would it take to recover when he ﬁnishes with the fastest heart rate than when he ﬁnishes with the slowest heart rate?

Create a strategy Multiply the result from part (a) by 7.

Apply the idea Recovery time = 37 × 7 = 259 seconds

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Multiply the heart rates Evaluate

c Find the median.

Create a strategy The median is the middle score in an ordered set of scores. In a stem-and-leaf plot, the scores are already ordered.

Apply the idea There are 23 scores in the data (the number of leaves gives the total count of scores). To find the middle score, we add 1 and then halve. This is the 12th position. Total scores = 23 Median position = 23 + 1 2 = 12

Add 1 to the total scores and then halve Evaluate

Count from the bottom to find the 12th score. The median is 178. Stem

Leaf

0 0 4 6 8 8th 9th 10th 11th 12th

7 scores

Key: 16 | 1 = 161

Idea summary In a stem-and-leaf plot, the ﬁrst digit or digits in each value become the ‘stem’ and the last digit becomes the ‘leaf’.

Dot plots In a dot plot, each dot represents one data point. The dot plot displays the number of goals scored in a game, where each game is one dot.

1 2 3 4 Goals scored

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We can use a dot plot to ﬁnd measures of centre and spread: • The mode is the score with the most dots. In the above, there are two modes, 1 and 4. • The range is the smallest score with a dot above, subtracted from the highest score with a dot above. In this case, it is 5 − 0 = 5 • The median is the middle score. However, you should remember that each dot represents a score. In the above dot plot, the middle dot is above the score of 3, so the median is 3. • The mean is the sum of all the scores divided by the number of scores (the number of dots). In this case it is

(to one decimal place).

Example 3 The number of ‘three-pointers’ scored by a basketball team in each game of the season is represented in the dot plot. A ‘three-pointer’ is worth 3 points. 0

1 2 3 4 5 Three-point shots

a What was the total number of points scored from three-pointers during the season?

Create a strategy Multiply the number of dots with their respective three-point shots and sum the products. Then, multiply the sum by 3.

Apply the idea Looking at the dot plot, we have: Total no. of points = (0 × 4 + 1 × 3 + 2 × 1 + 3 × 4 + 4 × 1 + 5 × 2 + 6 × 4) × 3

Multiply the number of dots by threepoint shots, add them, then multiply by 3

= 55 × 3 Evaluate inside the brackets = 165

Evaluate

b What was the average number of points scored from three pointers each game of the season? Round to two decimal places if necessary.

Create a strategy Divide the total number of points by the total number of games.

Apply the idea We have the 19 dots representing the total number of games, and we found 165 as the total number of points in part (a). Remember to include the games with zero ‘three-pointers’ scored. Average no. of points = = 8.68

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Divide 165 by 19 Evaluate

c Find the median.

Create a strategy The median is the middle score in an ordered set of scores. In a dot plot, the scores are already ordered.

Apply the idea There are 19 scores in the data (the number of dots gives the total count of scores). To find the middle score, we add 1 and then halve. This is the 10th position. Total scores = 19 Median position = 19 + 1 2 = 10

Add 1 to the total scores and then halve Evaluate

Count from the bottom to find the 10th score. The median number of three-point shots is 3. 8 scores

10th 9th 0

4 2 3 Three-point shots

Idea summary In a dot plot, each dot represents one data point, belonging to the value that it is placed above.

11.02 Practice questions Understanding

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

When might a frequency table help in simplifying the calculation of the mean of a data set, compared to working only with individual data points?

What type of data is best represented using a stem-and-leaf plot?

For a dot plot, describe how we determine the frequency of each score.

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4ac, 5, 6ab, 7, 9, 10, 12, 13

Fluency 4 Ex 1a

Ex 1d

4cd, 5, 6ac, 8, 9, 11, 13, 14

Complete the following frequency tables: a

Score Frequency Cumulative fx (x) (f ) frequency 1 7 2 4 3 7 4 8 5 8 6 8 Total

Score Frequency Cumulative fx (x) (f ) frequency 5 3 10 2 15 7 20 3 25 5 30 4 Total

Score Frequency Cumulative fx (x) (f ) frequency 34 119 36 125 38 160 40 143 42 185 44 137 Total

Score Frequency Cumulative fx (x) (f ) frequency 150 12 160 15 170 11 180 10 190 16 200 14 Total

State the mode of the following data sets: a

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4bc, 5, 6a, 7, 9, 10, 12–14

Score 3 4 5 6 7 8

Frequency 3 3 4 2 5 8

Mathspace Victorian Curriculum – Year 9 mathspace.co

Score 1100 1200 1300 1400 1500 1600

Frequency 125 129 140 151 153 149

Score 250 260 270 280 290 30

Frequency 28 27 26 28 26 24

6 Ex 1

For each scenario: i

Construct a frequency table for the results.

Calculate the range.

iii

Find the mode.

Calculate the mean, rounded to two decimal places.

Find the median.

Construct a column graph of the data.

Mr. Rodriguez recorded the number of pets owned by each of the students in his class. He found that 15 people had no pets, 19 people had one pet, 3 people had two pets and 8 people had three pets.

Mrs. Brown asked her students how many bedrooms they have in their house. Their responses were as follows:

Mr. Smith asked his students how many siblings they have. Their responses were as follows: 1, 1, 1, 2, 2, 1, 0, 3, 0, 1, 2, 1, 1, 2, 2, 3, 1, 2, 3, 3, 1, 2, 2, 1, 2

4, 4, 4, 3, 4, 3, 3, 5, 5, 4, 4, 3, 3, 4, 4, 4, 3, 4, 5, 4

The set of marks for a class of students is: 81, 81, 71, 81, 81, 61, 71, 93, 71, 58, 71, 58, 61, 93, 93, 71, 61, 61, 81, 58, 93

Organise the data into a frequency table.

Find the total number of students in the class.

Find the number of students who will get a Distinction grade (80 < Score ≤ 90).

Find the number of students who will get a High Distinction grade (Score > 90).

Find the mean mark, rounded to one decimal place.

A cyclist measured his heart rate immediately after ﬁnishing each event in which he competed. The results are recorded in a stemand-leaf plot, but two values are missing: a b

9 Ex 2

Leaf 16 3 5 9 17

He can remember no number appears twice. What is the smaller missing number?

5 8

18 2 4 7 8

If both missing numbers sum to 367, what is the second missing number?

A particular genre of electronic dance music has a beat per minute (BPM) range of between 160 and 200. The BPM’s of 23 songs are shown in the stem-and-leaf plot.

0 3 ?

19 0 1

Key: 12|3 = 123 Leaf

16 1 17

4 4 4 6 8 8

0 0 4 6 8

Calculate the mean BPM of the track selection, rounded to the nearest whole number.

18 0 2 3 3

Find the median BPM.

19 1

Find the modal BPM.

Key: 1|2 = 12 beats per unit

2 3 4 7 7 8

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For this data set: 49, 49, 49, 51, 48, 47, 50, 49, 51, 50, 51, 49, 52, 47, 50, 49 51, 48, 51, 51, 50, 48, 49, 52, 49, 47, 47, 52, 52, 49, 47, 48 52, 48, 51, 51, 49, 52, 48, 48, 52, 50, 48, 47, 52, 52, 52, 48 a

Construct a dot plot that represents the data:

Summarise the data using measures of centre and spread.

A group of people were surveyed on the number of websites they visit on a daily basis. The results are displayed in the dot plot.

Number of websites visited daily

13 Ex 3

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Complete the frequency table based on the dot plot. Number of websites visited daily (x) 0 1 2 3 4 5 6 Total

Frequency ( f )

Cumulative frequency

What is the mean? Round your answer to two decimal places.

What is the median?

Consider the dot plot showing students scores: a

Complete a frequency table for the data.

How many students scored above 20?

How many students scored below 30?

What was the mode for the scores?

Find the mean score. Round your answer to two decimal places.

10 20 30 40 50 Score

The number of ‘three-pointers’ scored by a basketball team in each game of the season is represented in the dot plot. A ‘three-pointer’ is worth 3 points. a

What was the total number of points scored from threepointers during the season?

What was the mean number of points scored from threepointers each game of the season? Round your answer to two decimal places.

What was the median number of points scored from three-pointers each game of the season?

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1 2 3 4 5 Three Point Shots

The given dot plot shows the number of students in each section of the school band: a

Which section has the least members?

Find the number of people in the band.

How many more people are there in the Strings section than in the Brass section?

What is the mean number of musicians per group?

Reasoning 15

Band members

Strings Woodwind Brass Percussion

16, 17

15–18

The number of hours spent on homework per week by 30 students was collected. 2, 3, 2, 4, 3, 3, 4, 4, 3, 2, 4, 5, 6, 5, 4, 3, 2, 3, 4, 5, 6, 5, 4, 3, 2, 2, 4, 5, 6, 4. Explain how a frequency table could help to understand this data.

Alison and Sarah both found the average distance travelled in Sydney from the stem- and-leaf plot. Alison found the average distance travelled was 38.24 km, while Sarah found the average distance travelled was 41 km. Who is correct and why?

Sydney 1

0 3 6

5 5 6

0 3 6 8 8 9

6 7 8 1

5 6 9

Key: 1|2 = 12 kilometres traveled 17

State whether the following is true or false about a dot plot. If false, explain why it is incorrect: a

A dot plot can effectively illustrate a correlation between two variables.

Dot plots can reveal multiple modes in the data set.

The mean, median and mode can all be determined directly by looking at a dot plot.

The range of the data cannot be determined from a dot plot.

Dot plots are only helpful for small data sets.

Here is a set of results: 10, 11, 11, 27, 28 Three new results are added to the data: 15, 15, 15 Without calculating the answer, describe the impact these new results will have on the range, mode, and median.

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Problem-solving 19–22 19

19–22

20, 21, 23

The level of mercury (ng/L) in 40 ﬁshing waters was tested and recorded. The results are given in the following stem-and-leaf plot: Leaf 9

10 1 11

3 5 5 6 2 4 4 5 5 6 8 9 9 1

2 3 3 5 6 7 7 7 8 8 8 8 9

12 2 2 2 2 2 3 4 6 9 Key: 1 |2 = 12

What was the second highest reading?

In how many places was a reading of 117 or higher recorded?

What was the 11th lowest score?

The safe level of mercury is recommended to be 109 ng/L or lower i

How many ﬁshing waters have safe levels of mercury?

What percentage of the places have safe levels of mercury?

A group of adults are asked: “How old were you when you passed your driving test?”. The responses were: 22, 17, 17, 17, 19, 21, 17, 22, 21, 18, 18, 17, 18, 22, 18 a

Plot the responses on a dot plot.

Find the following: i

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Range

Mode

iii

Median

How many people passed their driving test on or after their 19th birthday?

There was originally an outlier of 45. What would be the effect on the range, mode and median if it had not been removed?

Christa is a casual nurse. She used a dot plot to keep track of the number of shifts she did each week for a number of weeks: a

Over how many weeks did Christa record her shifts?

For how many weeks did she work 5 shifts?

How many weeks did she work less than 6 shifts?

If Christa works at least 6 shifts a week, she buys a weekly train ticket. What proportion of the time did she buy a weekly train ticket?

Mathspace Victorian Curriculum – Year 9 mathspace.co

4 5 6 7 8 Shifts per Week

At a national painting competition, the total number of paintings each of the 25 participating artists completed in a week is summarised in the frequency table: a

Determine the median number of paintings completed.

Determine the mode.

Find the mean number of paintings completed.

Total number of paintings completed

Number of artists

1 2 3 4 5 6 7 8 Total

3 4 3 2 5 3 4 1 25

The frequency table shows the number shown when a die is rolled. Carl rolls the die one more time and the new mean is 3. What was the number that Carl rolls? Length of hold (minutes) 1 2 3 4 5 6

Frequency 14 7 6 12 8 3

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INVESTIGATION

Opinion Polls Investigate how data is collected and presented on pressing environmental topics, such as land clearing, wind farms, and single-use plastics while assessing the validity of conclusions drawn.

Objectives To explore and interpret public opinion data on key environmental issues To critically examine methods of data collection and data presentation techniques To evaluate the reasonableness and implications of inferences made based on the survey results

Opinion polls In the world around us, data is continuously being collected through polls and surveys on topics relevant to our lives, shaping public opinion and inﬂuencing decisions. Research and analyse • Choose a website from the given resources or other website approved by your teacher • Find a recent public opinion survey related to an environmental topic of interest (e.g., land clearing, wind farms, single-use plastics) The independent think-tank, ‘The Australia Institute’, often conducts research on a variety of topics, including environmental issues. Their website may provide interesting survey results and reports. ‘Essential Media’, runs regular surveys on public opinion regarding various issues, including the environment. The Essential Report archives these survey results. The ‘Lowly Institute’ conducts annual polls on Australian attitudes towards a range of international matters, often including climate change and other environmental issues.

Investigate

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Examine the way the data is displayed. Identify if it’s categorical or nominal data. What types of graphs or charts are used? Are there any measures of central tendency (like averages) presented?

Investigate the methodology of the survey. How was the data collected? What was the sample size? Were there any potential biases in the method of collection?

What conclusions or inferences are made based on the survey results? Do you think these conclusions are reasonable given the data?

Reﬂect on how the data was presented on the website. Was it easy to understand? Could it have been presented in a better or more clear way, especially considering the lessons on data displays?

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Discussion 1.

If you had the chance to compare this data set from your research with another on the same topic ( from a different source), what things would you look for in terms of similarities or differences in data or conclusions?

Create your own opinion survey for your classmates to respond to. Display the data using what you have studied in statistics thus far. Then, write a short summary and draw a conclusion from your poll results.

Did you know?

Statistics are vital in hospitals for improving patient care and treatment. By tracking recovery rates, doctors can adjust treatments based on what works best for specific conditions. Hospitals also use statistics to monitor infection rates, predict patient needs, and allocate resources more efficiently. Statistical analysis helps identify patterns in disease outbreaks, ensuring hospitals are prepared for potential emergencies. It also plays a role in clinical trials, where patient data is used to evaluate the effectiveness of new drugs or therapies. This data-driven approach helps medical professionals make informed decisions that improve outcomes and enhance the quality of care.

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11.03 Back-to-back stem and leaf After this lesson, you will be able to… • understand what back-to-back stem-and-leaf plots are and how they differ from regular stem-and-leaf plots. • construct back-to-back stem-and-leaf plots. • interpret and compare two sets of data using back-to-back stem-and-leaf plots. • solve problems involving back-to-back stem-and-leaf plots.

Back-to-back stem-and-leaf plots Back-to-back stem-and-leaf plots are very similar to stem-and-leaf plots. However, they compare and display 2 sets of data. We call this a bivariate data display because it displays two sets of data that we want to analyse together. Group A

6 8

This is a back-to-back stem-and-leaf plot.

Group B 7

Key: 3|1 = 13

Both Group A and Group B share a common stem.

8 9

Key: 1 |0 = 10

Group A’s data is read in reverse. The stem (in the centre) is always first, and the leaves are oriented in increasing order from the stem outwards. So the lowest number in group A is 13, and the largest number in group A i s 58. Group B’s data is read as per a usual stem-and-leaf from left to right.

Back-to-back stem-and-leaf plots make it easier to identify similarities, differences and patterns between two sets of data.

Example 1 The test scores of 12 students in music and French are listed: • Music: 79, 59, 74, 94, 51, 71, 93, 84, 69, 61, 86, 86 • French: 62, 71, 64, 82, 83, 99, 87, 89, 66, 73, 59, 76 Display the data as a back-to-back stem-and-leaf plot.

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Create a strategy Start by sorting both data sets in ascending order. Then, create a stem-and-leaf plot with the stem representing all the digits before the ones place, and the leaf representing the ones place.

Apply the idea Let’s arrange the data sets in ascending order: • Music: 51, 59, 61, 69, 71, 74, 79, 84, 86, 86, 93, 94 • French: 59, 62, 64, 66, 71, 73, 76, 82, 83, 87, 89, 99 Here’s the stem-and-leaf plot for both subjects: French

Music 9

Key: 9|5 = 59

Key: 5 |1 = 51

Reﬂect and check Sorting the data in ascending order can help prevent errors when constructing the stem-and-leaf plot.

Example 2 The back-to-back stem-and-leaf plot shows the batting scores of two cricket teams, the Grasshoppers and the Cicadas. Grasshoppers

Key: 2|1 = 12

Music 5

Key: 1 |6 = 16

a What is the highest score for the Grasshoppers?

Create a strategy

Apply the idea

Find the largest leaf of the largest stem in the Grasshopper’s scores.

To ﬁnd the highest score for the Grasshoppers, look for the largest stem, which is 5, and the largest leaf with it, which is 7. So, the highest score for the Grasshoppers is 57.

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b What is the highest score for the Cicadas?

Create a strategy

Apply the idea

Look for the largest leaf with the largest stem in the Cicada’s scores.

To ﬁnd the highest score for the Cicadas, look for the largest stem, which is 6, and the largest leaf with it, which is 5. So, the highest score for the Cicadas is 65.

Example 3 The back-to-back stem-and-leaf plot shows the length (in minutes) of a random sample of phone calls made by Sharon and Tricia. Sharon

Tricia 3

Key: 5|1 = 15

Key: 1 |2 = 12

Determine if each statement is true. a Sharon did not make a 13-minute phone call.

Create a strategy

Apply the idea

Sharon’s stem-and-leaf plot is on the left, so look We can see from her stem-and-leaf plot that for 3∣1 on her plot. she has 3∣1 which means she made a 13-minute phone call. So, the statement is incorrect. b Sharon’s median is higher than Tricia’s median.

Create a strategy Since both of them made 12 phone calls, ﬁnd the average of their 6th and 7th values.

Apply the idea Sharon’s 6th value is 27, and her 7th value is 38. So her median is Tricia’s 6th value is 32, and her 7th value is 34. So her median is

Sharon’s median (32.5) is lower than Tricia’s median (33), which makes the statement incorrect.

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c The mean is greater than the median of both groups.

Create a strategy Find the mean or the average minutes of Sharon and Tricia.

Apply the idea Get the average Evaluate Get the average Evaluate Sharon’s mean (34.33) is greater than her median (32.5). Tricia’s mean (35.67) is greater than her median (33). So, the statement is correct.

Idea summary Back-to-back stem-and-leaf plots are constructed similarly to a typical stem-andleaf plot. They are used to compare the measures of central tendency and spread between 2 sets of data.

11.03 Practice questions Understanding

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

How many sets of data does a back-to-back stem-and-leaf plot display?

What is the actual value of the 3 numbers circled?

Lemomade sales Week 1

Week 2 8

Key: 5|1 = 15 3

Key: 2 |1 = 21

For what purposes is a back-to-back stem-and-leaf plot useful?

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Fluency Ex 1

4–8, 10, 12

4, 6, 7, 9, 11

The test scores of 12 students in history and maths are listed: • History: 81, 65, 72, 92, 57, 75, 95, 88, 77, 63, 80, 86 • Maths: 60, 79, 68, 90, 85, 97, 89, 91, 64, 71, 55, 82 Display the data as a back-to-back stem-and-leaf plot.

Two friends have been growing sunﬂowers. They have measured the height of their sunﬂowers to the nearest centimetre, with their results shown: • Oliver: 15, 25, 19, 15, 34, 44, 38, 27 • Valentina: 7, 45, 14, 22, 43, 16, 21, 26, 29, 49, 8, 35, 2 Display the data as a back-to-back stem-and-leaf plot.

Ex 2

Ex 3

The back-to-back stem-and-leaf plot shows the batting scores of two cricket teams, the Wildcats and the Thunderbolts:

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Thunderbolts

What is the highest score for the Wildcats?

What is the highest score for the Thunderbolts?

Which team has the higher median?

Key: 7|3 = 37

Key: 3 |2 = 32

The back-to-back stem-and-leaf plot shows the length (in centimetres) of a random sample of paper chains made by Layla and Alina:

Layla

Alina

Wildcats

Who made the 14-centimetre paper chain?

Who has the higher median?

Is Layla’s mean greater than her median?

Is Alina’s mean greater than her median?

The weight (in kilograms) of a group of people aged below 30 and over 30 were recorded and presented in the back-to-back stem-andleaf plot as shown: a

Find the mean weight of the group of 30 and overs.

Find the mean weight of the group of under 30-year-olds.

Which group is heavier? Explain your answer using data.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Key: 7|1 = 17

Key: 1 |5 = 15

Under 30 years of age

30 years and over

Key: 6|5 = 56

Key: 6 |3 = 63

The back-to-back stem-and-leaf plot shows the number of books read in a year by a random sample of university and high school students: a

Find the lowest score for the University students.

Compare the medians of both groups of students.

For which group is the mean greater than the median?

University Students

High School Students 7

5 6 7 Key: 4|1 = 14

The back-to-back stem-and-leaf plot shows the number of pieces of paper used over several days by Charlie and Dylan’s students: a

Did Charlie’s students use 32 pieces of paper on any day?

Who’s class had the higher median?

For which group is the mean greater than the median?

The back-to-back stem-and-leaf plot shows the number of desserts ordered at Hotel A and Hotel B over several days: a

Find the lowest score for Hotel A.

Which hotel’s median is higher?

For which group is the mean greater than the median?

Key: 1 |2 = 12

Charlie’s students

3 4

2 3

Dylan’s students

Key: 4|0 = 4

Key: 1 |3 = 13

Hotel A

Hotel B

Key: 2|1 = 12

Key: 1 |5 = 15

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The back-to-back stem-and-leaf plot shows the amount of cash (in dollars) carried by a random sample of students and teachers:

Teachers

Students 7

Which group carried more cash?

Find the median amount of cash that the students carried.

5 4 1

2 6 8

Find the median amount of cash that the teachers carried.

8 5 4

3 4 4 6 6 8 9

9 8 2 2 2 1

3 4 6

9 7 4 3

8 5 2

3 1

Which group has a greater range in the amount of cash?

Key: 7|0 = 7

Reasoning 13

Key: 2 |6 = 26

13, 14

Elizabeth is analysing the test scores of two classes, Class A and Class B, using a back-to-back stem-and-leaf plot. Based on the plot, which class appears to have a higher average test score? Justify your answer.

13, 14 Test Score Class A

Class B 3

4 0 1 5 6 8 9 4

5 0 2 4 6 8

8 7 1 0 6 0 1 3 4 5 6 7 9 6 5 3 0

0 1 2 4 5 7

8 7 6 4 2 0 8 7 4 2 0 9 Key: 9|3 = 39

Adal and Dado are arguing over who performed best overall in their exams this year. They have drawn up a back-to-back stem-and-leaf plot to compare their results. Who has performed better? Justify your answer using data.

Key: 3 |5 = 35

Adal’s results

Mathspace Victorian Curriculum – Year 9 mathspace.co

Key: 2|1 = 12

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Dado’s results

Key: 0 |4 = 4

Problem-solving 15–17 15

Find the median score for: i

India

Sri Lanka

Sri Lanka 9

Find the mean score of: i

15–19

The back-to-back stem-and-leaf plot shows the batting scores of two cricket teams, India and Sri Lanka: a

15–18

India

Sri Lanka

If a score of 95 was to be added to Sri Lanka:

Key: 4|3 = 34

Key: 3 |1 = 31

Is 95 within Sri Lanka’s usual score range? Why or why not?

How does the added score affect the mean and median for Sri Lanka?

iii

Which measure of centre was least affected by the addition of the outlier score of 95?

The back-to-back stem-and-leaf plot shows the test scores of a school’s two Year 7 classes, A and B.

Test score

What is the highest score in Class A?

Class A

What is the highest score in Class B?

Find the mean score of class A, correct to two decimal places.

Find the mean score of class B, correct to two decimal places.

Calculate the overall mean of the Year 7 students, correct to two decimal places.

The back-to-back stem-and-leaf plot shows the results of a survey conducted on the price of concert tickets locally and the price of the same concerts at an international venue: a

What was the most expensive ticket price at the international venue?

At the international venue, what percentage of tickets cost between $90 and $110 (inclusive)?

Class B

Key: 9 |6 = 69

Key: 6 |4 = 64

Local

International

Key: 4|6 = 64

Key: 6 |1 = 61

At the local venue, what percentage of tickets cost between $90 and $100 (inclusive)?

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A basketball coach is analysing the scores, of players, on his team, Team Amazing, and an opposing team.

Calculate the probability that a randomly selected player from Team Amazing has a score greater than 70. Leave your answer as a fraction.

Team Amazing

Team Incredible 5

8 7 5 Compare the probability of selecting a 9 4 1 player with a score greater than 70 from Team Amazing to the probability of selecting Key: 0|6 = 60 a player with a score greater than 70 from Team Incredible. Which team has a higher probability?

9 Key: 5|1 = 51

A local library conducted two reading challenges, one in summer and one in winter, to encourage children to read. The back-to-back stem-and-leaf plot displays the number of books each child read during the two challenges.

The goal of the challenge was to read 47 books. a

What percentage of the children in summer achieved the goal?

What percentage of the children in winter achieved the goal?

Was the summer or winter challenge more successful?

Determine the median of the combined number of books read across both challenges, rounded to the nearest whole number.

What is the mode of the number of books read across the summer and winter challenge, and what does it represent in the context of the reading challenge?

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Reading challenge Summer

Winter

Key: 9|1 = 19

Grouped histograms

• understand grouped frequency tables and how they are used with continuous data. • construct and interpret grouped histograms. • calculate the measures of centre and spread in a grouped frequency table. • understand the advantages and disadvantages of grouped frequency tables. • solve problems related to grouped frequency tables and histograms.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Key: 1 |2 = 12

After this lesson, you will be able to…

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Grouped frequency tables Grouped frequency tables can be helpful when collecting continuous numerical data or when the data is more spread out. It shows the number of values within a class interval or group. We can ﬁnd the frequency of a group by adding all the frequencies of the results contained in that group. Each class interval must be the same size. This grouped frequency table shows the time (minutes) taken, x, for 126 people to complete a 5 kilometre walking track: Class Interval 45 ≤ x < 50 50 ≤ x < 55 55 ≤ x < 60 60 ≤ x < 65 65 ≤ x < 70

Frequency 9 7 20 30 60

This is continuous data, so we use a class interval that captures the full range of scores possible.

The ﬁrst class interval includes the walking times for 9 different people. Each of their times fall within 45 to 50 minutes, including 45 but not including 50. The class centre is the midpoint or mean of each class interval. The class centre for the ﬁrst class interval would be: 45 + 50 Class center = 2 = 47.5 Class Interval 45 ≤ x < 50 50 ≤ x < 55 55 ≤ x < 60 60 ≤ x < 65 65 ≤ x < 70

Class center 47.5 52.5 57.5 62.5 67.5

Frequency 9 7 20 30 60

To ﬁnd the measures of centre and spread in the grouped frequency table above: • The modal class in a grouped frequency table is the group that has the greatest frequency. If there are multiple groups that share the greatest frequency, then there will be more than one modal class. The modal class is 65 to 70. • The range, in grouped frequency tables, can be found by calculating the difference between the highest and lowest possible values. The range is 70 − 45 = 25. • The median is the middle value of ordered data. In this case, the position of the median is between the 63rd and 64th position when the numbers are arranged in ascending order. The median would be found in the 60–65 class interval. The median would be the class centre of this interval, which is 62.5. • We also use the class centres to calculate the mean. The mean is found by ﬁrst multiplying the class centre of each group of scores by its corresponding frequency, then adding our results, and dividing it by the total frequency. The mean is approximately 62.46, rounded to two decimal places. The disadvantage of a grouped frequency table is that the data becomes less precise, since we are observing grouped multiple data points together rather than individual data points.

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Example 1 Consider the table below. Score 1–4 5–8 9–12 13–16 17–20

Frequency 2 7 15 5 1

Note: This is discrete data so we display the intervals as discrete values, 1–4, then 5-8. It implies that there are no scores possible between 4 and 5.

a Use the class centre (midpoint) of each class interval to determine an estimate for the mean of the sample distribution. Round your answer to one decimal place.

Create a strategy To ﬁnd the midpoint of each class, ﬁnd the average of the two end points.

Apply the idea For the ﬁrst class, we get the midpoint (or class centre) to be for each class, we get the following midpoints: Score 1–4 5–8 9–12 13–16 17–20 Total

Class center 2.5 6.5 10.5 14.5 18.5

Frequency 2 7 15 5 1 30

. Using the same process

Class center × Frequency 5 45.5 157.5 72.5 18.5 299

To ﬁnd the mean, we multiply each class centre by the frequency, add them together, and divide by the number of scores. The number of scores can be found by adding the frequencies.

Mean =

Sum of the column (Class centre × Frequency) Sum of the frequency column

299 = 30 = 10.0

Use the formula Substitue the values Evaluate and round

b Find the modal class of the data.

Create a strategy Choose the class that has the highest frequency.

Apply the idea Looking at the table, we can see that the modal class is the group 9–12, since it has the highest frequency.

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Example 2 The scores of a basketball team are collected over a season. 65, 41, 53, 76, 37, 48, 68, 34, 59, 72, 45, 50, 32, 78, 62, 69, 56, 76, 75, 47 a Determine a set of ﬁve class intervals that could be used to separate this data.

Create a strategy

Apply the idea

Identify the maximum and minimum values in the The minimum value is 32 and the maximum data set and ﬁnd appropriate class intervals. value is 78. Each class interval must be the same size.

Five class intervals could be: 30–39, 40–49, 50–59, 60–69, 70–79.

b Construct a grouped frequency table for the data.

Create a strategy

Apply the idea

Sort the data, and determine which class For 30–39, there are 3 values: 32, 34, 37 intervals they belong to. Then, count the number For 40–49, there are 4 values: 41, 45, 47, 48 of values in each class interval. For 50–59, there are 4 values: 50, 53, 56, 59 Note: This is discrete data, it is not possible to For 60–69, there are 4 values: 62, 65, 68, 69 achieve a score of 39.5, so the classes are set For 70–79, there are 5 values: 72, 75, 76, 76, 78 as 30–39, 40–49 etc. Arrange these into a grouped frequency table. Scores 30–39 40–49 50–59 60–69 70–79

Frequency 3 4 4 4 5

Idea summary A grouped frequency table shows the number of values within a set class interval. We can ﬁnd the frequency of a group by adding all the frequencies of the data contained in each group. Each class interval must be the same size. The class centre is the mean of the highest and lowest possible scores in the group. The measures of centre and spread of grouped frequency tables include: • The modal class is the group that has the greatest frequency. • The mean can be found by ﬁrst multiplying the class centre of each group of scores by its corresponding frequency, then adding our results, and dividing it by the total frequency. • The median is the middle value of ordered data.

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Grouped histograms A histogram is a visual representation of a frequency table. It displays the frequency distribution of continuous data. In a histogram, the class frequency is the height of the rectangle.

Exploration Histograms and column graphs look very similar. The ﬁrst graph is a histogram, the second is a column graph. 16 14 Frequency

12 10 8 6 4 2 0

100

110

120

130 140 150 Speed (km/h)

160

170

Waste generated in Australia by industry 2016/17 financial year

Amount of waste (millions of tonnes)

25 20 15 10 5 0

Households Construction Commercial Coal powered and local and and electricity councils demolition industrial generation

Compare the data type used for each graph.

What do you notice about the axes for the histogram and column graph?

Are you able to change the order of the columns in either of the graphs?

The key features of a histogram are: • The horizontal axis is a continuous numerical scale (like a number line), where the class intervals are placed on either side of the columns. • The vertical axis is the frequency of each data value or group of values. • There are no gaps between the columns because the horizontal axis is a continuous scale. It is possible for a class interval to have a frequency of zero, but this is not the same as having gaps between each column. • When creating a histogram, leave a half-column-width gap between the vertical axis and the ﬁrst column. 648

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This frequency distribution table and histogram shows the time taken, for 73 runners, to complete a 10 kilometre race: Class interval

Class centre

Frequency

45 ≤ x < 50

47.5

50 ≤ x < 55

52.5

55 ≤ x < 60

57.5

60 ≤ x < 65

62.5

65 ≤ x < 70

67.5

Frequency

Note: This is continuous data so we use inequality notation to capture the full range of scores. The median, which is the middle score, is the 37th in the list of 73 runners. Adding the frequencies for the first three intervals (9 + 7 + 20 = 36) shows that the 37th score falls within the 60–65 minute interval. Since the data is grouped into intervals, the exact median value cannot be determined directly, we call this the median class.

Distribution of running times for a 10 km race 73 runners

35 30 25 20 15 10 5 0

The mode is the most common class interval (tallest column). The modal class is 60–65.

50 55 60 65 Running time (minutes) 70 – 45 = 25 Range

The mean can be calculated in a similar way as we did for the frequency table. First multiply the class centre, of each group of scores, by its corresponding frequency. Then sum those up and divide by the total frequency. The mean is approximately 58.80. The range is the difference between the highest and lowest score.

The class centre (midpoint) is not typically used in the histogram itself but can be useful for statistical calculations, like estimating means.

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Example 3 Some people were asked how many of their high school friends they remained in contact with after high school. The results were grouped into categories, presented in the frequency distribution table: Note: The score is a count of people which is discrete. It is not possible to have a score of 9.5 people.

Score 0–9 10–19 20–29 30–39 40–49

Frequency 7 10 7 1 2

a Construct a histogram for the data.

Create a strategy Examine and use the frequency table to determine how many times each range of scores occurred.

The vertical axis as the frequency of each data value or group of values.

Frequency

Apply the idea

Half-column-width gap between the vertical axis and the first column.

10 9 8 7 6 5 4 3 2 1 0

There are no gaps between the columns.

0 10 20 30 40 50 No. of friends

The horizontal axis is a continuous numerical scale, where intervals are placed on either side of the columns.

b Find the modal class of the data.

Create a strategy

Apply the idea

Examine the graph or use the frequency table to Based on the given frequency table and the assess which range of scores occurs the most graph from part (a), the modal class is 10–19 as (has the highest frequency). this is the range of scores that occurs the most.

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Example 4 The heights of students in a class were recorded. The frequency table shows the ﬁndings. Height (in cm) 140.9 < h ≤ 150.9 150.9 < h ≤ 160.9 160.9 < h ≤ 170.9 170.9 < h ≤ 180.9 180.9 < h ≤ 190.9

Frequency 7 3 10 2 3

Note: Heights are continuous data, so we write the classes using inequalities to capture the full range of scores.

a What could be the height of the tallest student in the class?

Create a strategy

Apply the idea

Identify the highest class interval value in the last row.

190.9

b How many students in the class are taller than 160.9 cm?

Create a strategy

Apply the idea

Identify the total frequency in the corresponding We need to sum the frequencies from the class intervals. following class intervals: 160.9 < h ≤ 170.9, 170.9 < h ≤ 180.9, 180.9 < h ≤ 190.9 10 + 2 + 3 = 15 So there are 15 students who are taller than 160.9 cm. c Construct a histogram for the data.

Create a strategy Examine and use the frequency table to determine how many times each range of height (in cm) occurred.

Apply the idea

Frequency

0. 15 9 0. 16 9 0. 17 9 0. 18 9 0. 19 9 0. 9

Half-column-width gap between the vertical axis and the first column.

There are no gaps between the columns.

The vertical axis as the frequency of each data value or group of values.

10 9 8 7 6 5 4 3 2 1 0 Height (in cm)

The horizontal axis is a continuous numerical scale, where intervals are placed on either side of the columns.

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d If an additional column is added to the right side of the histogram, what would the new class interval be?

Create a strategy Recognise from previous rows that the lower class interval value uses a “less than” symbol (<) while the higher class interval value uses a “less than or equal to” symbol (≤). This is because we want to include the value from the previous class interval, but exclude it in the next class interval. Also notice that the end values have a difference of 10.

Apply the idea 190.9 < h ≤ 200.9

Idea summary Histograms differ from column graphs: • They are for numerical continuous data, not categorical data. • They have a numerical scale on the horizontal axis, where the class intervals are on either side of the columns. • They do not have gaps between each column. Measures of centre and spread for data presented in a histogram can be calculated in a similar way to those from frequency tables.

11.04 Practice questions Understanding 1

Explorer 1–2

Adventurer 1–3

Complete the frequency table based on the data set: 36, 50, 66, 46, 78, 63, 58, 39, 84, 40, 81, 45, 86, 51, 68, 43, 64, 67, 37, 68, 48, 65, 60, 82, 75

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Trailblazer 1–3 Class 30–39 40–49 50–59 60–69 70–79 80–89

Frequency

For each survey, state whether the data found should be grouped or ungrouped when constructing a frequency table: a

A survey conducted of 1000 people, asking them how many languages they speak.

A survey conducted of 1000 people, asking them how many different countries they know the names of.

A survey conducted of a class of 30 students, asking them how many pets they have.

A survey conducted of a class of 30 students, asking them how many minutes they spend on their phone each week.

Identify if the statements are true or false. a

The modal class is the one with the least frequency.

The class centre is found by calculating the range of the class interval values.

The median in a grouped frequency table is found by identifying the middle position and ﬁnding the corresponding class centre.

The mean in a grouped frequency table is found by ﬁrst multiplying the class centre of each group of scores by its corresponding frequency, then adding our results, and dividing it by the total frequency.

Fluency 4 Ex 1a

4–6, 8a, 9, 11–16

4–7, 8b, 10, 11, 13–17

For the frequency table: a

Complete the table.

Calculate an estimate for the mean. Round your answer to two decimal places.

Class 1–9 10–18 19–27 28–36 37–45

Class centre

f 8 6 4 6 8

State the modal class of the data set.

Class 30–39 40–49 50–59 60–69 70–79 80–89

Frequency 8 5 4 3 3 2

For the frequency table:

Score 1–5 6–10 11–15 16–20 21–25 26–30

Frequency 20 15 8 4 3 2

Ex 1b

4–7, 8b, 10, 11, 13–17

Estimate the mean by calculating the class centre for each row. Round your answer to one decimal place.

State the modal group of scores.

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Ex 2

For the frequency table: a

Estimate the mean. Round your answer to one decimal place.

State the modal group of scores.

Score 1≤x<6 6 ≤ x < 11 11 ≤ x < 16 16 ≤ x < 21 21 ≤ x < 26 26 ≤ x < 30

Frequency 18 14 9 5 3 1

Construct a grouped frequency table for the data. a

46, 54, 35, 23, 24, 28, 26, 11, 19, 17, 32

83, 68, 39, 42, 86, 66, 64, 76, 63, 43, 65, 83, 63, 67, 49, 51, 32, 55, 38, 65, 41, 73, 35, 36, 74

For the set of scores: 30, 67, 24, 51, 49, 53, 36, 24, 57, 66, 26

Determine a set of ﬁve class intervals that could be used to analyse this data.

Construct a grouped frequency table for the data.

For the set of scores: 12, 59, 61, 27, 58, 18, 76, 27, 52, 19, 13, 56, 71, 31, 73, 60, 41, 17, 22, 68, 57, 15, 40, 19, 76, 44, 60, 55, 36

Ex 3

Determine a set of seven class intervals that could be used to analyse this data.

Construct a grouped frequency table for the data.

A survey was conducted to collect data on the number of hours a group of students spent studying per week, rounded to the nearest hour. The collected data was then organised into a frequency distribution table.

Create a histogram to represent the data.

Find the modal class of the data.

In product testing, the number of faults detected in producing a certain machinery is recorded each day for several days. The frequency table shows the results. a

Construct a histogram to represent the data.

What is the lowest possible number of faults that could have been recorded on any particular day?

Hours 0−4 5−9 10−14 15−19 20−24 Number of faults 0–3 4–7 8–11 12–15

Frequency 12 20 15 8 5

Frequency 10 14 20 16

If the range of a set of continuous data is 28, ﬁnd the size of each class interval if we group the data into: a

4 equal width class intervals.

7 equal width class intervals.

If one of the class intervals for height is 10 < height ≤ 19, ﬁnd the next equal-width class interval.

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As part of a fuel watch initiative, the price of petrol, p, at a service station was recorded each day for 21 days. The frequency table shows the ﬁndings.

Frequency 4 6 5 6

Find the highest price that could have been recorded.

How many days was the price above 130.9 cents?

Create a histogram for the petrol prices over the 21 days.

If an additional column is added to the right side of the histogram, what would the new class interval be?

The histogram shows the number of hours that students in a particular class had slept for the night before. 12 11 10 9 8 7 6 5 4 3 2 1 0

6 8 Time (hours)

Construct a histogram to display the data shown in the frequency table.

Which groups represent peaks in the data set?

18, 19

How many students are in the class?

How many students had at least 8 hours of sleep that night?

What percentage of students had less than 6 hours of sleep?

The frequency table shows the price (to the nearest dollar) of the most recent book that seventy-two university students bought.

Reasoning 18

Price (in cents per litre) 120.9 < p ≤ 125.9 125.9 < p ≤ 130.9 130.9 < p ≤ 135.9 135.9 < p ≤ 140.9

Frequency

Ex 4

18, 19

Price ($) 0–19 20–39 40–59 60–79 80–99 100–119 Total

Frequency 6 20 12 6 19 9 72

18–20

The frequency table shows the data distribution for the length of leaves (in millimetres) collected from a species of tree in the botanical gardens. a

How many leaves less than 60 mm were collected?

What was the most common interval of leaf length collected?

Are leaves more likely to be at least 60 mm in length?

Can we conclude from the table that there were no leaves collected with a length less than 5 mm? Explain your answer.

Leaf length 0 ≤ x < 20 20 ≤ x < 40 40 ≤ x < 60 60 ≤ x < 80 80 ≤ x < 84

Frequency 5 11 19 49 43

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The frequency table displays the number of books read by a group of readers in a month. a

Create a histogram to represent the data from the frequency table.

Determine whether each of the statements about the data is accurate and justify your reasoning.

The majority of readers in the group read between 5 and 10 books in a month.

Most readers in the group read less than 15 books in a month.

iii

Only a small number of readers in the group read 20 or more books in a month.

Salvia and Lee create two different histograms for the same data set. The data collected shows the years of life expectancy after patients were diagnosed with lung cancer. Salvia’s Histogram

60 Frequency

Frequency

15 10 5 0

a b

Lee’s Histogram

40 30 20 10 0

5 10 15 20 25 30 Years

20 30 Years

Explain which histogram is more useful to analyse the data set. Lee claims that their histogram shows that 51 patients lived 30 years after being diagnosed with lung cancer. Is this correct? Explain your answer.

Problem-solving 21, 22

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Frequency 12 18 8 5 2

Number of books read 0≤x<5 5 ≤ x < 10 10 ≤ x < 15 15 ≤ x < 20 20 ≤ x < 25

21–23

The frequency table shows the average time spent travelling to work for 50 people: a

Construct a histogram to display the data shown in the frequency table.

State whether the statements about the data are accurate.

21–23 Commute time (minutes) 0 ≤ Time < 20 20 ≤ Time < 40 40 ≤ Time < 60 60 ≤ Time < 80 80 ≤ Time < 100 Total

Frequency 14 16 10 7 3 50

The data shows that most people travel to work by car or by walking, since most travel times are fairly short, and only a few people travel by bus or train.

The data suggests that people prefer a shorter commute to work. A majority live within 40 minutes travel, and in general the longer the commute the less people there are in that category.

iii

The data suggests that people don’t care too much about how far away from work they live. Roughly equal portions of people live less than 40 minutes away and more than 40 minutes away.

The data shows that everyone lives within an hours travel from their work, with the peak amount of people living between 20 and 40 minutes away.

Mathspace Victorian Curriculum – Year 9 mathspace.co

The histogram shows the speed of balls delivered by cricket bowlers. 16 14

Frequency

12 10 8 6 4 2 0

110

120

130 140 Speed (km/h)

150

160

170

Complete a frequency table for this set of data.

What percentage of balls were bowled at less than 130 km/h? Round your answer to one decimal place.

A battery manufacturer tests their batteries in a portable music player, to ﬁnd out how long the batteries last under constant use. The histogram shows the data from their tests. 35 30 25 Frequency

100

20 15 10 5 0

400 500 600 700 800 900 1000 1100 1200 1300 Battery Life (mins)

Create a frequency table for this set of data.

What percentage of batteries last longer than 10 hours? Round your answer to one decimal place.

If the average length of a song is 4 minutes, what percentage of batteries will last for an average of 250 songs or more? Round your answer to one decimal place.

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11.05

Shape of data

After this lesson, you will be able to… • identify and describe the features of symmetrical data distributions. • distinguish between positively and negatively skewed data distributions. • analyse the implications of skewness on the mean, median, and mode. • interpret histograms and stem-and-leaf plots to determine the skewness and clustering in data. • recognise and explain the presence of clusters in data sets. • identify clusters in data and distinguish between unimodal, bimodal and multimodal data.

Symmetry and skew Data may be described as symmetrical or skewed. This is the shape that symmetrical data would look like. It is similar to the shape of a bell. This is often referred to as a normal distribution.

There are a few features that are related to symmetrical data: • The data tends to be around a central value. • Roughly 50% of scores will be above the mean and 50% of scores will be below the mean. • The mean, median and mode should roughly coincide. This is the shape of data that is positively skewed. Notice the tail at the positive end. There are a few features that are related to positively skewed data: The mean is usually greater than the median. The median is usually greater than the mode. A negatively skewed graph has this general shape with the left side stretched out. There are a few features that are related to negatively skewed data: The mean is usually less than the median. The median is usually less than the mode.

Interactive exploration Discover this concept in action online

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This is the general shape of a frequency histogram with a symmetrical shape.

Percentage

30 25

Even though the data is not exactly symmetrical, the dark line shows that a symmetrical curve can roughly be drawn over the histogram following the data.

20 15 10 5

To describe shape we don’t need the data to be exactly the same, we are looking to describe what it is most like.

Frequency

75 80 85 90 95 100 105 110 Weight

This is the general shape shown over a histogram of positively skewed data.

Frequency

Score

This is the general shape shown over a histogram of negatively skewed data.

Score

Example 1 State whether the scores in each data display can be best described as positively skewed, negatively skewed or symmetrical.

Frequency

10 9 8 7 6 5 4 3 2 1 0

14 15 Score

Create a strategy

Apply the idea

Check where the bulk of the data sits and look at the general shape of the distribution.

The scores are roughly even in both the high and low end, so the distribution is symmetrical.

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10 11 12 13 14 15 16 17 18 19 Score

Apply the idea The scores have higher-frequency dot stacks on the left (the lower scores), so the distribution is positively skewed.

Stem

Leaf

0 4

2 7

3 8

0 4 7

0 0 3 4 6 7 7 9

0 1

2 2 5 6 8 9 2 4 5 5 6 7 9 9 2 2 3 5 6 8 8

Key: 1|2 = 12 score

Create a strategy

Apply the idea

If most of the scores have lower values, the data is positively skewed. If most of the scores are centred in the middle of the range, the data is symmetric. If most of the scores have higher values, the data is negatively skewed.

Most of the scores have higher values, so the distribution is negatively skewed.

Idea summary A distribution is said to be symmetric if its left and right sides roughly mirror images of one another. A data set that has a positive skew has a longer tail of values to the right of the data set. The mass of the distribution is concentrated on the left of the ﬁgure. A data set that has a negative skew has a longer tail of values to the left of the data set. The mass of the distribution is concentrated on the right of the ﬁgure.

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Clusters In a set of data, a cluster occurs when a large number of the scores are grouped together within a small range. Clustering may occur at a single location or several locations.

Frequency

If the data in a histogram has two clear peaks, then the shape is called bimodal.

Score

If the shape of data in a histogram has more than 2 peaks (modes), then it is described as multimodal.

Example 2

Frequency

The percentage of faulty computer chips in 42 batches were recorded in the histogram.

14 12 10 8 6 4 2 0

a Which of these makes this statement true? The distribution is: A

Unimodal

Bimodal

1 2 3 4 5 6 Faulty chips (%)

Multimodal, but not bimodal

Create a strategy Count the number of peaks in the distribution.

Apply the idea Looking at the histogram, there are two peaks. So the distribution is bimodal. The correct answer is option B.

b Which of these are the modal classes? Select all that apply. A

0–1

1–2

4–5

5–6

2–3

3–4

Create a strategy The modal classes are those with the highest frequency.

Apply the idea The highest frequency is 12, and the classes with this frequency are 1–2 and 3–4. So the correct answers are options B and D. 11.05 Shape of data mathspace.co

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Example 3

Frequency

For the data shown in this column graph: 45 40 35 30 25 20 15 10 5 0 1

6 Score

a Is there any clustering of data? A

Yes

Create a strategy

Apply the idea

Check to see if the data is ‘bunched’ up in any particular range.

Looking at the column graph, there are more data values (scores with higher frequencies) on the left side of the graph. The correct answer is option A.

b Where does the clustering occur? A

8 to 11

1 to 4

5 to 7

Create a strategy

Apply the idea

Which of the regions has a signiﬁcantly higher number of scores?

The highest frequencies are found in the classes 1 to 4. So the correct answer is option B.

c What is the mode?

Create a strategy

Apply the idea

Remember that the mode is the score that occurs most often.

Looking at the column graph, we can see that the highest frequency can be found in score 3.

d Describe the distribution of the data.

Create a strategy

Apply the idea

If the scores are even in both the high and low end, the distribution is symmetrical. If the right tail is longer, the distribution is positively skewed. If the left tail is longer, the distribution is negatively skewed.

The right tail appears to be longer than the left. So the data is positively skewed.

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Idea summary To determine the modality of a data distribution: • If there is a single mode, the data is unimodal. • If there are two modes, the data is bimodal. • If there are more than two, the data is multimodal.

11.05 Practice questions Understanding

Adventurer 1–5

Trailblazer 1–5

What is a feature of a distribution that is symmetrical?

Describe what positive skew and negative skew mean in the context of a data set.

How can clusters be identiﬁed within a data set?

How do we describe the shape of a histogram with two distinct peaks?

A set of data is strongly positively skewed. If the median is 60, is the mean less than, equal to or greater than 60?

Fluency

6–18

Identify if the graph is positively skewed, negatively skewed, or symmetrical.

10 9 8 7 6 5 4 3 2 1 0

b Frequency

10 9 8 7 6 5 4 3 2 1 0

0 5 10 15 20 25 30 35 40 45 50 Class

25 28 31 34 37 40 43 46 49 52 55 Class

10 9 8 7 6 5 4 3 2 1 0

6 10 14 18 22 26 30 34 38 42 46 Class

Frequency

a Frequency

Frequency

Ex 1

Explorer 1–5

10 5

8 9 Score

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f Frequency

Frequency

8 9 10 11 12 13 14 15 16 17 Score

4 5 Score

Frequency

15 10 5

4 5 Score

2 3 4 5 6 7 Score

Identify if each graph is unimodal or bimodal. 120 110 100 90 80 70 60 50 40 30 20 10 0

b Frequency

130

Frequency

d Frequency

120 110 100 90 80 70 60 50 40 30 20 10 0

130 120 110 100 90 80 70 60 50 40 30 20 10 0

0 10 20 30 40 50 60 70 80 90 100110120 Length (cm)

130 Frequency

664

20 Frequency

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

0 10 20 30 40 50 60 70 80 90100110120 Length (cm)

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20 30 40 50 60 70 80 90 100 110 120 Length (cm)

130 120 110 100 90 80 70 60 50 40 30 20 10 0

10 20 30 40 50 60 70 80 90 100 Length (cm)

The stem-and-leaf plot shows the ages of a group of people on a bus. a

b 9

Find the: i

Median age

Difference between the lowest age and the median

iii

Difference between the highest age and the median

The mean age, rounded to two decimal places

Leaf 1

Key: 1|4 = 14 years old

Is the data positively or negatively skewed?

The stem-and-leaf plot shows the number of hours worked per week by a group of people. a

Is there any clustering of data? If yes, in what interval?

Find the mode(s).

Leaf 0

1 2

Key: 5|2 = 52 hours

Leaf

For the stem-and-leaf plot: 5

Is there any clustering of data? If yes, in what interval?

Find the mode(s).

Describe the shape of the data.

5 6 7 8 2

9 11

The percentage of faulty mobile phones in 118 batches were recorded. a

Is the data unimodal, bimodal, or multimodal but not bimodal?

Find the mode of the data.

Key: 1|2 = 12 20 18 16 14 12 10 8 6 4 2 0

Frequency

Ex 2

1 2 3 4 5 6 7 8 9 10 Faulty mobile phones (%)

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The number of peanuts in mixed nut packets were sampled and recorded in the stem-and-leaf plot.

Leaf

Describe the clustering of data.

3 6 8

2 2

0 0 3

4 6 8 8 9

0 2 4 5 6 9

0 4 5

7 1

8 8 4 5 9

2 3 5 5 6

Key: 5|2 = 52 13

For the given column graph: a

Is there any clustering of data?

Where does the clustering occur: 1 to 4, 5 to 7, or 8 to 11?

Find the mode.

Describe the shape of the distribution of the data. 45 40 35 Frequency

30 25 20 15 10 5 0

6 Score

The shoe sizes of all the students in a class were measured. a

Is there any clustering of data?

Where does the clustering occur?

Find the mode of the shoe sizes.

Describe the shape of the distribution of the data.

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Frequency

Ex 3

8 9 Shoe size

The dot plot shows the temperatures recorded over a period of time. a

Is there any clustering of data? If yes, in what interval?

Find the mode of the temperatures.

Describe the shape of the data.

For this histogram representing students’ heights in centimetres: a

Does the histogram represent grouped data or individual scores?

Estimate the value of the mean, rounded to one decimal place.

Describe the shape of the distribution.

40 41 42 43 44 45 46 47 48 49 Temperature 20

Frequency

15 10 5

110 120 130 140 150 170 160 Score

For the dot plot, describe the shape of the distribution. 1 2 3 4 5 6 7 8 9 Score

For the dot plot: a

Is there any clustering of data? If yes, in what interval?

Find the mode(s).

Describe the shape of the data.

Reasoning

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Sample

19, 20

Explain how a positively skewed distribution might impact the mean, median, and mode.

The ages of people were collected across 5 suburbs in a certain town. What factors might lead to this data set having one or more clusters?

Problem-solving 21

21, 22

A company records the number of sales made by its salespeople over a month. The data is heavily skewed to the right. What can be inferred about the sales performance of the team?

A teacher wants to compare the test scores of two different classes. One class has a symmetrical distribution, while the other has a positively skewed distribution. How might the teacher interpret and compare the performance of the two classes based on these distributions?

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INVESTIGATION

Statistical reports in the media Investigate how the media links claims to displays, statistics and representative data.

Objectives To analyse and interpret statistics presented in media articles To evaluate the consistency between the claims made in articles and the supporting data or statistics provided

Statistics in the media Statistics often pop up in news stories to help explain things or make a point. But it’s good to be curious and ask: “Is that statistic telling the full story?” In this investigation, we’ll look at how the media uses statistics to make claims. Choose one of the suggested media articles that can be found in Investigation: Statistics in the media or choose an article approved by your teacher. Read the article and access any links that cite studies or evidence for the claims in the article. Use the ﬁndings from the article to answer the investigation questions.

Article from ABC News: Intergenerational Report: Six charts that show what Australia could look like in 40 years Article from East Asia Forum: Australian reliance on Chinese exports an economic reality Article from The Guardian: Australia’s air pollution hotspots hit disadvantaged people harder Article from SBS News: How multicultural is your suburb?

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Investigate 1.

Identify the primary statistics or data presented in the article. How do the statistics relate to the main topic or argument of the article?

Do you believe the claims or conclusions made in the article align with the statistics provided? Why or why not?

From which source(s) were the statistics derived? Research the source(s) online. Are they credible and reliable? Provide reasons for your answer.

How were the statistics visually represented in the article (graphs, charts, infographics)? Was this representation clear and effective in conveying the data?

Based on what you’ve learned in previous lessons (like data displays, opinion polls, etc.), critique the article’s use of data. Is there anything you would change or present differently?

Discussion 1.

To what extent do you believe media outlets have a responsibility to accurately represent statistics? How might the misrepresentation or misunderstanding of statistics impact public opinion or decision-making?

Consider another current event or topic you are passionate about. How could you use statistics to communicate or inﬂuence an audience about this issue? What data would you need, and how would you ensure its accurate representation?

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11.06

Compare data sets

After this lesson, you will be able to… • compare and contrast two or more data sets using measures of spread, central tendency, and data shape. • interpret histograms, stem-and-leaf plots, tables, and column graphs to draw comparisons between data sets.

Compare data sets When we compare data sets, we should ask a few questions: • How do the spreads of data compare? Is one set of data more spread out than the other? • How does the skew of data compare? Is one more symmetrical? • Is there a big difference in their measures of centre (mean, median or mode)?

Exploration Consider the following histograms that show the height of students in two basketball teams. We know that one graph represents a team made up of Year 12 students and the other represents a Year 8 team. Team A:

Team B: 12

Number of Students

6 4 2 0

155 160 165 170 175 180 185 190 195 200 Height (cm)

10 8 6 4 2 135 140 145 150 155 160 165 Height (cm)

Are there the same number of students in each team? Does it matter?

What are the similarities and differences in terms of measures of spread, central tendency and shape of data?

Which team do you think corresponds to the Year 12 team, and which team do you think corresponds to the Year 8 team?

It is important to be able to compare data sets because it helps us make conclusions or judgements about the data. For example, suppose Jim scores 50% in a geography test and 60% in a history test. Based on those marks alone, it makes sense to say that he did better in history. But what if everyone else in his geography class scored 40%, while everyone else in his history class scored 80%? Now we know that Jim had the highest score in the class in geography, and the 670

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lowest score in the class in history. With this extra information, it makes more sense to say that he did better in geography.

Example 1 The weights (in grams) of female and male fish caught in a tag and release program were recorded and presented in the stem-and-leaf plot shown. Female

Stem

Male

Key: 1|6 = 61

Key: 6|2 = 62

a What is the mean weight of the group of male fish?

Create a strategy Use the formula: Mean =

Apply the idea We are focusing on the stems and the leaves under “Male” on the right side of the stem and leaf plot. Substitute the values

Evaluate the addition = 70.3 g

Evaluate the division

b What is the mean weight of the group of female fish?

Create a strategy Use the formula: Mean =

Apply the idea We are focusing on the stems and the leaves under “Female” on the left side of the stem and leaf plot. Substitute the values

Evaluate the addition = 62.9 g

Evaluate the division

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c Based on the mean, which group is heavier? A

Female fish

Male fish

Create a strategy Compare the mean weights of the two groups.

Apply the idea The group with the higher mean weight will also, on average, have individual fish that are heavier. The male fish had a mean of 70.3 g which were heavier than the female fish who had a mean of 62.9 g. So the correct answer is option B.

Example 2 The table shows the scores of Student A and Student B in 5 separate tests. Test 1 2 3 4 5

Student A 79 74 86 89 98

Student B 77 83 72 94 97

a Find the mean of the scores of Student A.

Create a strategy Use the formula: Mean =

Apply the idea We add the scores obtained by Student A divided by the number of tests. Substitute the values

Evaluate the addition

Evaluate the division

b Find the mean of the scores of Student B.

Create a strategy Use the formula: Mean =

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Apply the idea We add the scores obtained by Student B divided by the number of tests. Substitute the values

Evaluate the addition

Evaluate the division

c Which student did better overall on their tests?

Create a strategy We can compare the students by using the mean values calculated in parts (a) and (b).

Apply the idea Student A has a higher mean, 85.2, and did slightly better than Student B.

d What is the highest score overall? Which student (A or B) obtained that score?

Create a strategy

Apply the idea

Refer to the score columns and ﬁnd the highest score. Then, ﬁnd the student that obtained that score by referring to the column heading.

Based on the table, the highest score is 98, which was obtained by Student A.

e What is the lowest score overall? Which student (A or B) obtained that score?

Create a strategy

Apply the idea

Refer to the score columns and ﬁnd the lowest score. Then, ﬁnd the student that obtained that score by referring to the column heading.

Based on the table, the lowest score is 72, which was obtained by Student B.

Do the answers to parts (d) and (e) impact the answer to part (c)?

Create a strategy Determine the difference between what is required in part (c) and what is required in parts (d) and (e).

Apply the idea In part (c), we are asked to determine which student did better overall, meaning their average score across all five tests. In contrast, parts (d) and (e) ask us to find the highest and lowest individual scores among all 10 tests. Therefore, the answers to parts (d) and (e) do not impact the answer to part (c).

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Example 3 The following column graphs show the season results of two soccer groups, Group A and Group B, and the number of games (frequency) in which they scored a certain number of goals (scores). Group A

4 Frequency

Group B

3 2

2 3 4 5 Score - Group A

2 3 4 Score - Group B

a What is the mode for Group A?

Create a strategy

Apply the idea

Examine the column graph for Group A and assess which score occurred the most.

Based on the column graph for Group A, the score that occurred the most is 3.

b What is the mode for Group B?

Create a strategy

Apply the idea

Examine the column graph for Group B and assess which score occurred the most.

Based on the column graph for Group B, the score that occurred the most is 4.

c What is the range for Group A?

Create a strategy Use the fact that the range is the difference between the highest and lowest scores.

Apply the idea Based on the column graph for Group A, the highest score is 5 and the lowest score is 1. Range = 5 – 1 =4

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Subtract the lowest score recorded from the highest Evaluate

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d What is the range for Group B?

Create a strategy Use the fact that the range is the difference between the highest and lowest scores.

Apply the idea Based on the column graph for Group B, the highest score is 6 and the lowest score is 0. Range = 6 – 0 =6

Subtract the lowest score recorded from the highest Evaluate

e Which group scored the lowest total number of goals during the season? A

Group A

Group B

Create a strategy To ﬁnd the total number of goals scored by each group, we need to multiply each score by its frequency and then add the values.

Apply the idea We can create frequency tables for both Group A and Group B to easily ﬁnd and compare the total number of goals scored between the two groups. Frequency table for Group A: Scores (x) 1 2 3 4 5

Frequency ( f ) 2 4 5 4 2 Total

fx 1×2=2 2×4=8 3 × 5 = 15 4 × 4 = 16 5 × 2 = 10 51

Frequency table for Group B: Scores (x) 0 1 2 3 4 5 6

Frequency ( f ) 3 3 0 1 4 3 3 Total

fx 0×3=0 1×3=3 2×0=0 3×1=3 4 × 4 = 16 5 × 3 = 15 6 × 3 = 18 55

Comparing the total number of goals scored between the two groups, the group that scored the lowest during the season is Group A. So, the correct answer is option A.

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Which group had more varied results? What might be a reason for this?

Create a strategy

Apply the idea

Examine both column graphs and determine which data set is more spread out.

The range of goals scored for Group B is larger than for Group A, indicating a wider spread of results. There are many possible reasons why Group B has more varied results. For example, Group B might have a wider range of skill levels among its players, leading to a larger spread of scores.

Idea summary By comparing the measures of central tendency in a data set (the mean, median and mode), the measure of spread (the range) and the shape of the data, we can make comparisons between different groups and draw conclusions about our data.

11.06 Practice questions Understanding

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

What are three statistical measures we can use to compare data sets?

Can two data sets have the same mean but different medians?

Which data set has the lowest score?

676

9 8 7 6 5 4 3 2 1 0

Graph 2 5 4 Frequency

Frequency

Graph 1

3 2 1

3 4 Score

6 7 Score

Give two reasons using a data visualisation like a graph can make comparisons of data sets easier.

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Fluency 5 Ex 1

Ex 2

Key: 1|6 = 61

Which group takes less time on average to get to school?

The back-to-back stem-and-leaf plot shows the length (in minutes) of a random sample of phone calls made by Jenny and Joanne.

Jenny did not make a 13-minute phone call.

Jenny’s median is higher than Joanne’s median.

iii

The mean is greater than the median in both groups.

Is Jenny’s mean greater than her median?

Is Joannes’s mean greater than her median?

Find the total number of goals scored by both teams in Match C.

Find the total number of goals scored by Team 1 across all the matches.

Calculate the mean number of goals scored by: ii

Find the mean of the scores of: i

Alice

Key: 6 |2 = 62

Match A B C D E

Team 1 2 4 5 3 3

Team 2 3 2 2 5 4

Subject Maths Science History English

Alice 88 90 85 82

Bob 72 75 80 85

Team 2

The table shows the test scores of two students, Alice and Bob, in 4 different subjects. a

Joanne

Key: 1|6 = 61

The number of goals scored by Team 1 and Team 2 in a football tournament are recorded in the table.

Key: 6 |2 = 62

Jenny

Team 1

High school

Calculate the mean time taken to get to school for the group of high school students.

i 8

Calculate the mean time taken to get to school for the group of primary students.

Determine whether the following statements are true or false.

5, 6, 8, 10–12, 14–16

Primary

5, 6, 8, 10–13, 15–17

The time taken to get to school (in minutes) by a group of students was recorded and presented in a stem-and-leaf plot. a

5–11, 13, 15, 16

Bob

Find the mean of the combined scores of Alice and Bob.

Find the highest score overall and which student obtained that score.

Find the lowest score overall and which student obtained that score.

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The points scored by each player in a basketball game are displayed: • Team X: 10, 12, 15, 18, 20, 15, 12, 10, 10, 8, 5 • Team Y: 14, 16, 10, 20, 24, 12, 15, 8, 5, 5, 4 a

Which team scored more points?

Find the mean points of the teams, rounded to one decimal place: i

Team X

Team Y

In a Mathematics test out of 100, the students in two classes scored the results displayed: • Class A: 72, 66, 92, 76, 77, 63, 67, 91, 66, 77, 66, 73, 80, 68, 68, • Class B: 84, 70, 53, 82, 65, 85, 81, 87, 64, 81, 86, 79, 69, 82, 77 a

Find the median mark for: i

b 11

Class A

Class B

Using the medians, which class did better on the test?

The monthly rainfall for two cities across a year is displayed: • City A: 122, 120, 121, 67, 79, 24, 8, 46, 27, 67, 80, 128 • City B: 51, 66, 33, 78, 79, 92, 78, 99, 41, 60, 26, 56 a

Find the range in monthly rainfall for: i

b 12

City A

City B

Which city had more consistent rainfall throughout the year?

Harry is tracking the growth of two different types of sunflower plants. He records the number of sunflowers that he picks from each plant for 10 days. • Plant A: 15, 7, 8, 10, 9, 15, 6, 6, 12, 15 • Plant B: 11, 10, 9, 9, 12, 10, 11, 10, 9, 10

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Find the mean daily growth for Plant A. Round your answer to one decimal place.

Find the mean daily growth for Plant B. Round your answer to one decimal place.

Find the range of daily growth for Plant A.

Find the range of daily growth for Plant B?

Which plant shows more growth on average?

Which plant shows a more consistent daily growth?

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The stem-and-leaf plot shows the running times (in minutes) of two track teams, Team A and Team B. a

Find the fastest running time in: i

Team A

Team B

Key: 1|6 = 61

Key: 6|2 = 62

Perth

Adelaide

Perth

1 0

5 6 7 8 9

0 1

Key: 1|6 = 61

2 5 8

Key: 6|2 = 62

Adelaide

Find the mean temperature of: Perth

Adelaide

Which city would be described as being warmer? Is this supported by your answer to part (b)?

The back-to-back stem-and-leaf plot shows the number of books read in a year by students from two different schools, Kinderhaus and Learning Tree: a

Did any student from Kinderhaus read 12 books in a year?

Which school’s students had the higher median number of books read?

9 8 7 6 5 4 3

Find the highest temperature recorded in:

i c

Which team is quicker? Is this supported by your answer to part (b)?

i b

Team B

The stem-and-leaf plot shows the temperature (in degrees Celsius) recorded in two cities, Perth and Adelaide, over 10 days. a

Team B

Find the mean running time of: i

Team A

Kinderhaus

Learning Tree

Key: 1|6 = 61

Key: 6|2 = 62

Is the median number of books read greater than the mean number of books read in both groups?

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The graphs show the performance of two hockey teams, Team X and Team Y, in various games, along with the number of goals they scored in each game. 5

5 Frequency

Frequency

Ex 3

3 2 1

3 4 5 Score - Team X

2 3 4 Score - Team Y

Find the mode for Team X.

Find the mode for Team Y.

Find the range for Team X.

Find the range for Team Y.

Which team scored the lowest total number of points during the season?

Which team has the most varied results?

The pulse rates of two groups are given: • Group 1: 82, 85, 88, 65, 73, 89, 79, 90, 76, 68, 88, 65, 63, 62, 88, 82 • Group 2: 75, 88, 74, 73, 80, 76, 67, 81, 71, 83, 89, 62, 63, 80, 71, 78 a

Find the mean pulse rate of Group 1, rounded to two decimal places.

Find the mean pulse rate of Group 2, rounded to two decimal places.

What is the range for Group 1?

What is the range for Group 2?

Which group has the greater spread?

Reasoning 18

19, 20

18–20

18, 21

The weights of two groups of turtles are measured, in kilograms, to determine whether they might belong to the same species. The results are presented • Group 1: 56, 60, 58, 59, 62, 55, 57, 61, 56, 59 • Group 2: 62, 64, 60, 65, 66, 62, 61, 64, 67, 63 Do you think the two groups of turtles are from the same species? Explain your answer.

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Two Biology classes, each with 15 students, sit a 10-question multiple choice test, each with four possible answers (only one of which is correct). Their class results, out of 10, are: • Class 1: 3, 2, 1, 1, 2, 5, 2, 2, 2, 3, 2, 1, 5, 5, 4 • Class 2: 10, 6, 9, 8, 10, 9, 7, 7, 6, 8, 8, 10, 8, 9, 8 a

Complete the following table, rounding your answers to one decimal place when necessary: Mean

Median

Mode

Range

Class 1 Class 2 b 20

Which class was more likely to have studied for their test? Explain your answer.

The median salary in Company Alpha is $60 000 with a mean salary of $65 000 and the median salary in Company Beta is $60 000 with a mean salary of $55 000. Which company is most likely to have some very high paying jobs? Explain your answer.

The table shows the ages of employees working in two different departments of the same company. Department A Department B a

25 45

30 50

Copy and complete the table, rounding your answers two decimal places where appropriate:

40 60

45 45

35 50

Department A

40 55 Department B

Mean Median Mode Range

Which department seems to have older employees? Justify your answer.

Problem-solving 22, 23, 25

23–26

24–27

The prices of sales at two locations of a shop are recorded in the histograms. Location A

13 12 11 10 9 8 7 6 5 4 3 2 1 0

Location B

Sales

13 12 11 10 9 8 7 6 5 4 3 2 1 0

Sales

35 55

5 10 15 20 25 30 35 40 45 50 55 Price

What is the shape of the distribution of the histogram of Location A?

Which location made more sales?

Which location had a higher average sale price?

11.06 Compare data sets mathspace.co

681

An online shopping website records the user ratings for two similar products in the column graph. 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0

30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0

0.5

1.0

1.5

2.0 2.5 3.0 User rating

3.5

4.0

4.5

5.0

3.5

4.0

4.5

5.0

Product B

0.0

0.5

1.0

1.5

2.0 2.5 3.0 User rating

Product A

Product B

According to the website’s rating system, a perfect rating is 5.0. Which product scored more perfect ratings?

According to the website’s rating system, a positive rating is greater than 3. Which product scored more positive ratings?

The ages of employees at two competing fast food restaurants on a Saturday night are recorded. Some statistics are given in the table. Barry’s Burgers Frank’s Fries

682

0.0

Find the median user rating for: i

Product A

Mean 18 18

Median 17 19

Range 8 4

If the data for each restaurant was represented using a histogram, what would the likely shape of the histogram for Barry’s Burgers be?

Which restaurant likely has the oldest employee on the night the data is recorded?

Which restaurant has the most employees of a similar age?

Which statistical piece of evidence supports your answer to part (c)?

Which restaurant has an older workforce?

Which statistical piece of evidence supports your answer to part (e)?

Mathspace Victorian Curriculum – Year 9 mathspace.co

The hours of sleep per night for two people over a two week period are given: • Person A: 10, 9, 6, 6, 7, 8, 5, 6, 9, 9, 7, 5, 10, 9 • Person B: 7.5, 7.5, 8, 7.5, 7, 8, 7.5, 7, 7, 7, 7, 7, 8, 8 The table shows the measures of central tendency and spread for Person A and Person B. Mean 7.6 7.4

Person A Person B

Mode 9 7

Range 5 1

Which person seems to be the least consistent in their sleep habits?

Which statistical piece of evidence supports your answer in part (b)?

Which person had the most sleep over the 14 nights?

Which statistical piece of evidence supports your answer in part (d)?

A stationery shop kept a weekly record of the number of pens and notebooks sold. 25

20 Number Sold

Median 7.5 7.5

15 10 5 0

Mon Tue Wed Thu Fri Pens

15 10 5 0

Mon Tue Wed Thu Fri Notebooks

What is the shape of the histogram of the number of pens sold?

Find the percentage of pens sold on Thursday. Round your answer to two decimal places.

Calculate the mean number of notebooks sold per day. Round your answer to one decimal place.

Calculate the mean number of pens sold per day. Round your answer to one decimal place.

Which is the better selling product? Justify your answer.

11.06 Compare data sets mathspace.co

683

A basketball coach is analysing the free throw shooting performance of two basketball teams, Team X and Team Y. The data shows the number of successful free throws each player made during a season: • Team X free throws: 8, 9, 7, 6, 9, 8, 7, 10, 7, 6 • Team Y free throws: 9, 8, 10, 8, 7, 9, 8, 10, 7, 8

684

What is the lowest and highest successful free throws for both Team X and Team Y?

Calculate the mean number of successful free throws for both Team X and Team Y.

Calculate the median number of successful free throws for both Team X and Team Y.

Calculate the probability that a randomly selected player from Team X made 8 or more successful free throws.

Calculate the probability that a randomly selected player from Team Y made 8 or more successful free throws.

Based on the calculated means, modes, and probabilities, as well as the spread of data, what can you infer about the performance of Team X and Team Y? Which team’s players appear to have a more consistent performance in terms of free throw success?

Mathspace Victorian Curriculum – Year 9 mathspace.co

INVESTIGATION

Plan and conduct statistical investigations Objectives To plan and conduct a statistical investigation

Identify a problem Statistical investigation A statistical investigation is a process of transforming raw data into useful information that can tell us more about a subject and allow us to make recommendations and possibly make predictions of future outcomes. It is a cyclic process that involves several stages: • Identify a problem: clarify the problem and pose questions that can be answered with data. • Collect data: design and implement a plan to collect or obtain appropriate data. • Analyse data: select and apply appropriate techniques to analyse the data. • Interpret and communicate the results: interpret the results of analysis in a way that relates to the original question.

Identify a problem The first step is to pinpoint the desired information. This involves coming up with questions that, if answered, would lead to meaningful information that would allow us to draw a conclusion and to make recommendations. For example, suppose you were in charge of the school’s funds that have been set aside for the development of a new sports field, but aren’t sure which type of field (e.g. cricket pitch, basketball court) would be of greatest benefit to students. To investigate this issue, you would need to ask questions such as: • “What is the most popular sport among students?” ( you want to construct a type of field that would satisfy the majority of students) • “Are there enough funds to construct the students’ preferred type of field?” ( you can’t construct a type of field that you can’t afford) • “How long will it take to construct?” (there’s no point constructing the students’ preferred type of field if it takes 10 years to construct, at which point none of those students will be around to enjoy it) With answers to these questions, you would then be able to decide which type of field would benefit students the most.

Investigation: Plan and conduct statistical investigations mathspace.co

685

Investigate 1.

The military’s research and development team would like to ﬁnd out both the ideal type of uniform for its soldiers to wear in combat and the ideal type for them to wear in parades. Pose some questions that would need to be answered by the team in order to draw a conclusion and to make recommendations.

A restaurant owner would like to know how many chefs, waiters, cashiers and managers to hire for his new restaurant and would like to know how many staff to roster on during each time of the day. Pose some questions that would need to be answered by the investigation.

Collecting data Once we have identiﬁed a problem and posed statistical questions, we need to collect data to answer them. Before we do the actual collecting, we have to decide on how we will collect the data, the type of data we will collect and the sources from which we will collect them. Primary data Collecting primary data involves collecting the data yourself by interviewing or observing others or conducting experiments. When collecting primary data, it is important to ensure that the data to be collected can be organised easily. For example, when creating a questionnaire, it would be better to include questions that are not open-ended, but rather have a limited number of options from which participants can choose their answers. This way, the answers collected can be easily tallied and organised. Compose a question that is not open-ended, along with its response options, that can be asked to investigate: 1.

the most popular subject among students at a school.

the average income of teachers at a school.

the average mark of students in your class in last term’s maths test.

Secondary data Using a secondary source involves gathering data that has already been collected or generated by others. This could involve gathering data from books or the internet. It is important that the data to be collected are from a reliable source and not from some obscure website or outdated book, otherwise the data may not be accurate.

686

Mathspace Victorian Curriculum – Year 9 mathspace.co

Investigate Determine whether the data to be gathered to investigate the following would be from a primary or secondary source. Also, state the method (e.g. questionnaire, interview, observation, experiment) if the data is to be primary. State the source (e.g. books, newspapers, internet) if the data is to be a secondary. 1.

The most popular subject among students at school

The average daily temperature in Sydney over the last month

The number of traﬃc accidents in the country each year

The number of visitors to the local library in an afternoon

The average daily temperature in your home over the past week

The number of goals scored by the Socceroos since the last World Cup

Organise and analyse the data Organise the data After collecting the data, we arrange it into a form that gives structure and order to the data. A common way of accomplishing this is to use a table or a graph. Besides displaying the data in a graph, it may also be beneficial to summarise the data using statistical quantities such as the mean, median, mode and range. Analyse the data Next, we examine and interpret the data. This may involve identifying trends and patterns from the graph, and identifying how those trends and patterns change over time or across categories (such as across different populations). From these trends, we can then draw conclusions and possibly make predictions about future outcomes. Interpret and communicate The final step is to put everything in a written report. The report should address the aim of the statistical investigation and the questions it sought to answer, detail the data collection method, thoroughly discuss the findings, list and explain the reasoning behind the conclusions, and, if appropriate, include recommendations for the future. It should also include the tables and graphs used in the analysis.

Investigation: Plan and conduct statistical investigations mathspace.co

687

Investigate Suppose the Roads and Traffic Authority (RTA) has tasked you with investigating the number of vehicles travelling past the front of your school on an average day in order to determine whether there is any need to implement new measures to manage traffic flow. 1.

Pose some questions that would need to be answered in order to draw a conclusion and to make recommendations.

Will you be using a primary or secondary source to collect the data?

What method (e.g. questionnaire, interview, observation, experiment) will you be using to collect the data?

What type of data will you be collecting?

Collect the data and record it in a frequency table.

Display the data with an appropriate graph. Summarise the data by calculating the mean, median, mode and range.

Analyse the data and identify any trends. When is traﬃc ﬂow highest? When is it lowest?

Write a report of your inquiry. Be sure to include your tables and graphs from the previous steps.

Discussion

688

How could the RTA use your findings to manage traffic flow? Provide specific examples based on your investigation.

Research other methods that the RTA uses to manage traﬃc ﬂow in high-travelled roads during busy times of the day.

What other factors would the RTA need to research in order to implement new measures for managing traﬃc ﬂow in front of your school?

Mathspace Victorian Curriculum – Year 9 mathspace.co

11 Chapter review 1

For each survey question, classify the data as ordinal, nominal, discrete or continuous. a

How long did you sleep for last night?

How many people live in your house?

How many languages do you speak at home?

How long did you spend on homework last night?

After the government decided to increase the minimum retirement age, a news poll selected a group of people to ask their opinions on the changes. Should each of these groups be included to avoid sampling bias?

Only people who are employed

People in the community who have a wide variety of views, excluding politicians and policymakers

People in the community who have a wide variety of views, even if they are not directly affected by the changes

Members in the community not impacted by the changes

A local doctor’s clinic is trying out a new way to help patients get better quicker. a

What type of info should the clinic collect to see if this new way works?

How can the clinic make sure the info they get is not biased?

The goals scored by a football team in their matches are represented in the dot plot: a

Complete a frequency table for this data.

Calculate the mean, median, mode, and range of the goals scored. Round your answer to two decimal places.

Make a statement for each of the measures of centre and how it connects visually to the dot plot.

1 2 3 4 Goals scored

Consider the frequency distribution table: Score (x) 20–24 25–29 30–34 35–39 40–44 45–49 50–54

Frequency ( f ) 9 18 37 14 9 6 3

Cumulative frequency (cf )

Complete the table.

Calculate the total frequency.

Describe the shape of the distribution.

Approximately one third of the scores recorded are greater than what score?

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689

The data shows the running times (in minutes) of two track teams, Ace Rockets and Thunders: • Ace Rockets: 36, 37, 41, 41, 43, 47, 49, 50, 52, 56 • Thunders: 30, 32, 34, 36, 38, 42, 44, 55, 57, 62

Find the mean score of each team.

Display the data as a back-to-back stem-and-leaf plot.

The back-to-back stem-and-leaf plot shows the number of tasks completed over several days by Selma’s and Sirin’s students: a

Did Selma’s students complete 7 tasks on any day?

Whose class had the higher median?

For which group is the mean greater than the median?

Selma’s students

2 4

2 3

Sirin’s students

Key: 2|0 = 2 8

Key: 2|3 = 23

This histogram shows the speed of balls delivered by cricket bowlers: 16

Construct a grouped frequency table for this set of data.

What percentage of balls were bowled at less than 130 km/h? Round your answer to one decimal place.

14 Frequency

12 10 8 6 4 2 0

100

110

120

160

170

Consider the grouped frequency table showing the length of the phone calls made by Pauline in a month: Length of call (minutes) 0<t≤1 1<t≤2 2<t≤3 3<t≤4 4<t≤5 5<t≤6 6<t≤7

690

130 140 150 Speed (km/h)

Number of calls 5 7 2 9 5 9 8

Mathspace Victorian Curriculum – Year 9 mathspace.co

Pauline’s phone company charges phone calls as if they were at the upper boundary of the interval. For example, a phone call lasting 1 minute and 48 seconds would be charged as if it were a 2 minute call. a

Create a histogram for this set of data.

Find the total number of minutes for which Pauline will be charged for phone calls this month.

The histogram shows the number of crime novels in a bookshop for different price ranges rounded off to the nearest $5:

Novel genres

20 15 Frequency

Describe the shape of the distribution of the data.

10 5 0

Consider the data shown in the graph:

Are there any outliers?

Is there any clustering of data? If so, in what interval?

State the mode.

Describe the shape of the distribution.

35 Frequency

5 10 15 20 25 30 35 Price

25 20 15 10 5 1 2 3 4 5 6 7 8 9 10 11 Score

Consider the two histograms which show the grade distributions in two university courses: 20 18 16 14 12 10 8 6 4 2 0

Grades in a university course (Section 1, N = 104)

Number of individuals

80 Section 1

20 18 16 14 12 10 8 6 4 2 0

Which distribution has a clear mode?

Which distribution would be described as “uniform”?

Which distribution has the highest range?

Compare the mean and median of Section 1.

Grades in a university course (Section 2, N = 104)

80 85 Section 2

Chapter 11 review mathspace.co

691

The stem-and-leaf plot shows the ages of people on a bus to enter through the gates of a concert in the ﬁrst 5 seconds. a

Find the mean age, rounded to two decimal places.

After the data was collected, one more person boards the bus. After they board, the new mean age becomes 25 years.

Leaf 1

692

6 Key: 1 |4 = 14 years old

Pulse rate before exercise

Pulse rate after exercise

Find the mean pulse rate:

Before exercise

After exercise

What is the mode pulse rate after exercise?

Find the range of pulse rates:

Is the data positively or negatively skewed?

Ten participants had their pulse measured before and after exercise with results shown in the stem-andleaf plot:

How old is the person that just boarded the bus? c

Before exercise

After exercise

What can you conclude about exercise and pulse rate from the measures of centre and spread?

A battery manufacturer tests their batteries in a portable music player, to ﬁnd out how long the batteries last under constant use. The frequency table shows the data from their tests: a

Construct a histogram to represent the data.

What percentage of batteries last 10 hours or longer? Round your answer to one decimal place.

If the average length of a song is 4 minutes, what percentage of batteries will last for an average of 250 songs or more? Round your answer to one decimal place.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Key: 6|1|2 = 12 and 16

Battery life (mins) 400 ≤ x < 500 500 ≤ x < 600 600 ≤ x < 700 700 ≤ x < 800 800 ≤ x < 900 900 ≤ x < 1000 1000 ≤ x < 1100 1100 ≤ x < 1200 1200 ≤ x < 1300 Total:

Frequency 3 9 13 32 22 27 6 4 1 117

A pet store recorded the sales of different breeds of dogs over a year. The data showed a symmetric distribution for the sales of Labradors, but a positive skew for Beagles. How might the store owner interpret the popularity and sales trends of these two breeds based on this information

The following table shows the number of units of blood donated per week over 10 weeks in two states: VIC SA

3986 1355

3738 1213

3949 1275

3909 1397

4130 1181

4079 1252

3894 1372

4079 1247

3711 1501

3871 1175

What was the average number of units donated per week in Victoria and South Australia?

Using the average units donated per week for each state, what is an estimate for the number of units collected in Victoria and South Australia over a year? Assume that there are 52 weeks in a year.

A rough estimate of the units of blood currently required annually per state is 3% of the population. If the population of Victoria is 5 640 900 and the population of South Australia is 1 659 800. Will each state meet the required number of blood donations?

One study suggests that demand for blood is expected to increase by 25% in four years time due to factors such as prolonged ageing. How many units of blood annually will Victoria require in four years’ time?

Chapter 11 review mathspace.co

693

Answers

16 a

b b π

17 a 12

1.01 Rational and irrational numbers

c 5

d 8

18 Understanding 1 a iv

b ii

c i

Reasoning

d iii

19 Yes. For example

20 Yes. For example

3 4 a Rational b Rational c Rational d Rational

21 a Irrational

e Rational f Irrational g Irrational h Rational

i Irrational j Rational k Irrational l Rational

π is the exact value, so it is the most accurate. We can ﬁnd the difference between π, 3.14 to ﬁgure out which one is closer.

Fluency 5 a True

b False

c True

d False

e True b

, Rational

, Irrational smaller,

b Rational

c Irrational

d Irrational

e Irrational

f Rational

g Rational

h Irrational

−4

9 a b c d

−3

−2

−1

−2

Since the difference between π and

d 0.08, Rational

7 a Rational

23 a

−1

is closer to π than 3.14. and

is between

b Her answer should be less than 5.5 as 26 is much closer to 25 than it is to 36. 4

Problem-solving 24 a c = 13

25 26 a 12 square units

−3

−2

−1

b 13.0384 square units c 7.96%

10 a 7

b 11

c 8

d 15

11 a C

b C

c A

d B

12 a 49 < 59 < 64

b 7<

13 a 16 < 22 < 25

14 a

15 4

694

, and

6 a 4, Rational c

b Rational

Mathspace Victorian Curriculum – Year 9 mathspace.co

27 a 2.6 cm b It’s diﬃcult to measure small lengths like 1 cm accurately. It’s also diﬃcult to draw an angle of exactly 90°. So, your answer to (a) might have been incorrect due to inaccuracies in the drawing. c Since it’s easier to accurately measure larger lengths, we could enlarge the spiral. For example, if we enlarge the spiral by a factor of 2, we would start with a right triangle with sides of 2 cm. Then the hypotenuse of the cm. We would sixth triangle would be 2 × measure the hypotenuse of this triangle and divide by 2 to estimate the value of .

12 The order of operations has not been applied properly. The division should have been performed before the addition. The correct answer is −9.

2 2

2 4 2 3 2 2

2 5 2 6

2 1 2 17

2 7 2 16 2 8 2 15 2 9 2 10

2 14 2 11 2 12

2 13

13 Yes. The distributive property of multiplication over addition is applied in both expressions. is not a

14 The side lengths must be rational number. Problem-solving 15 60 minutes 16 14 people

1.02 Operations with integers

17 a 11th ﬂoor

Understanding 1 a False

b True

c False

d False

b 4

18 $156

Chapter 1 review

2 A

1 a Rational b Rational c Rational d Irrational

3 C

e Rational f Irrational g Irrational h Rational

4 A

2 a

−4

−3

−2

−1

Fluency 5 a −1

b −8

c 21

d −19

e −25

f 6

g 4

h 59

6 a −34

b 63

c 42

d 53

e −36

f 120

g −85

h 87

7 a −15

b 6

c 72

d −9

e 5

f −5

g 30

h −5

8 a 17

b −2

c 18

d −10

f −9

g 42

h 1

e −58 9 a Reasoning

10 Orion because he evaluated the expression from left to right. 11 The mistake occurred in step 1, where addition was performed before multiplication. The correct working step should be: 1

(−2) + 12 × 3 − (−4) = (−2) + 36 − (−4)

3 a D

b B

4 a

c 5 a 4

b 7

6 a The perimeter is

c 2 , so it is irrational.

b The area is 7, so it is rational. 7 a 7 × 4 = 28 m2

b 27.3861 m2

8 a −2

b 9

c −12

d 20

9 a −23

b 65

c −36

d 80

10 a 2

b −6

c 7

d 17

11 a 18

b −96

c −29

d 20

12 a

13 The error is in step 2. After the ﬁrst step, 15 was added to 7 to get 22, but multiplication should be done before addition.

= 34 − (−4)

= 34 + 4

= 15 + 14 − 9

= 29 − 9

= 20

= 38

15 − (−7) × 2 + (−9) = 15 + 7 × 2 − 9

Answers mathspace.co

695

14 $600

Reasoning

15 $25 000

9 The multiplication law requires the bases to be the same. Since the bases in the expression are different, we cannot simplify it using the multiplication law.

2.01 Multiplication law Understanding 1 a 35 = 3 × 3 × 3 × 3 × 3 b 34 = 3 × 3 × 3 × 3 c 35 × 34 = (3 × 3 × 3 × 3 × 3) × (3 × 3 × 3 × 3)

= 39

d When you multiply the same bases you add the indices. b 24

2 a 5×5×5 c a×a×a×a×a

d b

10 Marvin’s solution is correct. When multiplying terms with the same base, we add the indices: (b2) × (b3) × (b7) = b(2+3+7) = b12. 11 a (−2)3 = −2 × −2 × −2 = −8. −23 = −(2 × 2 × 2) = −8. So these expressions are equal. b (−2)4 = −2 × −2 × −2 × −2 = 16. −24 = − (2 × 2 × 2 × 2) = −16.

So these expressions are not equal.

3 a Trent’s statement is incorrect because the expanded forms of the terms do not match. To show this, let’s expand each term: 22 = 2 × 2 = 4, 32 = 3 × 3 = 9. So, 22 + 32 = 4 + 9 = 13, while 52 = 5 × 5 = 25. Therefore 22 + 32 is not equal to 52 since 13 is not equal to 25. b No, because multiplication index law applies to multiplication with the same base. In 22 + 32, the operation used is addition and the bases are different.

c • If a negative number is raised to an odd power, the result will be negative. • If a negative number is raised to an even power, the result will be positive. Problem-solving 12 a a2 × a9 = a11 b c1 × c7 = c8 c b5 × b6 = b11 d 2a2 b2 × 3a2 b = 6a4 b3 e 3mc2 × −1bm6 × 2b3 c3 = −6b4 c5 m7

Fluency

f 7

4 a 2

c 2

d 5

h 710

b 3

e 3

f 2

g 3

i (−2)8

j 29

k 38

5 a y5

b x9

c a12

d a5

f y

g u

h b13

i a3

j x3

k d3

l e−2

6 a 5y5

b 3y4

e b

f 21a

c 4x9 6

d 6a10 10

g 72u

7 a 6x3 y5

b 20c7 d7

4 9

10 11

h 15y4

d −6u v

e 12t11 u6

f −25a5 b6 c10

g −15w6 x8 y10

h 27a5 b5 c8

b b 3

y 4

6 c v 25

a2 b3 c6

13 a5 cm2 14 a 5, x = 9, y = 1, x = 8, y = 2, x = 7, y = 3, x = 6, y = 4, x = 5, y = 5. b There would be an inﬁnite number of possible solutions, x = 11, y = −1, x = 12, y = −2 etc.

2.02 Division law

d c 2

f x 5

2 a False, when dividing like bases, you subtract the indices. b False, the bases must be the same. c True

696

a1 b1 c2 =

Understanding

c 14a b

3 8 a a 4

a1 b2 c4 ×

Mathspace Victorian Curriculum – Year 9 mathspace.co

d True e True f F alse, different bases and we don’t subtract the bases. g True h False, the coeﬃcients should be divided, not subtracted.

b 76

c 911

d 32

e 6

f 2

g (5)

h 22

i (−8)7

4 a 51 = 5

which simpliﬁes to: m0

b 24 = 16

c 4 = 64

d (−2)1 = −2

e 32 = 9

f (−1)5 = −1

g (−2)5 = −32

Now, let’s consider dividing m6 by itself:

According to the division law, we subtract the indices:

Fluency 3 a 37

13 Using the division law, we know that when dividing two numbers with the same base and different indices, we subtract the indices:

Now, we know that any non-zero number divided by itself is equal to 1:

So, we have: m0 = 1

5 a a4

b x2

c y6

d b4

e j7

f x3

g f3

h s9

i p3

j h9

k c5

l d6

6 a 4b4

b 3y11

c 6m11

d 4c5

e 5b8

f 5y3

g 4k5

h 8t2

Problem-solving

i 2k5

j 8g7

k 3p3

l 5l6

14 a b9 ÷ b4 = b5

7 a x6

b y6

c t5

d k14

e a5

f b5

g u5

h w2

b c4 d7

c m7 p10

e 5h j

f 3xy

3 3

g 20r s

h 4k4 l9

9 a x3 y4

b a 3 b3

c jt

d k7 m4

8 a f 2 g8 4 3

e n p

3 2

f c d

3 2

g a z

This shows that, using the division law of indices, m0 is equal to 1 for any non-zero value of m.

d wv7

b x5 ÷ x3 = x2

h q w

15 Many possible solutions. For example:

Reasoning 10 Jane’s solution is incorrect. According to the division law, when dividing terms with the same base, we subtract the indices. Also, when she divided the coeﬃcients, she took the reciprocal, but shouldn’t have. The answer should be

16 4a4 cm 17 18 a Answers may look like:

11 The division law is am ÷ an = am−n. The base has to be the same. 12 Becky is correct. We cannot simplify using the index laws, but we can still evaluate using the

b Answers may look like:

order of operations

Answers mathspace.co

697

2.03 Power of a power law

Understanding 1 a Yes

b Yes

c Yes

d No

e No 2 a False 32

3 a (p ) = p

b True 6

f (4m2)3 × (2b2)2 = 256m6 b4

b ( pm)n = pmn

g (0.1a3)2 × (10b4)1 = 0.1a6 b4 h [(2a2)3]2 = 64a12

Fluency 4 a 221

b 316

e 924

f 1125

c 718 g 230

h 560

13 33a−2c × 52b−3d

j (−3)

k (−2)

l 13

5 a j18

b w8

c t12

d a21

e h42

f p18

g a25

h b28

i c

j e

3 6

6 a a b

4 12

12 216p

i 7

d 56

k g

l w

b w x

e t10 u15

f a28 b8

g 4h10

h 27a6

i 25c14

j a5 b20 c15 k

m −32a15

n −27a6

7 a 222

b 314

c a43

d x12

e y5

f f 19

g 9t11

Reasoning 8 If we apply the index law (am)n to each expression, we get w6. If we go even more basic, (w2)3 = w2 × w2 × w2 = w6, and (w3)2 = w3 × w3 = w6

14 125a12 cubic units

2.04 Zero and negative indices Understanding 1 a 2 b

2 a Base: 10

Power: −7

b Base: 2

Power: −4

c Base: 13

Power: −10

d Base: −5

Power: −8

3 C 4 B 5 A

9 The equation simpliﬁes to a2x = ax , so we need a value that satisﬁes 2x = x2. 2

x = 2 is one possible value (x = 0 is another).

6 Less than 1 Fluency 7 a 1

b 27

c 1

d 1

e 1

f −1

g 1

h 7

i −1

j −9

k 32

l 1

n 1

o 1

p 1

8 a 0

b 7

c 10

d 0

10 The student correctly simpliﬁed the numerator, (z4)3 to get z12, but incorrectly divided by z3. Instead of dividing 12 and 3, the student should have subtracted the powers. The correct simpliﬁcation is z12−3 = z9. Problem-solving

9 a

11 a (abc)3 = a3 b3 c3 2 33

6 9

b (2m n ) = 8m n

698

Mathspace Victorian Curriculum – Year 9 mathspace.co

2−1

105

104

103

102

100 000 10 000 1000 100 c

33 27

3−1

101 100 10−1 10 3−2

3−3

b 16

c 3

d 2

f 16

1 815

12 a

e an

b 37−1

c 5−1

d 4−7

e −15−3

f 10−5

g (−24)−10 h 25−3

−11

−8

13 a 3−1

19 a 32y8

b 60y5

c 40x

−2

d 72x−7

e 6y

f −4y11

g 2x−1

h 21x−3

20 a 36x−4

b 100x6 y−8

d 250xy−6

e 81x−10 y16

Reasoning 21 a 0 b

k 16−12

l (−45)−7

b −5

c −3

d −3

e −11

f −3

g −4

h −5

−1

b 10

c 2

−7

d 35

e 9−12

f 822

g 7−12

h 100−1

−2

−23

i 13

j −7

14 a −2 15 a 5

10 a 7

11 a

and dividing by zero is

impossible. 22 The expression 2x0 represents the number 2 times x raised to the power of 0. Any non-zero number raised to the power of 0 is equal to 1. This means that 2x0 simpliﬁes to 2. The expression (2x)0 represents the entire quantity 2x raised to the power of 0. When a non-zero base is raised to the power of 0, the result is always 1. This means that (2x)0 simpliﬁes to 1. 23 The correct value is −1 because the negation applies to the base, not to the exponent. 24 The −1 index only applies to the term directly in front of it. In this case it only applies to the a,

i 2

j 6

k 19

l −824

16 a 1

b 16

c 27

i 49

k 125

b 28a2

Problem-solving

h 25m3

26 a

17 a 20y6 e 8h7 i

k 3x8

not the 5. The correct working is

25 The expression x−n is the same as

due to

the negative index law. This law can be applied to any value of x, including negative values.

b −1 3

c a b × 2a b 2 −4 7

n 3x8

2 −4

= 2ab

−1

3 −8

d (gh ) g = g h o

p e

18 a

Answers mathspace.co

699

b 2200

27 a

c 4y8

d 36x7

9 a 2

b 0.000 73

c 7

d 4

e 734 000

f −5 b 8 × 10−5

10 a 6 × 109 6

c 1.8 × 10

d 4 × 105

f 5 × 10−5

e 3 × 10

b 4.21 × 10−2

11 a 8.3 × 105

cm2

c 5.39 × 10

30 4

12 2 × 10−6 m

2.05 Scientific notation

13 5 000 000 m 14 7.12 × 1020 litres

Understanding

15 a 3.8 × 10−3 seconds

1 C

2 A −8

3 a 10

b 10

c 10

d 10

b 2.2 × 10−8 g

c 1.67 × 10 L

d 6.3 × 107 L

e 3.4 × 10−10 m

f 8 × 1011 bytes

−2

g 5 × 10

bels

h 1.2 × 105 pascals

Reasoning

Fluency 4 a Largest: 7.27 × 10−7

Smallest: 2.22 × 10−7

−9

Smallest: 5.13 × 10−9

c Largest: 1.25 × 102

Smallest: 1.25 × 10−6

b Largest: 8.31 × 10

−4

d Largest: 9.37 × 10

e Largest: 7.88 × 102 9

Smallest: 6 × 10−9 Smallest: 2.93 × 10−2 8

f Largest: 4.7 × 10 × 10 Smallest: 4.7 × 10−8 × 108 5 a 4.5 × 101

b 1.28 × 102

c 2 × 103

d 8.84 × 105 7

e 8.4626 × 10 0

f 2.3456 × 104

g 6.14 × 10

h 9.7 × 109

6 a 6 × 10−1

b 2 × 10−3

−4

c 7.4 × 10

d 1 × 10−4

e 6.76 × 10−4

f 3 × 10−5

g 5.932 × 10−5

h 9.2 × 10−2

7 a 3 000 000 000 c 9 000

b 41 300 d 897 000

e 503 000

f 3014

g 8 004.5

h 13 008 000

8 a 0.0528

b 0.007

c 0.000 004 7

d 0.000 000 213

e 8.29

f 0.000 0347

g 0.0972

h 0.000 064

16 Extremely large and small numbers have many zeros so are diﬃcult to read and write. When given in scientiﬁc notation these numbers are much shorter to write, and a scientist can see at a glance the magnitude of the number by looking at the index of ten. 17 a i 4.75 × 107

18 a When a number is correctly expressed in scientific notation, the number part, a, must be such that 1 ≤ a < 10. Here a > 10. b When a number is correctly expressed in scientific notation, it is of the form a × 10n, where n is an integer. Here n is not an integer. c When a number is correctly expressed in scientific notation, the number part, a, must be such that 1 ≤ a < 10. Here a < 1. d When a number is correctly expressed in scientific notation, it is of the form a × 10n. Here the base is 8, not 10. Problem-solving 19 a 2 300 000

b 1.38 × 108

20 a 3 × 10−1 km/s

b 3 × 105 km/s

c 10 times

Mathspace Victorian Curriculum – Year 9 mathspace.co

ii 4.75 × 107

b The answers are both 4.75 × 107 because 1.35 × 103 is very small compared with 4.75 × 107.

700

d 7.49 × 10−4

21 a 8.63 × 10−23

b 7.2%

3.01 Algebra review

22 a 3.14 × 10−4 kg

b 2.76 × 10−3 cm

Understanding

Chapter 2 review 8

1 a 7

b (−2)

e c

8+h

d b

c 3

2 a 4 = 64

2 a True

b False

c True

e False

f False

g False

3 a a9

b (−5)20−t c x3 y5

d 2a

b (−2) = 16

b j

5 a 9

10 15

c x y z

b −1

c 8

d 1

7 a

9 a x10

b 3y

c 9c

d −2x

e 9c + 2d

f 5x + 5y

g 5x + 7y + z

h 4x + 7 − 2y

i x2 + 11x + 10

j 3ab + 5bc l 7a2 b + 4ab2

k x − 16

6 a 25

3 a 13x

8 a

h d 3t6

m 8x + 5

n 5ab + 10cd 2

o 4xy + 2x y − 2xy − 7 p −4x + x2 4 a 45u

b 28r

c 6xy

d 24rs

3 5

e 15u v

f 60pq2

g 15cd

h −25ef

i 30xy

j 36x2 y2 z2

k −54a3

l 20a2 b2

7 5

n −18w4 x5 y5

7 8

p −15e14f 7

m 20a b o 42c d

10 a 16

d 5 a

c 3

d −4

d 8.9 × 10−4

g 4

b 38 500

i 3r

j −2

6 a 15

j 35fg

d 0

f 18ab

i 3

k 9c

m 3a3

n 3u6 v5

o 15x5 y4

p 7a6 b5

11 a 2.34 × 102

b 1.13 × 105

c 8 × 10−1 12 a 60 000 000 c 1.4

d 0.004

e 0.0336

f 0.000 003 005 −7

13 a 8 × 10

b 6 × 10

d 3 × 10−8

c 2 × 10

14 a 3.51 × 10

b 2.35 × 10−2

15 a 5.1 × 10−3 seconds

b 3.04 × 103 m

c 5 × 10 bytes e 2.3 × 100 bels 16 a8 m3 6

−8

d 8.9 × 10

7 a 8c2 2

17 P ÷ P = P 18 315 m 19 20 3 × 104 mm

d False

Fluency

4 a 1 =1

d −1

f −3y and 5y

b (−6) = 36

c 26 = 64

e −9

f 5x

b x and y c 5

1 a 6 6

Answers mathspace.co

701

Reasoning 8 a 3a and 4b are not like terms. We can only add or subtract like terms. b Yes, if a = 1 and b = 1 the expression is true. 9 In algebra, “like terms” refer to terms that have exactly the same variable factors, i.e., the same variables raised to the same power. Even though x2 and x have the same variable, x, they are raised to different powers.

multiplying two or more numbers together. As long as we multiply every term inside the bracket by the term outside the bracket, then we maintain equivalence of the expression. 3 a 3a(2a + b) = 3a × 2a + 3a × b b 6m2 v(3 − 2m2 + v) = 6m2 v × 3 − 6m2 v × 2m2 + 6m2 v × v c 4ab(3b + 2c − 6a) ≠ 4ab × 3 + 4ab × c − 4ab × 6

10 This would only be the case for a = 0

Fluency

11 In 7a + 5a × 2, if we do the multiplication ﬁrst we get the correct answer of 10a. If we do the addition ﬁrst the answer will be 24a, which is incorrect. By following the order of operations, we ensure that we do not get two different answers to the same problem.

4 a 3 (2x + 4) = 6x + 12

12 Lawrence subtracted both the coefficients and the powers of the variable. However, when dividing by a monomial, the coefficients should be divided and the powers should be subtracted. Marika, on the other hand, divided both the coefficients and the powers when she should have divided the coefficients and subtracted the powers.

b 2 (−4s + 5) = −8s + 10 c 5 (3p + (−6)) = 15p − 30 d −2 (3g − 7) = −6g + 14 b m2 + 7m

5 a ab + 7a c y + y2

d 2y + 12

e 4m2 + 40

f m2 − 3m

g 4y2 − 10y

h 8m − m2

i 5t2 + 70t

j 4b2 + 32b

l −w2 − w

k −15a + 20a 2

m −16t − 12t

n −11mn − 19mp

3 5

o 56s t − 63s 6 a 7y3 + 28y

p 6wy2 + 6w2 y b −15y3 + 18y2

d 24u5 + 30u7

Problem-solving

c 18a + 36a

13 (6b + 2c) cm

e −15x3 + 6x6

f −2s4 − 5s13

14 (10x − 2y) cm

15 a Middle box 7a + 2b. Top box 11a + 3b. b Middle box 2x − 4y. Bottom left box 3y. Bottom right box −x − 3y. 2

c 20x + 36xy − 8x

b 3xy2

19 (8.5 − 2x) + x + x + (8.5 − 2x) + x + x + (11 − 2x) + x + x + (11 − 2x) + x + x = 39

3.02 Distributive law Understanding 1 Dortha is correct because the rule of exponents dictates that the proper way to multiply a3 and a3 is by adding the exponents, which gives a6. 2 Commutative property of multiplication states that the order of numbers doesn’t matter when

Mathspace Victorian Curriculum – Year 9 mathspace.co

b 6a2 − 30ab + 18ac d 2a4 + 8a3 − 14a2 b h2 + 14h + 81

8 a 11y + 75 c x + 6x + 10

17 $(10x + 5y + z)

702

16 36a cm

18 a xy

7 a 2f 2 − 16fg − 14fh

d 27c3 + 6c2 + 21c

e 32c − 12c − 20c

f 17y2 − 52y

g 10y2 − 57y

h 39x − 54y

i 26x2 + 58xy

j 8x − 14

k 6x − 9

l 16u2 + 20uv

m −7n2 − 147n

n −5n2 − 50n

Reasoning 9 a i Leon did not multiply both terms. ii Jill used addition for the ﬁrst term instead of multiplication. iii Claire did not change negative and positive signs.

iv Beryl forgot the negative when multiplying the second term. b −6(4x − 25) = −24x + 150 10 a i x + 1

e m2 + m − 30

f 10m2 − 61m − 63

g 21w2 + 53w + 30

h 30y2 − 3y − 27

i −28y2 − 64y − 36

6 a a2 − 9

ii x + 2

b x2 − 4

c y2 − 25

b x + 2(x + 1) = 15 + 2(x + 2) c 17, 18, 19 d No, because the equation used to solve for x in part (a) has only one unknown, x, to the power of one. Therefore there is only one solution. 11 15π + 10

d 4x2 − 25

e 16x − 36

f 25x2 − 4

g 9x − 49

h 9y2 − 16

i 81m2 − 64

j 144x2 − 1

7 a x2 + 14x + 49

b x2 + 8x + 16

c x2 − 6x + 9

d x2 − 10x + 25

Problem-solving 12 2 (x − 12) + 9 = 3x − 8 13 a A = 2 + (x + 1) (x + 5) A = x + 7 + x (x + 5)

e x + 12x + 36

f x2 − 4x + 4

8 a 9a2 + 24ab + 16b2 2

c 4x − 28xy + 49y

c The variable x represents a length. Lengths in geometry cannot be negative, as they measure size or distance.

14 a If w metres represents the width of the garden, then the area will be w(2w − 1) = 2w2 − w square metres.

b 4a2 − 12a + 9

b A = x + 6x + 7

d 36p2 + 36pq + 9q2 2

e 16m − 40mn + 25n f x2 − 4xy + 4y2 h

9 a ab − 12a + 6b − 72 b k2 + 5k − 2kr − 10r c q2 t + 9q2 − 7t − 63

b 8w + 12

d x3 + 5x2 y + xy + 5y2

c 11 × 21 metres

e 3m3 − 3m2 p − 4mp + 4p2 f 24g3 + 20g2 h3 + 18gh + 15h4

15 −2a4 + 8a2 + a

g 21a4 − 18a3 b2 − 7ab2 + 6b4

16 a A(B + C) = (x + 1) (2x + 3)

h 35x2 y4 − 60x7 y + 28xy3 − 48x6

b A(B + C) = 2x2 + 5x + 3 17 a 40x2 − 14x

Reasoning

b The range of possible surface area is between 26 cm2 and 3860 cm2.

3.03 Binomial expansion Understanding 1 a 64

b 25

c 9x2

d 144y2

2 a Yes

b No

c No

d Yes

3 a i A = y2

ii B = 3y

iii C = 6y

iv D = 18

b y + 9y + 18 4 a False

b True

c True

Fluency 5 a x2 − 6x − 72

b x2 + 8x + 12

c v2 + 21v + 110

d x2 − 11x + 28

10 The difference in the number of terms arises because in (x + 2) (x + 2), the middle terms (2x and 2x) combine to give 4x, resulting in three terms. In contrast, in (x − 2) (x + 2), the middle terms (2x and −2x) cancel each other out, leaving only two terms. 11 a We can determine Jane’s bedroom by multiplying (x − 1) by (x + 1). This means that the area of Jane’s bedroom is (x − 1)(x + 1) = x2 − 1. Since x2 > (x2 − 1) for all values of x, then Marvin has the larger bedroom. b Marvin’s bedroom is larger by 1 m2. 12 Yes, the results of each expansion are x2 − 8x + 16 and 16 − 8x + x2 respectively. This occurs because when a negative number is multiplied by itself, it becomes positive; and when expanding a perfect square, where the

Answers mathspace.co

703

two expressions are identical, the negative signs negate each other and yield the same result.

c 5ab = 5 × a × b = a × 5b

13 Not correct, as (a + b)2 does not equal a2 + b2. The square of a binomial has three terms. Rahul forgot to include twice the product of x and 9 in his answer. The answer should be x2 + 18x + 81.

After crossing out 5b, the remaining factor is a.

14 Anette made a mistake when using the area model calculate the area of the top-left square. (10x) (10x) = 100x2, but Anette put 100x instead. Her ﬁnal answer should have been 100x2 + 20x + 30x + 6 = 100x2 + 50x + 6

After crossing out 5b, the remaining factor is a2.

Problem-solving

d 5a2 b = 5 × a × a × b = a2 × 5b

e 5ab2 = 5 × a × b × b = ab × 5b After crossing out 5b, the remaining factor is ab. f 120a2 b2 = 2 × 2 × 2 × 3 × 5 × a × a × b × b

15 a 3y + 24y + 45

b 8x − 64x + 96

c −x2 − 11x − 30

d −4y2 − 40y − 36

= 24a2 b × 5b

e −9y + 9y + 378

f −48x2 + 72x − 27

After crossing out 5b, the remaining factor is 24a2 b.

g 24x2 + 40x − 16

Fluency

16 a (t + 2)

b (t + 11)

c (t − 5)

d (t + 6)

e (t + 12)

f 2 (t + 9)

g 2 (v − 3) h 2 (t + 8)

b 8x − 48x + 96x − 64 2

c 2x + 10xy − 8xz + 3x + 15y − 12z 3

d 6t + 25t + 16t − 15

19 x2 + 7x + 12 cm2 20 n + 8 3

21 4x + 48x + 140 m

Understanding

d y2

b 2t (t + 1)

c 3y ( y + 2)

d −m (m − 19)

e −y ( y + 2)

f v (8 − v) b 5(a − 2)

c 5(3 + 4a)

d −2 (9a − 8)

e 2a(2b + 3)

f m (11k − n)

g −2 (6s − 5)

h h (8j − 9g)

i kt (16j − 5)

j −7 (5m − 7n)

7 a −3(a + 2)

3.04 Algebraic factors

c b2

f bx

e 4m

6 a 4(a + 3)

18 4x2 + 20x + 25 square units

b x

5 a y ( y + 5)

17 a x3 + 9x2 + 27x + 27 3

4 a x

b 2(2 − 3a)

c −5(c + 6)

d −4(2 + 3d)

e −b(4 + 5a)

f −a(bc + 5a)

g −w (w + 6)

h −9mq (5n + 8)

1 a 2

b 5

c 4

d 6

8 a y ( y + 4)

b k (k − 9)

e 8

f 6

g 9

h 20

c x (3 − x)

d z2 (1 + 6z2)

2 a −12x + 5, No 2

c 20x − 12x , No

b −12x + 20x2, Yes 2

d −12x + 20x , Yes

3 a 10ab = 2 × 5 × a × b = 2a × 5b After crossing out 5b, the remaining factor is 2a. b 15a2 b = 3 × 5 × a × a × b

= 3a2 × 5b

After crossing out 5b, the remaining factor is 3a2.

704

Mathspace Victorian Curriculum – Year 9 mathspace.co

e z2 (1 + 8z4)

f 5u (u − 3)

g w2 (3y − 8)

h −7 (3r − 2) (3r + 2)

i 4uv (11 − 2u)

j 5k2 t (1 + 8t)

k 7pq (7p − 4q)

l −2a2 (8 + 9b)

m −5w (6 + 5y)

n uv (9v − 10u)

9 a 4(x + 3 + 4xy)

b 3(x + 3 + 4yx)

c 2x(1 + 5xy + 4y)

d 4x(xy + 2y2 + 3z2)

e 5bc(ab + 5c + 20a) f 10c(3b2 + ab + 2c) g 2ab(ab − 1)

h 5y(x2 y + 3z2 + 5y + 12)

3.05 Factorise by grouping

Reasoning 10 The factor of −5 was not factored out of the −15x term correctly. The correct answer is −5x (x + 3).

Understanding 1 x+7

11 They are both correct. If you expand the answer each gave you will see that they both produce −5x + 10y. 12 Although he took out 2 as a common factor, he forgot to take out x as a common factor. The fully factorised form is 2x(x2 + 4x − 2)

2 18, 9 3 a p = 9, q = 10

b p = −9, q = 7

c p = −6, q = −4 Fluency 4 a (x + 2) (x + 3)

b (x + 2) (x + 5)

c (x + 3) (x + 3)

d (x + 2) (x + 6)

e (x − 1) (x − 2)

f (x − 2) (x − 4)

g (x − 2) (x − 3)

h (x − 3) (x − 3)

i (x + 8) (x + 3)

j (x + 9) (x + 8)

k (x − 3) (x − 1)

l (x − 5) (x − 12)

m (x + 5) (x − 6)

n (x + 5) (x − 4)

o (x + 6) (x − 9)

p (x + 7) (x − 10)

q (x + 5) (x + 8)

r (x − 5) (x − 7)

s −(x + 4) (x + 2)

t −(x − 3) (x − 4)

5 a ( y + 5) ( y + 7)

b (a − 4) (a − 3)

c ( p − 3) ( p + 6)

d (q + 4) (5 − q)

e (r − 3) (5 − r)

f (2r − s) (6r − rs)

b x(2x + 16)

g (t + u) (7t + 2u)

h ( y − z) (x − w)

c 2x(x + 8)

i (4w + 3x) (5y − z)

j ( y − 4) (8y − 3)

Problem-solving 13 x − 4 14 a 5n(4 + n), so the dimensions are 5n and 4+n b Length = 5n, Width = 4 + n c Each section is now n × (4 + n) and to fence this the perimeter is 8 + 4n units of fencing. 15 a

16 a 4x y − 4xy

6 a (x − 2y) (8 + z)

b (3 + x) (8 + y)

c (x + z) (7y + w)

d (x + 2) (x + 5)

e (x − 3) (x + 8)

f (5 + x) (10 + y)

g (12y + w) (x + z)

h (x + 5) (x + 8)

i (x − 5) (x + 10)

j ( p + 2) (2m + 3)

k ( p + 3) (5m + 2)

l (2x + 3z) (1 + 6y)

m (4x + 3z) (1 + 8y)

n (a + 8) (a2 + 1)

7 a (x + 4)2

b (x + 5)2

c (x − 2)

d (x − 10)2

e (8 + x)2

f (6 − x)2

b x−y c i xy(x − y) 17 a 60a3 b2 − 30a2 b3 c 40% 18 (t − 5) (3t + 4)

ii 4 (x − y)

Reasoning

b 60a2 b2(2a − b)

8 ( y − z) ≠ (z − y) so either a factor of −5x needs to be taken out of the ﬁrst group or a factor of −7 needs to be taken out of the second group.

9 We know that the factor pairs of 24 are (1, 24), (2, 12), (3, 8), (6, 4), (−1, −24), (−2, −12), (−3, −8) and (−6, −4). When we add these pairs up, the largest possible integer we can get is 25, so, k = 25.

Answers mathspace.co

705

Problem-solving

15 a 3 (x + 3)

b a (16b − c)

10 a 2 (x + 2) (x + 5)

b 4 (x + 2) (x + 4)

c −4 (2x + 1)

d −7x (5y + 3)

c 3 (x − 5) (x − 2)

d −2 (x − 2) (x + 11)

16 a 7c (c − 4)

b −f 2 g (3 − g)

e −3 (x − 2)2

f 4 (x − 9) (x + 8)

g 5 (x − 2) (x + 3)

h −4 (x − 5) (x + 2)

17 a 3x (6x + 1)

11 a x (x + 3) (x + 4)

b 2x (x + 5) (x + 3)

12 a (2x2 + 3z) (2y − 1)

b (3x2 − 5y) (x + 4y)

b Length = 6x + 1 Width = 3x

c (g + h) (3f + g + h)

d (2c − d) (c + 5d + 3)

13 a t + 7

b 4 (t + 7)

14 (4 + x) (3 − x) units2

d 3t (st + 3s2 + 10)

c 5 (k + 7jk + 3)

c Fencing = 12x + 1 18 a (x + 5) (x + 7)

b (x − 3) (7x − 2)

19 a (g + 2) (g + 6)

b (a − 2c) (3 + b)

c ( y + 5) (3x + 2) 20 a (x + 4) (x + 7)

Chapter 3 review 1 a 5y

b 8a b 42u2 v4

2 a 6rs 2 2

4.01 Midpoint

d −6c14 d5

c 80xy z

b (7a − 13b) (a2 − 3b)

21 a 3 (x + 5) (x + 2)

d −10a2 + a

c 4x + 7y

b (x − 8) (x − 6) d (x − 9)2

c (x + 4) (x − 8)

3 a

4 a −2

d 5t

Understanding 1 One

5 Total cost = 2000 + 50x + 75y 6 a 2x2 − 5x

b M (10, 6)

3 a Yes

b No

b −11ab − 9ac

c −4g2 − 8g 7 a 6f 3 + 18f

d 21c3 d4 − 63c3

Fluency

b −16a3 + 8a6

4 a M (5, 7)

2 a M (2, 5)

d 5g − 20fg + 40g 8 a −5j2 − 70j

b 23n + s

c 15y + 5

d 7c2 − 40c

c M (−3, 2)

d M (−3, −2)

e M (−2, 1)

f M (2, −4)

5 a M (8, 6)

 13 d M  4, −   2

c M (−5, 9)

9 4 (x − 8) + 13 = x + 14

10 a b2 + 10b + 21

b c2 − 19c + 88

e M (−9, −4)

f M (−5, 2)

c 3d − 13d − 30

d 15e2 − 2e − 24

g M (2, −6)

h M (−9, −7)

e f 2 − 81

f 144g2 − 100

11 a h + 6h + 9 2

b 2

c 16c + 40cd + 25d

12 a rs + 7r − 8s − 56 b p3 + 13p2 w + pw + 13w2 13 14 a t

6 a i M (9, 7)

b 2y2

11 10 9 8 7 6 5 4 3 2 1

y B M A

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14

706

Mathspace Victorian Curriculum – Year 9 mathspace.co

f i M (−3, 2)

b i M (−3, 4) ii

8 6 M

4 2 2 4 6 8

y 2 A −12−10−8 −6 −4 −2 −2 M −4

2 4 6 8

y 4 x

−7 −6 −5 −4 −3 −2 −1 M −2

1 2 3 4

−4 −6 −8

−8

−10

h i M (−4, −5)

d i M (5, −2) 8 6 4 2 −2

−2 −4 −6 −8 −10 −12

y B

1 −12 −10 −8 −6 −4 −2 −1

x 2

6 M

y x 2

−2 −3

−4 M A

−5 −6 −7

7 $29 million

e i M (5, 7) 9

−6

g i M (−3, −2)

c i M (−2, −3)

−10 −8 −6 −4 −2 −1 A −2 −3 −4

x −12−10−8 −6 −4 −2 A −2

5 4 3 M 2 1

8 63 years

7 6

10 a 207 203

b 268 291

Reasoning

3 2 1

x 1

11 a The diagonals bisect one another. b i

c In a parallelogram, the diagonals bisect each other, meaning the midpoints of the diagonals coincide (are the same point).

Answers mathspace.co

707

Since the midpoints of AC and BD are both , ABCD is a parallelogram. 12 a T (8, 8)

19 a = −6 and c = 0 or a = 18 and c = 12 b k = −11

20 a M (3, 7)

b S (6, 1)

c Parallelogram, because the diagonals bisect each other. 13 The mean of two numbers is halfway between the two numbers. The mean of the x-coordinates is the midpoint of the x-coordinates and the mean of the y-coordinates is the midpoint of the y-coordinates.

21 (0, 5.5) 22 b b=0

23 a

4.02 Gradient Understanding

Problem-solving

1 A

14 a i B (7, 7) ii

y 8 B

6 M

x −2 −2

2 A

b Positive gradient

3 a Zero gradient

b Undeﬁned

4 a Increasing

b Decreasing

5 a Negative

b Positive

6 a The line has a gradient of zero. b The line has an undeﬁned gradient.

b i B (−5, −7)

7 C, B, A, D

y A −6

2 a Negative gradient

Fluency

−8

18 Q (−6, 1), R (−8, 4), S (−10, 7)

−4 M

8 a i 2

ii 6

iii

b i 12

ii 6

iii 2

c i 14

ii 7

iii 2

−2

d i 12

ii 4

iii 3

−4

e i −12

ii 4

iii −3

−6

f i −6

ii 2

iii −3

2 x

−2

−8

9 a 2

10 a i

15 a A (5, 6)

b A (6, −5)

c A (5, 6)

d A (5, −6)

e A (−5, 6)

f A (4, −3)

g A (−4, −5)

h A (6, 22)

i A (−3, 3)

j A (−7, −4)

16 The centre of a circle is always the midpoint of its diameter. So we can use the midpoint formula to ﬁnd the coordinates of point B. So B (8, −1). 17 (4x, 3y)

708

Mathspace Victorian Curriculum – Year 9 mathspace.co

1 −5 −4 −3 −2 −1 ii 3

x 1 2 3 4 5

b i

6 7

16 a y = 27

2 1 −2 −1 −1

4 7

f −

6 11

b 5

Reasoning

1 2 3 4 5 6 7 8

17 a

−2

−3

c Yes, the points A, B, and C lie on a straight line and the gradients of AB and BC are the same.

ii c i

Since both gradients are equal to

A, B, and C do indeed lie on a straight line.

−1

18 Because a vertical line has no run, so the

−1

formula for the gradient

−2

−4

19 They will get the same gradient. One characteristic of a line is that its gradient is constant all the way along it. So, you can choose any 2 points along a straight line and their gradients will be equal.

−5

ii −5 6 5 4 3 2 1

20 Line 2 is steeper (increases faster) because it has a larger gradient value than Line 1. The gradient of a line represents its steepness, and a larger gradient means that the line increases more rapidly in the vertical direction as it moves horizontally. In this case, for every unit increase in the horizontal direction, Line 2 increases by 5 units vertically, while Line 1 increases by only 3 units. Therefore, Line 2 is steeper and increases faster than Line 1.

−6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6

1 2 3 4 5 6

Problem-solving

11 a 0

b −4

c −3

d 3

f 0

g −2

21 a 3

i 1

j −1

22 m = 1

m −10

n −

13 As x increases, the value of y decreases. 14 As x increases, the value of y increases. 3 2

b −

b 15

c 20

23 T he gradient of 0 has to have y-values the same, this will have to be A(3, 5) and D(7, 5).

7 15

12 Undeﬁned

15 a

will have a

denominator of zero, and any fraction with a zero on the denominator is undeﬁned.

−3

d i

, the points

5 2

c −

3 5

The positive gradient needs to increase from left to right, so the x and y-values of the ﬁrst point need to be lower on the Cartesian plane than the x and y-values for the second point. So E(−3, 10) and F (4, 12). The negative gradient needs to decrease from left to right, so B(4, 6) and C(−1, 8).

Answers mathspace.co

709

b Positive

24 a y = 1

c The gradient is 20 metres per second.

b y=1

d The gradient, which is 20 metres per second, indicates that the car maintains an average speed of 20 metres per second throughout this period.

ii −1

c i 1 d −1

25 a Since the line has a positive gradient, it means that Jay’s distance from the starting point increases as time passes. In the context of the problem, this implies that Jay is moving forward towards his school along the straight track.

4.03 Distance between two points Understanding 1

in this context represents

b The gradient

Jay’s average walking speed during his journey to school. The rise corresponds to the change in distance, while the run corresponds to the change in time. A steeper gradient indicates a faster walking speed, and a ﬂatter gradient indicates a slower walking speed. 26 i a Positive

b 2

b 5

c In the context of the graph, this positive gradient means the object is moving away from the starting point at an average speed of 5 metres per second. b −3

iii a Negative

c In the context of the graph, this negative gradient means the object is moving back towards the starting point at an average speed of 3 metres per second. iv a Neither

100 Distance (metres) 90 80 70 60 50 40 30 20 10 Time (seconds) 1

710

d AB = 5 units

3 a AB = 14 units

b AB = 8 units

c AB = 6 units

d AB = 3 units

4 a True

b False

5 a i PQ = 3 units

c False

d True

Mathspace Victorian Curriculum – Year 9 mathspace.co

ii QR = 4 units

iii PR = 5 units b i PQ = 7 units

ii QR = 4 units

iii PR = 8.06 units c i PQ = 3 units

ii QR = 8 units

iii PR = 8.54 units d i PQ = 4 units

ii QR = 4 units

iii PR = 5.66 units 6 a AC = 3.61 units

b AC = 10 units

c AC = 13 units

d AC = 13.04 units

7 a AC =

units

8 a i

b AC = 4

b 0

c In the context of the graph, this zero gradient means the object is not moving, and its distance from the starting point remains constant over time. 27 a

b AB = 5 units

c AB = 3 units

Fluency

c In the context of the graph, this positive gradient means the object is moving away from the starting point at an average speed of 2 metres per second. ii a Positive

2 a AB = 10 units

units

3 2 1

−6 −5 −4 4 − −3 3 − −2 2 − −11 −11 −

x 1

−2 − 2 −3 − 3 −4 4

ii AC =

units

b i

y 5

4 3

−8 −6 −4 4 − −2 2 −11 −

−2 2

ii AC =

22 a i MN =

units

b AC =

units

c AC =

units

d AC =

units

e 8 units

units

ii NR =

units

ii QR =

units

23 23.1 units

b 15 units

units

b (7x, −2x)

b 2:5

10 a AC =

units

21 a 10x d 6.40 units

11 a 10 units

units

iii BC =

c Yes

b 25 units

c 11.40 units

ii AC =

• To show it has two 45° angles, check if two sides are equal since the triangle must be isosceles.

units

9 a 20 units

units

b • To show it has a 90° angle, check if the sides satisfy Pythagoras’ Theorem.

2 1

20 a i AB =

24 a i PQ =

units

iii PR =

units

b Yes ii

25 a i iii

c 13.89 units

d 10.20 units

b i

e 5.10 units

f 9.49 units

c Parallelogram

13 a 6.40 units

b 3.09 units

c 10.11 units

d 12.36 units

b 60 units2

e 8.67 units

f 0.91 units

12 a 4.47 units

b 6.71 units

14 a 9.90 units

c N

b 11.18 units

15 a AB = 3.61 units

units

b AM = 1.80 units

c N is closer because 17 a AC =

units

units units

c C is equidistant from A and B. d No. If we plot points A, B, and C in the Cartesian plane, we see that the points do not lie on the same line. Problem-solving 18 AK = 7.2 units 19 a i AB = 5 units iii BC = 8 units b Isosceles

27 a 50 units2

units

28 a 17 units

b 34π units

units c d= c 17 units

4.04 Gradient-intercept form

is smaller than b BC =

ii BD =

d Yes, because the distance between the centre and the point (20, 4) is the same as the radius.

Reasoning 16 a

units

26 a i AC =

ii AC = 5 units

. Understanding 1 m represents the gradient of the line and c represents the y-intercept of the line. 2 a The graph of the line will have an upward incline, moving from left to right. This means that as the x-coordinate increases, the y-coordinate also increases. b The graph of the line will have a downward incline, moving from left to right. This means that as the x-coordinate increases, the y-coordinate decreases.

Answers mathspace.co

711

c The graph of the line will be horizontal, having no incline. This means that regardless of the x-coordinate, the y-coordinate remains constant. 3 a m = 15, c = −10

b m = 7, c = 0 d m = −1, c = 6

c 4 a m = −4

b −4

Fluency 5 a i (0, 3)

ii m = 2

b i (0, 3)

ii m = 4

c i (0, 5)

ii m = −2

d i (0, 2)

ii m = −6

6 a y = 2x + 5 c

b y = −6x + 9

14 a y = 4x + 3 c y = 0.6x + 3 15 a y = −3x – 2

b Yes

16 a y = −3x + 8

b No

17 a i 2

ii y = 2x + 2

b i 2

ii y = 2x − 9

c i -6

ii y = −6x + 2

d i 7

ii y = 7x − 3

18 a

c y = −x – 1 y 4 3 2

g y = −8x − 3

7 a i m = −2

ii c = 9

b i m = −4

ii c = −8

c i m=8

ii c = 6

d i

ii c = −1

8 a i

19 a

e y = 8x − 4

−2

−1

ii c = 7

ii c = −8

c i m = 12

ii c = −9

d i m = 18

ii c = −12

e i m=4

ii c = −5

f i m = −1

ii c = 4

−4 −3 −2

−1

−1 −2

b y = −2x + 3 c

c y = −51

y 3

10 a y = 2x + 1

b y = 59 kilometres

11 a y = 3x – 4

b y = 146 metres

12 a y = 3x − 4

d y = 3x + 1

e y = −2x + 5

13 a y = 5x + 3

−2

b i m=9

9 a m = −2, c = 3

b y = 5x − 6

712

Mathspace Victorian Curriculum – Year 9 mathspace.co

−2

−1

1 −1 −2 −3

22 a The coeﬃcient of x is the same.

b All the graphs have the same gradient.

c The lines would be parallel on a Cartesian plane because they have the same gradient but different y-intercepts. This means they will not intersect each other, because they are always the same distance apart.

1 x −1

−1 −2

23 a b Line A. The gradient of line A is

20 a i y = −2x − 2

which has a larger magnitude than the gradient of line B.

ii y = −2 iii x = −1 iv

24 Line B. The gradient of line A is

, which has a

smaller magnitude than −1, the gradient of line B.

25 The gradient is −3. A negative gradient tells us that we draw a line going down or decreasing from left to right.

6 4 2

−10 −8 −6 −4 −2 −2

2 4 6 8 10

26 The equation y = 2x represents a line with a gradient of 2 and a y-intercept at (0, 0). The equation y = −2x represents a line with a gradient of −2 and a y-intercept at (0, 0).

−4 −6 −8 −10

The difference between the lines is y = 2x is increasing from left to right while y = −2x is decreasing from left to right.

b i ii y = 12 iii x = −15 iv

y 20 16 12 8 4 −20−16 −12 −8 −4 −4

x 4

8 12 16 20

−8

27 No, it is not possible for a vertical line to be represented by an equation in the gradientintercept form y = mx + b. This is because a vertical line has an undeﬁned gradient, and m in the gradient-intercept form represents the gradient of the line. Vertical lines can be represented by an equation in the form x = k, where k is the x-coordinate of any point on the line. Problem-solving

−12 −16

−20

29 a i m = 1

ii c = −13

b i

ii c = −3

c i

ii c = −7

d i m=1

ii c = 3

e i

f i

Reasoning 21 a The constant term is the same. b All the graphs intersect the y-axis at the same point. c Lines of varying gradient will each pass through same y-intercept of (0, 1).

Answers mathspace.co

713

30 a y = −x + 2

b a=8

y 4

31 y = x + 15

32 y = 28x + 85

4.05 Linear graphs

y = 4x

Understanding

−2

−1

1 B , plot 4 on the y-axis, then

2 To graph

rise 2 units up and run 3 units to the right, which gives the point (3, 6), then draw a line passing through the points (0, 4) and (3, 6).

d i

−1

y 4

3 You let y = 0, then you solve for x.

y = −x + 1

Fluency

4 a i x −1 0 1 2 y 8 9 10 11 ii

10 9 8 7 6 5 4 3 2 1

y=x+9

−9 −8 −7 −6 −5 −4 −3 −2 −1 −1

b i

−1

1 x −2

−1

e i

−1

−7 −4

−1

y 1

−1

−1 −2 y = 3x − 4

−3 −4 −5

y 4

f i

3 y = 3x + 3 2

x −2

−1

10 9 8 7 6 5 4 3 2 1

−1

c i

−1

−4

8 −1

714

Mathspace Victorian Curriculum – Year 9 mathspace.co

y = −2x + 8

x 1

5 a i

d i

−3

−1 −10 −19

−2

y 9 8 7 3x 6 y=− 2 +6 5 4 3 2 1

2 y = −3x − 1

1 x

−2

−1

−1 −2

b i

−3 −2 −1

6 a

−9 −7 −5 −3

−2 −1

−1

y 1

−1

−1 −2 −3 −4

y 1 −4 −3 −2 −1

−3

−2

−1

−2

−1

−3

−2

−4

−3

−5 −6 −7

y = 2x – 9

−8 −9

c (0, −2)

d (−2, 0)

e y = −5

f x=2

7 a i (−1, 0) iii c i

−3

ii (0, 1) 4

3 2

y 3

−4 −3 −2 −1 −1

−2

−4

x 1

−3 x −7 −6 −5 −4 −3 −2 −1 x y= +2 −1 3

b i

ii (0, −1)

Answers mathspace.co

715

iii

y 5

2 1

−4 −3 −2 −1 −1

−2

c i (4, 0)

ii (0, 4)

g i (0, 0)

3 2

−3

−4

h i (2, 0)

iii

2 1

−4 −3 −2 −1 −1

ii (0, 6) 7 6 5 4 3 2 1

−2 −3

−1

−4

x 1

−1 −2

ii (0, −3)

e i (1, 0) iii

8 a

y 2 1

−1

−1 −2 −3 −4

716

−2

−3

ii (0, −4)

f i (2, 0)

−4 −3 −2 −1 −1

−2

−4 −3 −2 −1 −1

2 1

iii

ii (0, 0)

iii

d i

−1

−4

iii

−1

−3

ii (0, 4)

Mathspace Victorian Curriculum – Year 9 mathspace.co

−3

−8 −6 −4

2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11

y x 1 2 3 4 5 6 7 8 9 10

c (0, −6)

e y = −2

d (9, 0)

9 a i

ii (0, 1)

iii

10 a

7 6 5 4 3 2 1

2 1 −3

x −2

−1

−2

−1

x 1

−1 −2

−1

−2

3 2 1

ii (0, −9)

b i (−2, 0) iii

−2

12 10 8 6 4 2

−1

1 −1 −2 −3

−10 −8 −6 −4 −2 −2

2 4 6 8 10

11 a

−4 −6 −8 −10 −12

y 2 1 x −1

iii

1 −2

d i (−2, 0) iii

x 1

−1 −2 −3 −4 −5 −6 −7 −8 −9 −10

−2

−1

1 −1

ii (0, −9)

c i (3, 0)

−3

−4

3 2 1

ii (0, 8)

−4 −3 −2 −1 −1

−2

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

x 1

−3 −4 x 2 4 6 8 10

Answers mathspace.co

717

c 4

iii

−4 −3 −2 −1 −1

−3 −4

−4

ii (0, 3)

d i

iii

3 2 1 −4 −3 −2 −1

−2

−3

−4 −3 −2 −1 −1

−2

x 1

−1

−4 −3 −2 −1 −1

−2

−3 −4

ii (0, −3)

12 a i 3 iii

4 3 2 1 −2

−1

13 a

x 1

−1 −2 −3 −4 −5 −6

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

b i −2 iii

5 4 3 2 1

ii (0, −1) 4

5 4 3 2 1

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

x 1 2 3 4 5

−4

c i 1

ii (0, −1)

Reasoning 14 Elisabeth was correct. She remembered to go downhill for her rise, as the gradient was negative.

718

Mathspace Victorian Curriculum – Year 9 mathspace.co

15 No, the graph of y = −4x + 1 does not rise from left to right. It falls from left to right because the gradient of the line is −4, which is a negative number. In a graph, a negative gradient means the line goes downhill as you move from left to right.

19 a W = 5 − L b

L 0 W 5

16 The graphs of y = 7x − 1 and y = 7x + 3 would be parallel lines on a Cartesian plane because they have the same gradient but different y-intercepts. This means they will not intersect each other, because they are always the same distance apart.

0 −1 −2 −3 −4 −5

17 y = 10x + 50 The graph of the corresponding line is:

10 y 9 8 7 6 5 4 3 2 1 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

Problem-solving

9 10

x 1 2 3 4 5 6 7 8 9 10

Money saved

d Yes, the graph should have endpoints. For the shape to be a rectangle, L needs to be greater than 0 but less than 5. That means W must also be less than 5 and greater than 0.

150

100

4.06 Equation of a straight line

50 Weeks 1

2 3 4 5 6 7 8 9

Understanding 1 a

b −4

f −1

2 a No

b Yes

It would take 10 weeks for Mary to save $150. 18 a i 2 iii

ii (0, 6) 10 8 6 4 2

4 a i m = −3 x 1 2 3 4 5

−5 −4 −3 −2 −1 −2 −4 −6 −8 −10

c Yes

d No

ii (1, 6)

b i

ii (−4, 5)

c i m = −9

d i m=1

ii (−8, −3)

Fluency

ii (0, −8)

b i 2 10 8 6 4 2

3 B

−5 −4 −3 −2 −1 −2 −4 −6 −8 −10

iii

c −3

5 a y = −4x − 24

b y = 4x + 19

c y = −3x + 15

d y = −9x − 33

e y = −2x − 15

1 2 3 4 5

Answers mathspace.co

719

6 a i y = 2x + 7

e i y = −8x − 6

y = 2x + 7

−4 −3 −2

−1

8 7 6 5 4 3 2 1

−1

−14

−30 −646

−9

−6

−45 −30

−1 −2

10 a y = −5x + 40 c y = −6x + 66 11 a Yes

5 4 3 2 1

b No

b y = 2x d c Yes

d No

Reasoning 12 Yes, it is possible. You can ﬁnd the gradient using the coordinates of the two intercepts using the formula

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

13 If the line is horizontal, the gradient m is 0. The formula becomes y − y1 = 0(x − x1), which simpliﬁes to y = y1. Since y1 is a single value, this is the equation of a horizontal line.

ii y = −4x + 9

b i

c i

. Once you have

the gradient, you can use the point-gradient formula or the gradient-intercept form to ﬁnd the equation of the line.

3x −1 y=− 2

7 a i −4

14 To sketch the graph of the line from the equation, begin by plotting the point (x1, y1 ) which was given in the equation.

8 4x − 7y + 40 = 0

Then, using the gradient from the equation, use the rise and the run to sketch more points on the line. Connect the points to get the full graph of the line.

9 a i y = −3x x

−3

−6

−9 −12 −48

15 Example answers: • y = 2x + 1

b i ii

−3

y f i y = 5x ii

b i

−28

−4

−2

−21

−9

−8

−7

−6

• y = 3x • y = −x + 4 All lines have different gradients and y-intercepts.

c i y = −8x + 90 ii

−15

210

d i y = −2x + 5 ii

−8

−4

−1

−7

−29

Problem-solving 16 5x − 4y + 44 = 0 17 a (4, 0)

b y = −4x + 16

18 a (1, 1)

b 3x − 10y + 7 = 0

19 a p = 17

b q = 12

c 9x − 2y − 129 = 0 20 18 square units

720

Mathspace Victorian Curriculum – Year 9 mathspace.co

4.07 Parallel and perpendicular lines

Reasoning

Understanding

19 T he ﬁrst two lines have the same gradient of 3, so they are parallel. The last two lines have the

1 a Their gradients, the values of m, are the same.

same gradient of

b All graphs have the same gradient or rate of increase.

, so they are

ﬁrst and third lines is

perpendicular. Since opposite sides are parallel and adjacent sides are perpendicular, the enclosure is a rectangle.

2 −1 and

, so they are parallel.

Additionally, the product of the gradients of the

20 y = 2x + 4

b One pair: y = 3x + 15 and

21 y = 2x + 4

Another pair: y = 3x + 15 and

Problem-solving 4 Perpendicular 22 a

6 D

23 a −1

b y = −x + 18

Fluency

24 a 6

5 D

7 a Yes

b No

c Yes

d Yes

e No

f No

g No

h No

25 y = 0.5x + 5

8 a Yes

b Yes

c No

d Yes

e No

f No

g No

h No

9 a −5

b y = −5x + 40

27 a

b k = 10

10 a y = −2x − 5

b y = 8x + 5

28 a

b (5, 10)

c y = −3x – 4

d y=2

e x = −7 11 a

c −8

d 1

12 a No

b No

c Yes

d Yes

13 a No

b No

c No

d Yes

14 a No

b No

c Yes

d Yes

e No

f Yes

g Yes

h No

29 c = −2 30 a −1

b 1

c −1

d Yes

31 a 0

b −1

c No

d Yes

e A trapezium 32 a Gradient of: b Yes

15 a

33 a i 16 a 17 a

c No

b b b

18 a c y = 2x + 6

c Yes

iii 1

iv 1

b A parallelogram. This is because the opposite sides have the same gradient, making them parallel. It is not a rectangle, as the adjacent lines are not perpendicular.

b d x = −8

e y = −8

Answers mathspace.co

721

b 6

4.08 Direct proportion

c a = 6b Understanding

d a = 42 e b = 10

1 a Gradient: 3

b Gradient:

2 a Yes

b No

c No

d No

3 a True

b True

c False

d True

e False

10 a 280

b 5

11 a 12 a 300 Water 270 240 210 180 150 120 90 60 30

Fluency 4

b 70

Rice 10 20 30 40 50 60 70 80 90 100 110 120 130 140

10 20

b 2.1

6 a No

b No

c No

7 a b = 9a c 8 a

d Yes

c For every 100 g of rice William uses, he adds 210 mL of water.

b y = 2x − 10

The unit rate of water with respect to rice is 2.1 mL/g.

13 a

9 8 7 6 5 4 3 2 1

Fish caught

Hours 1

2 3 4 5 6 7 8 9

r 10

b c 9 ﬁsh are caught every 4 hours

b d q = 15

14 a i 15

e r = 44 9 a

50 5

40 30

x 2

20 10 b 2

722

Mathspace Victorian Curriculum – Year 9 mathspace.co

b i

y 12 10 8 6 4 2

Amount saved

22 20 18 16 14 12 10 8 6 4 2

Jimmy

Ben 1

2 4 6 8 10 12 14 16 18

Days 3

c By comparing the constant of proportionality or unit rate. The higher unit rate would be the steeper graph, which represents the person who is saving more quickly.

ii c i

y 120

18 a y = 4x

100

x 10

Online

Newspaper

10 5

d i

c There were more online enquiries each day than there were newspaper enquries.

d The total number of enquiries, both online and newspaper, are directly proportional to the number of days advertised.

Problem-solving

5 x 2

19 m = kp 20 r = kn

21 a

15 a

b b

Reasoning 16 It increases by a factor of 7. 17 a

Days

Amount saved

21 28

Hours worked

Plates painted

18 16 14 12 10 8 6 4 2

x 2

Answers mathspace.co

723

22 a

Number of weeks

Number of books read

b Carl 23 a

Robots built 7 6

Derek

5 4 3

6 a 1.4 m/min

b y = 1.4x

c 8.4 m

d 9 minutes

7 a 0.5x + 1.5y = 5

b 4 sheets

8 a 135 mg

b 70 hours

9 a y = 45x + 150

b $420

10 a y = 16x – 1300

b $3660

11 a y = 100 − 9x

b 59.5 km

12 a y = 75x + 50

b 218.75 km

Reasoning

Yuri

13 a The cost of each text message. Time (hours)

b The amount of credit Paul purchased.

2 3 4 5 6 7 8 9

c $21.64

b Derek

14 The baseball is moving downwards instead of upwards at T = 4 seconds.

24 a 6

b 84 cm

25 a k = 0.82

b h = 0.82s

15 a The rate at which the water level is falling per minute.

c h = 1.64 cm 26 a A = 3C

b 28

b 21 grams

c The water level before any water is pumped out.

4.09 Model linear equations

16 a 49 represents the initial value of the heartbeat (i.e. resting heartbeat).

Understanding 1 a x+4=6

b The change in one minute of Anri’s heartbeat.

b x − 10 = 9 c 11x = 88

Problem-solving

17 a y = 10x + 340

b $700

e 4x − 13 = 21

18 a $63.50

b $5.50

f 8x + 11 = −5

c A ﬁxed cost such as a toll charge

g 6(x − 20) = −10

19 a 3.5 megabytes

b S = 8 − 0.5(D − 1)

c 15th day

20 a C = 2920R + 50

2 The independent variable.

b $0.21 Fluency 3 a y = 6.8x + 0.8

b $41.60

4 a 0.2 seconds

b y = −0.2x + 6

c 5.1 seconds

d x = 2.5 hours

5 a

Number of minutes passed (x)

c If the smart speaker is left on for 24 hours a day, the total running cost would be 24 × 365 × R = 8760R. Adding the initial cost of the smart speaker, we get the new total cost: 20

Amount of fuel left in 150 125 100 75 50 the tank ( y) b y = 150 − 5x

724

c 30 minutes

Mathspace Victorian Curriculum – Year 9 mathspace.co

Cnew = 50 + 8760R. To show that the cost does not triple, we can compare the new cost with three times the original cost.

Substituting the expressions for Cnew, we have:

7 a i x>1 ii

Cnew = 50 + 8760R

−5

−10

−5

−0.5

0.5

b i y<0

= 50 + 8760(0.21) = 1889.60

Three times the original cost is 3 × 650 = 1950.

−10

c i j<4

From the computation, we can conclude that the cost does not triple when the smart speaker is permanently left on. b m = 220 − a

21 a 188 bpm

−10

c Between 94 bpm and 159.8 ≈ 160 bpm

4.10 Linear inequalities

d i a ≥ −1 ii

−10

e i v>2 ii

Understanding

−10

f i b ≥ −12 ii

1 a i 25 26 27 28 29 30 31 32 33 34 35

−15

g i m>8

ii –33–32 –31 –30–29–28–27–26–25 b For x ≥ 29: i No

ii No

iii Yes

iv No

iii No

iv Yes

For x < −29: i No

ii No

2 The inequality symbol must be reversed. 3 a False

b True

c False

4 a x ≥ 14

b v≥0

c c>3

d x > −8

e b≤0

f g≥9

g s < 19.8

h y < −1.52

i x ≤ 8.2

h i a < 0.4 ii

−1

i i b ≤ 2.8 ii 0

j i s < 7.5 5

e m > −1

f b ≤ −2

g h < 11

h m>6

i m < −34.6

j x ≥ −1 b

c m ≥ −24

d y < −14

e a>4

f k ≤ 45

g y > 40

h x≤3

−10 −8 −6 −4 −2 0 2 4 6 8 10

l i x>5 ii

d u ≤ 2.3

k i y≤5 ii

b x ≤ −3

c e < −13

6 a

−5

Fluency

5 a y>2

d True

−5

0.5

1.5

−5

mi ii n i q>1 ii

−10

o i u≥8 ii 0

Answers mathspace.co

725

b No, washing 8 windows would only generate $16 which would give him a total of $38.10, which is not enough money to afford the equipment.

p i x≤6 ii

−10

−5

8 x≥3

c No, he gets paid for washing the whole window, not parts.

9 a q > 16

d Yes, if Ryan washed 9 windows he would earn enough to buy the equipment. If he washed more than 9 windows, he would have more than enough money.

b There is no greatest value. 10 p + 15.25 ≤ 32 p ≤ 16.75 11 A − 33.40 ≥ 32

Problem-solving

A ≥ 65.40 12 x + (63 + 73 + 98) ≥ 300 13 a 5h + 72 ≥ 379

b 14 rockmelons

25 a

x ≥ 66

26 Lachlan, James

b h ≥ 61.4

27 a x + 650 ≥ 950

c 62 14 a x + 488.30 ≤ 2118.40 b x ≤ 1630.10 15 a B ≤ 3.4

b $960

c Yes c $1630.10

28 a 5 031 kg

b 3

29 150 grams ≤ 0.18w + 0.03(1500 − w) ≤ 225 grams

16 W ≤ 4

b No

17 a No

b Yes

c Yes

d Yes

Least value is 700 grams and greatest value is 1200 grams

18 a Yes

b Yes

c Yes

d No

Chapter 4 review

19 a No

b No

c No

d Yes

20 a No

b Yes

c No

d No

Reasoning 21 x < 4 means x is any number less than 4. It doesn’t include the endpoint (4). 0 ≤ x ≤ 4 means x is any number between 0 and 4. It does include the endpoints, so x can also be 0 or 4.

1 a M (−1, 2)

b M (−2, −2)

2 a (−2, 0)

3 a A (−6, 6)

b A (8, 3)

4 a Positive

b Undeﬁned

c Negative 5 a −2

6 −9

b 25 − 2x < 31 Step 1: −2x < 6 Step 2: x>−3

8 a

23 The minimum should be at −9 instead of −8 and the dot at that point should be ﬁlled as −9 is included in the solution. 24 a No, washing exactly 9 windows would generate $18, which with the $22.10 he already has enough money to afford the equipment.

726

Mathspace Victorian Curriculum – Year 9 mathspace.co

b Undeﬁned d 0

22 a Step 1 is incorrect. Isabelle has subtracted 25 from both sides but the left-hand side should now be −2x not 2x.

d Zero

7 a 8.06 units units

b 9.43 units b

units

9 28.4 units 10 a i m = −4

ii c = 12

b i m = 12

ii c = 4

c i

ii c = −7

d i

11 a m = −3, c = 1

b y = −3x + 1

c y = −59 12 a

−3

−1

15 a No

b Yes

c No

d Yes

16 a No

b No

c Yes

d Yes

17 a y = −5x – 2

y 12

19 a y = 4x + 15

b y = −3x − 22

20 a (2, −3)

b 6x − 7y − 33 = 0

21 a Yes, y = 2x

b No

22 a

4 2 −3 −2

−1

5 4 3 2 1

x 1

−2

23 a $.04

24 a

0.5

1.5

Number of kilometres 150 110 left to drive ( y)

b y = 150 − 80x

a x≥9

1 2 3 4 5

c x > 5 26 a No

ii (0, −6)

iii

d b No

b i (5, 0) iii

−2 −3 −4 −5 −6

d No

b 12 pineapples

x 0 1 2 3 4

y First Difference Second Difference −1 −1 − (−1) = 0 1−0=1 −1 0 − (−1) = 1 0 2−1=1 2−0=2 2 3−2=1 5−2=3 5

2 a Linear

b Quadratic

ii (0, −3)

c Linear

d Quadratic

e Linear

f Linear

g Quadratic

h Quadratic

i Quadratic

j Quadratic

1 −2 −1 −1

c Yes

Understanding x

−4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

c 112.5 minutes

5.01 Quadratic relationships

2 1

b c<3

27 a 14 a i

c 281 g

Number of hours passed (x)

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

b $12.60

x 1

3 a

Step number

Number of square

b To go from one step to the next, we add the next whole number. First we add 1, then 2, then 3, and so on. c 15

Answers mathspace.co

727

Reasoning

Fluency 4 a No

b Yes

c No

d No

e No

f Yes

g Yes

h No

i No

j No

k Yes

l No

m Yes

n No

o No

p Yes

5 a Quadratic

b Linear

c Quadratic

d Neither

e Neither

f Quadratic

6 a Quadratic

9 To conﬁrm this, we could calculate the ﬁrst and second differences and check if the second difference is a non-zero constant. The second differences are zero so the table of values cannot have a quadratic relationship. 10 a y1 is a linear relationship as the highest power of x is 1. y2 is a quadratic relationship as it has an x-squared term and no higher powers.

b Linear

c Quadratic d Neither - notice the x-values are not increasing by the same amount each time 7 a

x −3 −2 −1 y 10 5 2 x

10 17 26

1 2 y 25 16

−2 −1

−1

−1 −6 −13 −22 −33

−3 −2 −1

−1

17 31 49

b y1 is a linear relationship because it increases by 3, for each increase in x by one unit. y2 is a quadratic relationship because the increase in y increases by 2 each time for each change in x by one unit (example: +1, +3, +5...). This would make the second differences all 2, which is a non-zero constant. c y1 has a minimum and is U-shaped, so it is a quadratic relationship. y2 has no minimum or maximum output and has a constant rate of change, so it is a linear relationship. 11 Many possible answers. For example:

8 a i T he first figure has 7 squares. The second figure has 10 squares. The third figure has 13 squares and so on.

−3

−2

−1

−3

−5

−3

If we ﬁnd the ﬁrst and second differences, we get:

The number of squares in each figure is increasing by 3. ii Number of squares = 19 iii Linear b i T he first figure has 1 hexagon. The second figure has 4 hexagons. The third figure has 9 hexagons and so on.

−3 −2 −1

−3 −5 −3

First differences

−10 −6 −2

Second differences

Since the second differences are all the same non-zero number, this relationship must be quadratic.

ii Number of hexagons = 25

If we were to plot these points on a graph and join them with a curve, we get a U-shaped graph with a minimum, so this is a quadratic relationship.

iii Quadratic

At each step, the increase goes up by 2 more, for example +3, +5, +7 and so on.

c i T he first figure has 3 squares. The second figure has 6 squares. The third figure has 11 squares and so on. At each step, the increase goes up by 2. Example: +3, +5, +7 and so on. ii Number of squares = 27 iii Quadratic

728

Mathspace Victorian Curriculum – Year 9 mathspace.co

14 12 10 8 6 4 2 −4 −3 −2 −1 −2 −4

x 1

12 a Sometimes true. For example, the ﬁrst table of values has all the y-values positive, but has non-zero constant second differences, so it is quadratic. The second table has all the y-values positive, but has constant ﬁrst differences, so it is linear. x

−2 −1

16 Many possible answer. For example, x

−2 −1

5.02 Graphs and features of quadratics Understanding 1 a Concave up c Concave down 2 a True

b Always true. A change in signs like this means there is a turning point, linear relationships do have not turning points. c Never true. If all of the y-values are the same, then it would be graphed as a horizontal line, so it is not a quadratic relationship. Also, it would have zero ﬁrst and second differences, so it is not quadratic. Problem-solving 13 Any pattern that has 4, 7, 12, and 19 elements in the ﬁgures in the correct order.

b False

b Concave up d Concave down c False

d True

3 The turning point lies right on the axis of symmetry. Knowing the axis of symmetry makes it possible to ﬁnd the turning point, and vice versa. Both elements help in understanding the shape and position of the graph. Fluency 4 a i Concave down

ii Maximum point

b i Concave up

ii Minimum point

c i Concave down

ii Maximum point

d i Concave up

ii Minimum point

5 a The x-intercepts are (4, 0), (−2, 0) and the y-intercept is (0, −8)

b The x-intercepts are (2, 0), (−4, 0) and the y-intercept is (0, 8). 14 a

Width (m)

Length (m)

Area (sq m)

144

216

225

c The x-intercept is (3, 0) and the y-intercept is (0, 9). d The x-intercept and the y-intercept are both (0, 0). 6 a i Concave up

ii (2, −25)

216

iii x = 2

176

144

iv The x-intercepts are at x = −3, x = 7. The y-intercept is at y = −21

b The largest possible area would be 225 square metres because the largest possible rectangular area, given a fixed perimeter, is in the shape of a square. This occurs when both the width and length are 15 m. This is also the maximum value for the quadratic relationship that models this data. c A = w(30 − w) or A = 30w − w2 15 a To find the y-values find the square of x-values, and then multiply the result by 4.

b i Concave down ii (2, 4) iii x = 2 iv The x-intercepts are at x = 0, x = 4. The y-intercept is at y = 0 c i Concave up ii (3, 0) iii x = 3 iv The x-intercept is at x = 3. The y-intercept is at y = 9

b Quadratic c y = 4x2

Answers mathspace.co

729

d i Concave down

9 y 8 7 y-intercept 6 5 4 3 2 x-intercept 1 x-intercept x

ii (0, −4) iii x = 0 iv There is no x-intercept. The y-intercept is at y = −4 7 a i (−2, −4)

ii x = −2

b i (−4, 16)

ii x = −4

c i (7, 0)

ii x = 7

d i (−4, 0)

ii x = −4

8 a i

−4 −3 −2 −1−1 −2 −3 −4

−3 −2 −1

27 12

12 27

27 24 21 18 15 12 9 6 3

c i

iii Equation for the axis of symmetry: x = 0 iv

27 24 21 18 15 12 9 6 3

−3

5 9 y 8 7 6 5 4 3 2 1

d i

x 1

iii Equation for the axis of symmetry: x = 0

730

0 18 16 14 12 10 8 6 4 2

x 1

16 14 12 10 8 6 4 2

Mathspace Victorian Curriculum – Year 9 mathspace.co

y-intercept

−4 −3 −2 −1 1 2 3 −2 x-intercept

−3 −2 −1

−4 −3 −2 −1−1 −2 −3 −4

iii Equation for the axis of symmetry: x = −1

x-intercept and x −4 −3 −2 −1 1 2 3 4 −3 y-intercept −6

b i

−3 −2 −1

−4 −3 −2 −1 −2

−6

−4 −3 −2 −1−3

−3 −2

−1

3.5

−2.5 −4 −4.5 −4 −2.5

4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

x 1

iii Equation for the axis of symmetry: x = 1

y 22 20 18 16 14 12 10 8 6 4 2 y-intercept

8 6 4 2

x-intercept −4 −3 −2 −1 −2

x-intercept x 2 3 4

−4 y-intercept −6

−4 −3 −2 −1

9 a i

−3 −2 −1

−14 −9 −6 −5 −6 −9 −14

y −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18

−4 −3 −2 −1 1 2 3 −2 −4 y-intercept −6 −8 −10 −12 −14 −16 −18

b i

−3 −2 −1

d i

22 20 18 16 14 12 10 8 6 4 2 −4 −3 −2 −1

−3 −2 −1

−7

5 10 y 9 8 7 6 5 4 3 2 1

x 1

iii Equation for the axis of symmetry: x = 0

x 1

10 y 9 8 y-intercept 7 6 5 4 3 2 x-intercept 1 x-intercept x −3 −2 −1−1 −2 −3

x 4

iii Equation for the axis of symmetry: x = 1 iv

−4 −3 −2 −1−1 −2 −3 y

iii Equation for the axis of symmetry: x = 0 iv

c i

−3

−2

−1

2.25

0.25

2.25

10 y 9 8 7 6 5 4 3 2 1 −4 −3 −2 −1−1 −2 −3

x 1

iii Equation for the axis of symmetry: x = 0

Answers mathspace.co

731

c i

10 y 9 8 7 6 5 4 3 2 x-intercept and 1 x

y −2

−1

−3 −2 −1

−1

−2 −3 −4 −5 −6

−4 −3 −2 −1−1 1 2 3 4 −2 y-intercept −3

10 a

x 1

−7 −8

−3 −4 −3

b 2 and −2 c (0, −4)

ii Axis of symmetry: x = 0 iii Turning point: (0, 0) iv x-intercept: (0, 0) y-intercept:(0, 0) d i

d Minimum

e Positive

11 a i 8

y 2

7 1

6 5

x −2

−1

3 2

ii Axis of symmetry: x = 0

1 −2

−1

iv x-intercepts: (−2, 0) and (2, 0) y-intercept: (0, 4)

ii Axis of symmetry: x = 0

Reasoning

iii Turning point: (0, 0)

12 Answers may vary. An explanation may be: The square of any number (positive or negative) is always greater than or equal to 0. Therefore 0 would be the minimum

iv x-intercept: (0, 0) y-intercept:(0, 0) b i 7

13 Always true.

5 4 3 2 1 −2

−1

iii Turning point: (0, 4)

x 1

All parabolas have a turning point, which has either the maximum value (if the parabola opens downwards) or the minimum value (if the parabola opens upwards):

ii Axis of symmetry: x = 0 iii Turning point: (0, −1) iv x-intercept: (−1, 0) and (1, 0) y-intercept:(0, −1)

maximum

2 1

−4 −3 −2 −1 −1 −2 −3 −4

732

Mathspace Victorian Curriculum – Year 9 mathspace.co

x 1

minimum

14 The y-value of the turning point is signiﬁcant in determining the range of a quadratic function because it represents the minimum or maximum value of the function, depending on whether the parabola opens up or down. For an upwardopening parabola, the y-value of the turning point is the lowest point on the graph, and the range of the function will be all y-values greater than or equal to this minimum value. For a downward-opening parabola, the y-value of the turning point is the highest point on the graph, and the range of the function will be all y-values less than or equal to this maximum value.

15 a i B

e 20 ﬂoors high

ii C

iii A

iv D

b D has a negative coeﬃcient so the parabola will be concave down. B has a y-intercept at (0, 4). A has the coordinate (1, 1). C has the coordinate (1, 2). 16 a C

b B

c D

d A

Problem-solving

35000

Amount charged

30000 25000 20000 15000 10000 5000

Number of floors 1

c y = 1000x2 d $64 000 23 a −54 °C b At h = 3 hours and h = 18 hours c Approximately 56 °C d Yes e Approximately at h = 4 hours and h = 17 hours

5.03 Basic quadratics

17 a (0, 9) b The height above the ground from which the tennis ball is initially thrown.

Understanding 1 Any equation where a > 1

c (3, 0) d The time it takes for the tennis ball to hit the ground. 18 a The parabola is concave down because the coeﬃcient of the x2 term is negative. It has a maximum value. b (2, 8) c The maximum value is 8. 19 a The parabola is concave down because the coeﬃcient of the x2 term is negative. It has a maximum value. b (2, 7) c The maximum value is 7.

Examples: y = 2x2, y = 1.5x2, y = 5x2 2 Any equation where 0 < a < 1 Examples: 3 The outputs of y = x2 are 0 and positive values, and the outputs of y = −x2 are 0 and negative values. The graph of y = x2 is concave up, and the graph of y = −x2 is concave down. The graph of y = x2 has a minimum value, but the graph of y = −x2 has a maximum value. The shape of graphs are the same, and they both have a turning point at the origin. The axis of symmetry for both graphs is the same.

20 (−10, 0)

4 a Vertical enlargement b Vertical enlargement

21 x = −2.5

c Vertical compression d Vertical compression

22 a

Number of ﬂoors

Amount charged

$1000

$4000

$9000

$16 000

$25 000

$36 000

Answers mathspace.co

733

Fluency

7 a y = 2x2, y = x2, y = 0.2x2

5 a 8

7 6

8 a y = −x2

y = x2

5 4 y = 3x2 3 2 1 1

9 a y = −2x2

y = 3x is vertically enlarged. y −4 −3 −2 −1 −1

x 1

d y = 2x2

c y=x

10 a y = 4x2

c y = −3x2

−4 −3 −2 −1

b y = 2x2

Reasoning

−2 −3 −4 y = −2x2 −5

y = −x2

−3

−2

−1

y = x2

−6

−7 −8

The outputs of

y = −2x2 is vertically enlarged. 6 a

y 8

3 2

1 2 x 3

−4 −3 −2 −1

is vertically compressed. y

x 1

−2 −3 −4 −5 −6 −7 y = −x2 −8

734

are

closer to the x-axis, and the parabola is less steep (wider).

• All three graphs have the same axis of symmetry.

−4 −3 −2 −1 −1

y = x . This means the outputs of

• All three graphs are vertically enlarged.

of the outputs of

12 Similarities:

7 y = x2 6

are

is vertically compressed.

Mathspace Victorian Curriculum – Year 9 mathspace.co

• All three graphs have the same x-intercept, (0, 0). • The graphs of y = 2x2 and y = −2x2 are enlarged by the same factor. • The graphs of y = 2x2 and y = (2x)2 are concave up. Differences: • The graph of y = (2x)2 is narrower than the graphs of y = 2x2 and y = −2x2. • The graph of y = −2x2 has been vertically reﬂected across the x-axis, and is concave down. The other two have not been reﬂected and are concave up. • The graph of y = −2x2 has a maximum value, but the other two graphs have minimum values. • The graph of y = (2x)2 has a a different enlargement factor than y = 2x2 and y = −2x2.

13 For any value of x, the result of x2 will be positive. Since the result of x2 is multiplied by −3, and a negative times a positive is negative, the value of y will always be negative. Problem-solving 14 a V = 30s2 15

b s = 8 cm 8 7 6 5 4 3 2 1

5 The connection between the y-intercept and the value of k in y = x2 + k is that the y-intercept is equal to k, because when x = 0, the equation becomes y = k. Fluency

−3 −2 −1

4 The entire parabola shifts below the x-axis so there are no longer x-intercepts.

6 a 6 units upwards

b 3 units downwards

c 1 unit up 7 a 4 units upwards

b 2 units upwards

c 3 units downwards x 1

8 a i

4 3 2 1

Area ≈ 7 units2 16 a c y = −4x2

d y = 10x2

17 a

b 6 countries

5.04 Vertical translations

3 Change:

x 4

y = x2 − 5

5 4 3 2 1

y = x2 + 3

y = x2

−4 −3 −2 −1 −1 −2 −3 −4 −5

2 Example answers: y = x2 − 4, y = x2 − 8

y = x2 1 2 3

ii 3 iii (0, 3) c i

Stays the same:

3 −4 −3 −2 −1

−3

2. Turning point: The turning point of both parabolas is still on the y-axis. For y = x2, the turning point is at (0, 0), while for y = x2 + 7, the turning point is at (0, 7). 3. Axis of symmetry: Both parabolas have the same axis of symmetry, which is the vertical line x = 0.

y = x2 1 2 3

iii (0, −5)

b i

1 Examples answers: y = x2 + 5, y = x2 + 10

1. Shape: Both graphs have the same parabolic shape, meaning they are both “U” shaped.

ii −5

Understanding

Vertical translation: The graph of y = x2 + 7 is a vertical translation of the graph of y = x2. The whole graph has been shifted 7 units upwards.

d 1 unit downward

−4 −3 −2 −1 −1 −2 −3 −4 −5

a unit downwards

−6 −9 −12

y = x2 − 12

ii −12 iii (0, −12)

Answers mathspace.co

735

d i

Reasoning

y 20

13 a y = −x2 + 3

ii 10 iii (0, 10) 9 a y = x2

b y = x2 − 11

d y = x2 − 8

10 a y = 2x2 + 3, The graph has been narrowed (vertically enlarged) by a factor of 2 and vertically translated up 3 units. b y = x2 − 1, The graph has been vertically translated down by 1 unit. c y = −5x2 − 4, The graph has been vertically reﬂected across the x-axis, narrowed (vertically enlarged) by factor of 5 and vertically translated down 4 units. d

, The graph has been vertically

reﬂected across the x-axis, widened (vertically compressed) by a factor of vertically translated up

16 a The vertical translation of 10 metres means that the initial height of the toy rocket is 10 metres above the ground. To include this in the equation, we need to add 10 to the original height equation: h = −4t2 + 30 + 10. The new equation becomes h = −4t2 + 40. b 40 metres c 3.16 seconds d 40

h (metres)

30 20 15 10 5

t (seconds) 1

17 a The coeﬃcient −2 in the equation h = −2t2 + 8 represents the effect of gravity on the water’s height, as it causes the parabolic arc to open downwards. The constant term +8 represents the initial height of the water above the fountain when t = 0. t 1

c The initial height of the rocket is given by the value of h when t = 0: h = 4(0)2 + 10 = 0 + 10 = 10 metres. This value corresponds to the y-intercept of the graph.

736

Problem-solving

unit.

12 a 46 metres 50 45 40 35 30 25 20 15 10 5

d y = x2 − 12

and

11 hmax = 20 metres and t = 2 seconds b

b y = 3x2 − 4

15 Yes, vertical translations always result in a new turning point. When a function is vertically translated, every point on the graph, including the turning points, is shifted up or down by a constant value. This means that the turning points will have the same relative position to the rest of the graph, but their coordinates will be different due to the vertical shift.

−5

c y = x − 49

14 a y = x2 c y=x −3

y = x2 1

d y = −x2

c y = −x − 1

y = 2x2 + 10

−4 −3 −2 −1

b y = −2x2 + 8

Mathspace Victorian Curriculum – Year 9 mathspace.co

b 8 metres c 2 seconds d The two fountains have similar parabolic arcs as both equations have the same coeﬃcient of −2, indicating the same effect of gravity on the water’s height. However, the constant term differs, with +8 for the ﬁrst fountain and +10 for the second fountain. This means that

the second fountain has an initial height of 10 metres, which is 2 metres higher than the ﬁrst fountain. Consequently, the maximum height of the water in the second fountain is also 2 metres higher than the ﬁrst fountain.

5 The relationship between the x-value of the turning point and the value of h in the equation y = (x − h)2, is that the x-value of the turning point is equal to h. Fluency

18 a b As we move further and further away from the hill (increasing x), the elevation (h) will continue to increase indeﬁnitely according to the equation , which is a parabola opening upwards. This indicates a problem with the model, as it does not realistically represent the elevation of a hill, which should have a maximum height and then decrease as we move further away.

6 a Right

b 1 unit left d 5 units left

8 a 1 unit right

b 3 units left

c 10 units right

d 5 units left

e 3 units left

g 6.5 units right

h 5 units right

9 a i

c i T he amount of water in the tank is affected by the vertical translation in the equation. In this case, the constant term 1500 represents the initial water level in the tank.

b i

1 Example answers:

x 1 2 3 4 5

iii (−1, 0) 7 6 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2

• y = (x + 2)

• y = (x + 5)2 ii 1 c i

• y = (x − 8)2 3 When comparing the graph of y = x2 and y = (x − 2)2, the shape and width of the parabolas remain the same, but the vertex shifts from the origin (0, 0) to the point (2, 0) due to a horizontal translation of 2 units to the right.

1 2 3 4 5

iii (1, 0) 7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2

ii 9

unit left

ii 1

Understanding

4 When the graph of y = −x2 is shifted left, the y-intercept changes. For example, if we shift y = −x2 two units to the left, the new y-intercept will be (0, −4).

7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2

5.05 Horizontal translations

2 Example answers:

d Left

c 1 unit right

b It takes approximately 22.36 minutes for the tank to be completely drained.

• y = (x − 3)2

c Right

7 a 3 units right

19 a 1500 cm

ii The rate the water drains is affected by the dilation in the equation. In this case, the coeﬃcient of the x2 term, −3, represents the rate at which the water drains.

b Left

x 1 2 3 4 5

iii (−3, 0)

Answers mathspace.co

737

d i 7 6 5 4 3 2 1

d i

−5 −4 −3 −2 −1 −1 −2

7 y 6 5 4 3 2 1 x −9 −8 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

x 1 2 3 4 5

ii x = −6 ii

11 a y = (x + 7)

10 a i

7 y 6 5 4 3 2 1 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

ii x = −1 b i

c i

d y = (x − 4.6)2

e y = − (x + 3)2

f y = − (x − 2)2

12 a The graph has been horizontally translated 6 units to the left. b The graph has been horizontally translated 9 units to the right.

d The graph has been horizontally translated 13 units to the left. e The graph has been horizontally translated unit to the left. x

2 3 4 5 6 7

f T he graph has been horizontally translated 0.3 units to the right. 13 y = − (x − 6)2 or y = − (x − 4)2 14 There are two values for D that solve the equation approximately 7.32 metres and −5.32 metres, however negative distance doesn’t make sense in this context, so the answer is 7.32 m.

iii (5, 0) 7 y 6 5 4 3 2 1

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

ii x = −2

b y = (x − 13)2

c The graph has been vertically reflected across the x-axis and horizontally translated 3 units to the right.

7 y 6 5 4 3 2 1

ii x = 5

738

x 1 2 3

iii (−1, 0)

−2 −1−1 −2 −3 −4 −5 −6 −7

iii (−6, 0) 2

iii

Reasoning x 1

iii (−2, 0)

Mathspace Victorian Curriculum – Year 9 mathspace.co

15 A quadratic of the form y = (x − h)2 always yields non-negative y-values because squaring any real number, whether positive, negative or zero, results in a positive value of zero. 16 The maximum y-value is 5. This is always the case because, for any value of h, (x − h)2 will always be zero or positive since it’s a square number.

17 Sample Answer:

For a given value of x to get the same y-value as y = x2 for the function y = (x − 3)2 we need to add 3 to the x-value. For example: For the quadratic equation y = x2 when x = 2, y = 4, to get a y-value of 4 for the quadratic y = (x − 3)2 we need x − 3 to be 2. So we add 3 in order to get the point (5, 4).

72 64 56 48 40 32 24 16 8

y (in dollars)

x (month) 1 2 3 4 5 6 7 8 9 10 11 12

Problem-solving

5.06 Turning point form

18 a There are two possible horizontal positions: 7 metres to the right of the shooter (positive direction) and 1 metre to the left of the shooter (negative direction). b The target should be placed at a height of 484 metres above the ground if the marksman wants to hit a target 25 metres away.

Understanding 1 a (−3, 4)

b (4, −2)

c (3, 3)

d (−3, 2)

2 a Right

b Down

c Left

d Up

3 D 4 a i A

ii C

iii B

b i B

ii A

iii C

19 a When t = 0, s = 4 km/hr. b When s = 9, t = −1, 5. We take positive values only so t = 5 hours. c s = 0 when t = 2 hours

Fluency 5 C 6 A

d Fastest speed occurs at the end of the journey at 6 hours s = 16 km/hr

7 C

16 s(km/hr)

8 a (x − 2)2 − 4

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

b (x − 5)2 + 1

d (x + 6)2 + 3

c (x + 4) − 1 9 a y = (x − 3)2 − 5

b y = −(x − 3)2 − 2

d y = (x + 2)2 + 1

c y = −(x − 5) − 1 2

f y = 2(x + 1)2 − 5

e y = −2(x + 6) + 3

t(hours) 1

10 a i (1, 5)

ii (0, 4)

iii x = 1

iv Concave down

20 a The maximum revenue occurs at the end of the range, so at month 12.

6 5

b R = (12 − 4) = 64, which is $64 000

4 3

c Sales reach 6000, set R = 6. So, 6 = (m − 4) . Solve for m we get two values as: ± + 4 = m the ﬁrst time the sales reach 6000 is in month 1.55, which we would say is month 2. d At month 4, the revenue is 0.

2 1 −2 −1 −1

b i (−2, 6) iii x = −2

x 1

ii (0, 2) iv Concave down

Answers mathspace.co

739

f i

iii x = −6

iv Concave down 8 y 6 4 2 x

3 2 1 −5

−4

−3

−2

c i (4, 2)

−9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18

−1

ii (0, 18)

iii x = 4 v

−1

iv Concave up

16 14

11 a i (3, 7)

ii (0, 25)

iii x = 3

iv Concave up

y 30

8 6

4 2

1 2 3 4 5 6 7 8 9

d i (−8, −4) iii x = −8 v

ii (0, 60)

iv Concave up

60 y 55 50 45 40 35 30 25 20 15 10 5 x −16 −14 −12 −10 −8 −6 −4 −2−5

e i (−0.5, −0.6)

b i (−1, 4)

5 4 3 2 1 −4 −3 −2 −1 −1

x 1

x 2

−2 −3 −4

Mathspace Victorian Curriculum – Year 9 mathspace.co

ii (0, −3)

iii x = 1

−1

y 6

c i (1, −1)

−2

ii (0, 1) iv Concave down

iv Concave up

1 2 3 4 5 6 7 8 9

iii x = −1

ii (0, −0.35)

iii x = −0.5

740

−2 −1

iv Concave down 4 y 3 2 1

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

x 1

d i (−4, 8) v

c The graph has been vertically enlarged by a factor of 2, horizontally translated 3 units to the right, and vertically translated 7 units up.

ii (0, 16)

iii x = −4

iv Concave up 18 16 14 12 10 8 6 4 2

e The graph has been vertically enlarged by a factor of 5, horizontally translated 0.3 units to the right, and vertically translated 0.8 units up.

x 1 2 3 4

−8−7−6−5−4−3−2 −1 −2

e i (0.2, −0.7)

d The graph has been vertically reﬂected across the x-axis (it’s now concave down), vertically enlarged (made more narrow) by a factor of 4, horizontally translated 1 unit to the left, and vertically translated 6 units up.

f T he graph has been vertically reﬂected across the x-axis (it’s now concave down), vertically compressed (made wider) by a

ii (0, −0.5)

iii x = 0.2

iv Concave up

factor of , horizontally translated 7 units to

y 0.8

the left, and vertically translated 3 units up.

0.6

13 The time it takes for the splash to reach its highest point is 2 s.

0.4 0.2

−0.8 −0.6 −0.4 −0.2 −0.2

0.2 0.4 0.6 0.8

−0.4

iii x = −4 v

The initial height of the platform is 2 m above the water.

−0.6

14 The company needs to sell 5 units to maximise its proﬁt.

−0.8

The maximum proﬁt is $300. The initial cost incurred by the company is $225.

f i

The maximum height of the splash is 10 m.

iv Concave down y 8 6 4 2 x

−9 −8 −7 −6 −5 −4 −3 −2 −1−2 −4 −6 −8 −10 −12 −14 −16 −18

Reasoning 15 An infinite amount of quadratic equations share a common turning point because there are an infinite amount of values for the leading coefficient, a. 16 Any situation that has three vertices and three points described to fit the provided diagram. For example, these could be waves on an oscilloscope. • Wave 1 starts at (−2, 0), has a maximum at (1, 9) and finishes at (4, 0).

12 a The graph has been horizontally translated 4 units to the left and vertically translated 3 units down.

• Wave 2 starts at (4, 0), has a minimum at (8, −8) and ﬁnishes at (12, 0).

b The graph has been vertically reﬂected across the x-axis (it’s now concave down), horizontally translated 2 units to the right, and vertically translated 5 units up.

17 a y = 2(x − 3)2 − 5 and y = 3(x − 2)2 − 5

• Wave 2 starts at (12, 0), has a maximum at (14, 2) and ﬁnishes at (16, 0). b Similarities: both equations have a minimum value of −5.

Answers mathspace.co

741

Differences: y = 2(x − 3)2 − 5 has its turning point at (3, −5), whereas y = 3(x − 2)2 − 5 has its turning point at (2, −5). The vertical stretch of y = 2(x − 3)2 − 5 has a scale factor of 2, and for y = 3(x − 2)2 − 5 the scale factor is 3.

6 a iv: y = − (x + 1) (x + 3)

18 The y-axis scale goes up by 0.5, not 1. So although the turning point is two grid boxes down, it is actually only 1 unit down. Also, from the turning point, moving one unit to the right we go down 1 grid box, which is only 0.5.

7 a i (1, 0) and (4, 0)

The correct equation is y = −0.5 (x − 4)2 − 1.

b iii: y = 2 (x − 1) (x − 3) c ii: d i: y = −4 (x + 1) (x − 3) ii (0, 4)

iii y = (x − 1) (x − 4) b i (−2, 0) and (3, 0)

ii (0, 6)

iii y = − (x + 2) (x − 3) c i (−2, 0) and (0, 0)

ii (0, 0)

iii y = x (x + 2) d i (2, 0) and (3, 0)

Problem-solving

ii (0, 12)

iii y = 2 (x − 2) (x − 3) e i (−5, 0) and (2, 0)

ii (0, 2)

iii 20

f i (−2, 0) and (2, 0)

ii (0, 8)

iii y = −2 (x − 2) (x + 2)

21 a y = −0.074(x − 8)2 + 4 b At 11 m from Karima, the ball will be at a height of 3.33 m, so Riley will not block it. c At 13 m from Karima, the ball will be at a height of 2.15 m, so will land in the goal.

8 a i (0, −8)

ii (2, 0) and (−4, 0)

iii x = −1

iv (−1, −9)

22 The width x of the garden that maximises the area is 3 m

2 −8 −6 −4 −2 −2

The maximum area is 36 m .

x 2

−4

The length of the garden is 12 m.

−6 −8

5.07 Factorised form

−10

Understanding 1 a Factorised form c Not factorised form 2 a (−1, 9)

b Not factorised form d Factorised form

iii

b (0, 8)

c (−4, 0) and (2, 0)

d x = −1

e Down

f Negative

3 a True

b False

b i (0, −12)

c False

ii (−3, 0) and (4, 0) iv y

−3 −2 −1 −2

−6 −8

4 a x = 1 and x = 4

b x = 3 and x = −7

−10

c x = 0 and x = 4

d x = 1 and x = −5

−12

e x = 5 and x = −6

f x = −3

742

−4

d True

Fluency

5 a y = 24

x 1

b y=0

Mathspace Victorian Curriculum – Year 9 mathspace.co

c i (0, −4) iii

ii (−1, 0) and (2, 0) iv

y 2

−3 −2

2 1

−1

−8

−3

−10

−4

−12

d i (0, 0)

ii (0, 0) and (2, 0)

iii x = 1

ii (1, 0) and (3, 0) iv (2, 1)

h i (0, −3) iii x = 2

iv (1, −4)

1 1

−3

−4

e i (0, −8)

ii (2, 0) and (4, 0)

i i (0, −10) v

x 3

−3 −4

−2 −1 −1 −2

−5 −6

f i (0, 1)

ii (−1, 0) and (1, 0)

iii x = 0

iv (0, 1) 4

x 1 2 3 4 5 6 7 8

j i (0, 4)

1 −4 −3 −2 −1 −1

ii (−4, 0) and (1, 0) iv

x 1

−2

−3

−4

iii x = 0

iii

g i (0, −12)

iv (3, 8) 8 7 6 5 4 3 2 1

−2

ii (1, 0) and (5, 0)

iii x = 3

iv (3, 1)

−2

−3

iii x = 3

−4 −3 −2 −1 −1

−2

−1 −1

−4 −3 −2 −1 −1

−6

−2

−4

−4 −3 −2 −1 −1

x 1

−5 −4 −3 −2 −1

ii (−2, 0) and (2, 0) iv (0, −12)

−1

x 1

9 y = 3 (x + 4) (x + 7)

Answers mathspace.co

743

Reasoning 10 1. Identify the coordinates of the x-intercepts of the function. Use the fact that the x-intercepts occur when y = 0. Based on the graph, the x-intercepts are (0, 0) and (5, 0). 2. Substitute the x-intercepts for x1 and x2 on the equation y = a (x − x1) (x − x2). Since the x-values of the x-intercepts are 0 and 5, we know that the factorised form will be: y = a (x − 0) (x − 5) 3. Use any other point on the graph to substitute for x and y and solve for a. To ﬁnd a: y = a (x − 0) (x − 5) Factorised form −4 = a (1 − 0) (1 − 5) Substitute (1, −4) −4 = a (1) (−4)

Evaluate the subtraction

−4 = −4a Evaluate the multiplication a = 1 Divide both sides by −4 4. The quadratic equation in factorised form: y = x (x − 5) 11 1. Since the quadratic equation is in factorised form, then the x-intercepts are (6, 0) and (−2, 0). 2. Find the y value of the y-intercept by substituting x = 0 into the equation. This means we have y = 12. 3. Determine the equation of the axis of symmetry by calculating the halfway point between the two x-intercepts. This means that the axis of symmetry is x = 2. 4. Find the coordinates of the turning point by substituting x = 2 into the equation to ﬁnd the x-coordinate of the turning point. This means that the coordinates of the turning point are (2, 16).

18 16 14 12 10 8 6 4 2 −3 −2 −1

This means they also have the same axis of symmetry at x = −0.5. However, they will have a different turning point as when you substitute x = −0.5 into the two equations you get different y-values. 13 The error is in Step 3. When identifying the x-values of the x-intercepts from the equation we must change the signs. The correct x-values of the x-intercepts are −2, 3. 14 For y = (x − 2) (2 − x), take −1 out of the second bracket. The equation becomes y = − (x − 2) (x − 2). We can see that the x-intercept for both brackets is (2, 0). Therefore, there is only one x-intercept. 15 • We can easily identify the x-intercepts or solutions to the equation. • When the equation is fully factorised (no common factors in the brackets), we can easily identity the concavity of the parabola. • Although not immediately obvious, it is simple to ﬁnd the y-intercept; we just need to substitute x = 0 and multiply the remaining factors. • We can also calculate the axis of symmetry easily becasue it is halfway between the x-intercepts. Problem-solving 16 17 We can factorise out a common factor of −2 from the ﬁrst bracket to get y = −2 (x − 1) (x + 3), and then graph.

x 1 2 3 4 5 6 7

12 They both have a positive scale factor, so are concave up and have a minimum.

744

For y = 4 (x − 2) (x + 3), the x-values of the x-intercepts are 2, −3 and the same for y = (x − 2) (x + 3). The ﬁrst equation has a scale factor of 4 but this does not affect the x-intercepts.

Mathspace Victorian Curriculum – Year 9 mathspace.co

9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1

x 1 2 3 4 5

18 a

21 Since the turning point lies on the axis of symmetry, we know that 12 is the average of 10 and x1. So then, we must have x1 = 14. If we substitute the coordinates of the turning point into the function, we get:

y 4 (0, 4) 3 2

20 = k (12 − 14) (12 − 10)

If we solve this equation we get k = −5.

(−1, 0)

(2, 0)

−1

5.08 Standard form

b Since y represents Burnell’s height above the water, he will enter the water when y = 0.

Understanding

In other words, the x-intercepts of the model represent where Burnell might enter the water. This will be at either (−1, 0) or (2, 0).

1 a y-intercept

b Narrower

c Maximum

d a and b

We also know that x represents Burnell’s horizontal distance from the springboard towards the pool. Since Burnell is jumping into the pool, the x-value when he enters the water must be positive. This means that Burnell will enter the water at the coordinates (2, 0), which is 2 metres horizontally from the springboard. The other x-intercept, (−1, 0) is a non-viable solution because it has a negative x-value. 19 a (50, 110)

2 3 a

−3 −2 −1

−5

16 27 40

Fluency 4 a i

−3 −2 −1

−3 −2

b i

metres (h) 7 6 5 4 3 2 1

−2 −3

18 16 14 12 10 8 6 4 2 −1

−2

−3

−2

−1

−20 −13

−8

−5

−4

−5

−8

Answers mathspace.co

745

y −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18 −20

b (1, 6)

b C

b 20 a The horizontal intercepts represent the times when the eraser is 0 m above Wilson’s hand. This will be exactly when Wilson tosses the eraser and when it returns to his hand. Since Wilson tosses the eraser at t = 0 and it takes 2 seconds for it to return to his hand, the horizontal intercepts are (0, 0) and (2, 0). Since one of the horizontal intercepts is at the origin, we have the vertical intercept is also (0, 0).

seconds (t) 1

d y = −6x (x − 2)

c i

−3

−2

−1

−30 −14

−4

−2 −10 −24

−3 −6 −9 −12 −15 −18 −21 −24 −27 −30

y 80 70 50 40 30 20 10 −12−10 −8 −6 −4 −2

−2

−1

−3

−13 −15

−9

c i (0, −8) iv

8 10

−10

−12

d i (0, 9)

−15

iv 5 a iv: y = x2 + 6x + 4

b i: y = x2 + 6x − 4

c iii: y = x2 + 4x + 6

d ii: y = −x2 + 6x + 4

6 a i Minimum

ii (0, 8)

iii x = 2

iv (2, 4) iv (−2, −5)

c i Minimium

ii (0, 1)

iii

e i (0, −64) iv

ii (0, 46)

iii x = 6

iv (6, 10) ii x = 1

iii (1, 4)

10 −8 −6 −4 −2 −10 −20 −30

−50 −60

−70

6 4 2

x 2

Mathspace Victorian Curriculum – Year 9 mathspace.co

ii x = 0

−40

iii (−0.5, 8.5)

−4 −3 −2 −1

d i Minimum

ii x = −0.5 18 16 14 12 10 8 6 4 2

ii (0, −9)

b i Maximum iii x = −2

746

−8

−10

−6 −4 −2 −2

−4

−5

x 2

−6

7 a i (0, 5)

−6 −4 −2 −2

iii (3, 1)

−1

x 2 4

ii x = 3 4

−3 −2

iii (−6, 3)

−3

ii x = −6

b i (0, 75)

y −3 −2 −1

d i

iii (0, −64)

y x 2

f i (0, 0)

ii x = 3

iii (3, 2.25)

2 −6 −4 −2

ii (1.5, −12.25) y x 2

−2

1 −4 −3 −2 −1

−8 −10 −12 −14

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

x 1 2 3 4 5

−1 −2 −3 −4 −5 −6 −7 −8 −9

iii

y 8 6 4 2 x

c i (−4, 0) and (1, 0)

−5 −4 −3 −2 −1 −1 −2 −3

x 1

and (2, 0) ii

f i

−2

7 6 5 4 3 2 1

ii (0, −9)

b i (−3, 0) and (3, 0)

iii

1 2 3

e i (−2.5, 0) and (0.5, 0) ii (−1, −9) iii

−6

−7 −6 −5 −4 −3 −2 −1 −2

−4

iii

18 16 14 12 10 8 6 4 2

x 2

8 a i (−2, 0) and (5, 0) iii

ii (−3, 0)

iii

4 y 3 2 1 −8 −6 −4 −2−1 −2 −3 −4 −5 −6 −7 −8 −9

d i (−3, 0)

ii (−1.5, 6.25)

−1

−2

Reasoning 9 a i T he parabolas both cross the x-axis at the same two points. x 1 2 3 4 5

ii One parabola is concave down and has a maximum value and the other parabola is concave up and has a minimum value. iii min = −6.25, max = 6.25 iv Both parabolas have the same axis of symmetry. b i T he parabolas both cross the x-axis at the same two points.

Answers mathspace.co

747

ii Both parabolas are concave up and have a minimum value. iii y = x2 + x − 2, min = −2.25 iv Both parabolas have the same axis of symmetry. 10 a i Minimum

axis of symmetry, as the equation is

iv The graphs of y = 2x2 and y = x2 have the same minimum value and axis of symmetry, but the graph of y = 2x2 is narrower. b i Minimum ii 0 iii The parabola has an axis of symmetry along the y-axis of the graph.

Derek forgot the 2 in the denominator. This error creates an incorrect axis of symmetry and y-value for the turning point. The turning point is (1, 8).

13 a y = −x2 + 8x − 7 b y = − (x − 1) (x − 7) c y = − (x − 1) (x − 7) = − (x2 − 7x − x + 7) = − (x2 − 8x + 7) = −x2 + 8x − 7  14 a Jackie could make $525 because the maximum proﬁt is $530 if each car wash costs $25. b

and y = x2 have the

c i Minimum

400 200

ii −7

iv The graph of y = x2 − 7 has the same axis of symmetry as y = x2, but a different minimum value. 11 a y = ax2 + (−2ah) x + (ah2 + k) b Turning point form and standard form are equivalent quadratic equations in different forms. The equations highlight different key features. We can find the turning point, (h, k), efficiently using turning point form. We can find the y-intercept, c, efficiently by identifying the constant using standard form. We can also find the axis of symmetry, , in standard form and use it to find the y-value of the turning point. In both forms, the leading coefficient, a, is the vertical enlargement or compression factor and tells us if the parabola will be concave up or concave down. 12 Johnny incorrectly used 0.5, when the a value of the quadratic is −0.5. This error creates an incorrect axis of symmetry and y-value for the turning point.

748

300

is wider.

iii The parabola has an axis of symmetry along the y-axis of the graph.

Mathspace Victorian Curriculum – Year 9 mathspace.co

P 500

same minimum value and axis of symmetry, but the graph of

Problem-solving

ii 0

iii The parabola has an axis of symmetry along the y-axis of the graph.

iv The graphs of

Derek used an incorrect equation to ﬁnd the

100

x 5

15 a

10 9 8 7 6 5 4 3 2 1

20 25 30 35 40 45 50

y (m)

x (min) 6

b Because the y-axis represents height in metres, we can look at when our graph has a y-value of 3. This occurs twice, when x is around 0.7 and 4.3. Because x is in terms of minutes, we can say that the ball will be 3 m above the ground around 0.7 and 4.3 minutes. 16 a The y-intercept is the height that the frisbee was thrown from. The frisbee was thrown from 240 m. b The x-intercept is the time it takes for the frisbee to reach the ground. The frisbee reaches the ground after 6 seconds. c 320 m

b i

17 a y = −5x2 + 10x + 3

b 2 seconds

c After 3 seconds. Because Bella starts at 3 m above the pool surface, we know that the pool surface is at a height of 0 metres. Therefore, 12 metres deep would be the same thing as y = −12. We can plug this into our equation and see that x = 3, meaning that this will happen 3 seconds after jumping. 18 a y = −16x2 + 32x + 3.25 b

20 18 16 14 12 10 8 6 4 2

2 1 −2

−1

x 1

−1

9 8 7 6 5 4 3 2 1 −3 −2

c 3.25 m

−1

ii x = 3, x = −3

19 y = x2 + 5x + 2

d i 8

5.09 Solutions to quadratics

Understanding

2 a Two solutions

b Two solutions

c No solution

d One solution

3 a Sometimes true

b Never true

c Sometimes true

d Always true

4 A and D Fluency 5 a i

9 8 7 6 5 4 3 2 1 −1

−3 −2 −1

1 They are all terms for the same feature of the graph.

−3 −2

−3 −4

sec 3

−2

c i

ii x = 0

metres

−2 −4 −6 −8

ii x = 2, x = −2 e i

30 27 24 21 18 15 12 9 6 3

−4 −3 −2 −1

x 1

ii No real solutions x 1

ii x = 0

Answers mathspace.co

749

f i

24 21 18 15 12 9 6 3

d i

6 5 4 3 2 1

−2 −1 −1 −2 −3 −4 −5 −6

−4 −3 −2 −1 −3 −6

2 3 4 5 6 7

ii x = 0, x = 4 ii x = 1, x = −1

e i

6 a i

9 8 7 6 5 4 3 2 1 −4 −3 −2

−1

ii x = 2 36 32 28 24 20 16 12 8 4

x 1

−2 −3

x 1

f i

−2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

2 3 4 5 6 7

−9−8−7−6−5−4−3−2 −1 −4

x 1 2 3

ii x = 0, x = −8 7 a

ii x = 3 c i

2 1 −5 −4 −3 −2 −1−1

y 2

y x 1 2 3 4 5

−2 −3 −4 −5 −6 −7 −8 −9 −10

ii x = −4, x = 2

−6 −5 −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18

Mathspace Victorian Curriculum – Year 9 mathspace.co

x 1 2 3 4

b x = −5, x = 3 8 a x = 2.00

b x ≈ 0.70, x ≈ −4.30

c No real solutions

d x ≈ −0.09, x ≈ −2.91

e No real solutions

f x ≈ 0.43, x ≈ 1.18

9 a B

750

−1 −1

ii x = −1 b i

9 8 7 6 5 4 3 2 1

b D

Reasoning 2

10 a The equation 2x = 5 can be rearranged to 0 = −2x2 + 5. In the table of values for y = −2x2 + 5 we can see that the values of y are both positive and negative, this means that there are two x-intercepts, and therefore two real solutions to the equation. b 6

5 4 3

14 a 7 m

b 5m

1 −4 −3 −2 −1 −1

13 Since the axis of symmetry is x = 3, the point directly opposite to (1, 0) on the other side of the axis of symmetry will also be a solution. The axis of symmetry divides the parabola into two symmetric halves. To find this other solution, you can reflect the point (1, 0) across the axis of symmetry where x = 3. The reflected point will have the same y-coordinate but will have an x-coordinate that is twice the distance from the axis of symmetry as the original point. So the other solution is (5, 0).

x 1

−2

c x ≈ 1.5, x ≈ −1.5

c Measurement cannot be negative so we can only consider values for x that are greater than 0. Problem-solving

11 a True. The turning point is the only point on a parabola that does not share the y-value with another point, because it is on the axis of symmetry. If there is only one point on the x-axis where y = 0, then it must be the turning point. b False. Doing the same thing to both sides of an equation, does not change the solution. 12 If we consider the graph of the quadratic function y = x2, we can see that it has a turning point at (0, 0), and therefore x2 = 0 has only one solution, x = 0.

15 a x = −4, x = −3 b x = −3, x = 3 16 6 m 17 5 cm 18 a

If k > 0 then the graph of the function y = x2 + k is translated (shifted) up and will not cross the x-axis. Therefore the equation x2 + k = 0 has no real solutions. If k < 0 then the graph is translated down. This means it will have two x-intercepts and therefore the equation will have two real solutions.

y y = x2 + 1 1 y = x2 −1

x 1

y = x2 − 1 −1

100 160 180 160 100

−140

200 180 160 140 120 100 80 60 40 20

x 1

c 5 seconds 19 a We know the distance at which the sprinkler reaches its greatest height, which means we can determine the turning point of the resulting graph. This means that turning point form will be the most eﬃcient to write. b c x = 0, x = 36 d

Answers mathspace.co

751

20 a The purple graph ( y1) is Neptune, the green graph ( y2) is Earth, and, the blue graph ( y3) is Mercury. The greater the value for gravity, the greater the absolute value of the coeﬃcient of t2, and the narrower the graph. Since 11.15 > 9.8 > 3.7, Neptune is the narrowest, then Earth, then Mercury. This makes sense because the greater the force of gravity, the faster an object will fall. b The x-coordinate of the x-intercept is the time in seconds it would take for the object to hit the ground. Neptune: (1.34, 0)

g x = 4, −10

8 a x = 0, −7

c x = 0, −5

d x = 0, −3

e x = 0, −5

g x = 0, 5

h x = 0, −1

9 a x = 5, −7

b x = −1

c x = −2, 5

d x = −4, 1

Mercury: (2.33, 0)

e x = −5, 10

21 For Garden A, the length and width are 7 m and 2 m while for Garden B, the length and width are 8 m and 3 m.

Earth: (1.43, 0)

b x = −3, 1

5.10 Solve algebraically

c d x = 3, 9

Understanding 1 A linear equation can be written in the form ax + b = 0, where a and b are constants. An example of a linear equation is 2x + 3 = 0. A quadratic equation can be written in the form ax2 + bx + c = 0, where a, b, and c are constants. An example of a quadratic equation is x2 − 5x + 6 = 0. 2 a True

10 a x = 1, 5

b False

c True

d True

b 7.071

c 11.180

d 4.583

e x = 2, 6 f x = −8, 12 11 a i (5, −25) iii

ii (0, 0), (10, 0)

x 2

−5 −10 −15

3 x = −12

−20

4 a 4.123

−25

Fluency

5 a c x = ±11 6 a y = ±5

iii

d b

c k = ±2

d x = ±11

e x = ±5

f x = ±8

g x = ±15

h x = ±10

7 a x = 11, −5

752

b i (−4, 16)

b x = ±5

c x = −7, 11

d x = 16, −2

e x = −2, 16

Mathspace Victorian Curriculum – Year 9 mathspace.co

ii (0, 0), (−8, 0) 20 16 12 8 4

−10−8 −6 −4 −2 −4 −8 −12 −16 −20

x 2 4 6 8 10

ii No x-intercepts

c i (−1, 12) iii

40 36 32 28 24 20 16 12 8 4

1 2 3 4 5

d i (6, 27)

ii (3, 0), (9, 0) 30 27 24 21 18 15 12 9 6 3

−3

−4 −3 −2 −1−1 −2 −3 −4

−3

10 y 9 8 7 6 5 4 3 2 1

−5 −4 −3 −2 −1−4 −8

iii

15 The equation can be rewritten as x2 = −4. Since the square of a real number is always nonnegative, there is no real number x that satisﬁes the equation. Therefore, the equation has no real solutions.

The graph of the equation is a parabola that opens upwards. It does not intersect the x-axis at any point, which means there are no real solutions for the equation x2 + 4 = 0. x 3

12 Between 3 seconds and 9 seconds 13 t = 14, t = 10 Reasoning

16 Ted made an error in Step 3. He only took the positive square root of both sides. To correct the error, he should have taken both the positive and negative square roots of both sides. 17 Samantha made an error in Step 3. She interchanged the numerator and the denominator on the right hand side when dividing both sides by 36.

14 The equation can be rewritten as x2 = 9. This equation has two solutions because there are two values of x which, when squared, make 9. One is the positive square root of 9 (x = 3) and the other is the negative square root of 9 (x = −3).

Problem-solving

19 a x(x + 4) = 0

4 y 3 2 1 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10

x 1

18 b 0 or −4

x 1

The graph of the equation is a parabola that opens upwards. It intersects the x-axis at two points: x = 3 and x = −3, which correspond to the two solutions of the equation x2 − 9 = 0.

20 a A = (x + 3)

b x = 2 or x = −8 c 5 cm 21 a h = 0.99 metre b The stone will be at the same height above the water as it was when it was thrown at t = 0 (when it was thrown) and t ≈ 0.88 seconds (when it starts falling back down). c 1.916 metres

Answers mathspace.co

753

5.11 Solve by factorisation Understanding 1 Factorisation cannot be used to solve all quadratic equations, as not all quadratics are factorable. Alternative methods like the quadratic formula or completing the square can be used for non-factorable equations. b The factorised form of a quadratic simpliﬁes solving it as the Null Factor Law means we solve for one or the other of the two possible factors to be equal to zero. Fluency 3 a x = 0, 1

b x = −3, 0

c x = −1, 1

d x = −1, 2

e x = −4, −2

g p = 0, −8

c m = 0, −5

b d m = 0, −5

e b = 0, 9 5 a x = 5, −7 c

f y = 4, 6

g m = 2, −9

h n = −12, 7

10 (−3, 0) and (2, 0) 11 a x = 0, −3

b x = 0, −6

c x = 0, −3

d x = 0, 18

e x = 0, −4

f x = −15, 0

Reasoning

2 a B, D

4 a x = 0, −7

e y = −2, −4

12 The equation (x − 2) (x − 2) = 0 has only one solution because both factors are the same. Using the null factor law, we set (x − 2) equal to zero and ﬁnd that x = 2 is the only solution. On the graph it will touch the point when x = 2 and y = 0, which is also the turning point of the parabola. 13 The quadratic equation cannot be factorised using integers because there are no two integers that multiply to 5 and add to −4. 14 The error is in Step 2. The correct factoring should be (x + 2) (x + 3). The numbers in the factors must add to 5 and multiply to 6, not the other way around. 15 The quadratic equation has the common factor of x, so x = 0. The linear factor will be ax + b. Solving ax + b = 0 gives the second solution

6 a m = 0, 14

Problem-solving 16 a

e m = 0, −2

1 7 f n = −4, −2

7 a m = −2, −6

b m = −3, −8

c y = −3, −2

d y = −2, −7

e x = −3, −5

f x = −3, −4

g x = −2, −1

h x = −6, −3

−10

8 a m = 5, −4

b m = 6, 1

−12

c m = 6, 5

d m = 14, 13

e x = −7, −6

f x = −7, 3

g x = 10, −5

h x = 11

i x = 2, −10

j x = 19, −6

k x = −8, −10

l x = 6, −8

17 a a = −10

b a=1

9 a m = 5, −2

b x = −4, 1

18 a x = 4, −7

b x = 8, −3

c x = 2, 10

d m = −7, 2

c x = 7, −4

d x = −9, 1

d y = 0,

h 2 −2

t 1

−4 −6 −8

b The kingﬁsher enters the lake at 1 second and exits the lake at 8 seconds c The kingﬁsher was underwater for 7 seconds

19 a t = 3, 9

754

Mathspace Victorian Curriculum – Year 9 mathspace.co

2 3 4 5 6 7 8 9

b 3 hours

c 6 hours

Chapter 5 review 1 a Yes

iii

b No

c No

y 2

d No

2 a Quadratic

b Neither

3 a i Concave down

ii (0, 15)

x −6 −5 −4 −3 −2

−1 −2 −4

iii x = 0

−6

iv x-intercepts: (−5, 0) and (5, 0) y-intercept: (0, 15)

−8

ii (4, −25)

b i Concave up iii x = 4

iv x = −5, −1

iv x-intercepts: (−6, 0) and (14, 0) y-intercept: (0, −21)

c i

4 a −45 °C

iii

ii (0, 0) 4

b At d = 4 days and d = 16 days

c Approximately 25 °C

d After 10 days

5 a y = 3x2

b y = 0.5x2

c y = −0.5x2

−4 −3 −2 −1 −1

d y = −x2

−2 −3

−4

7 a 5 units upwards

b 1 unit left

c 4 units right 8 a y = (x + 2)2 − 5

d 2 units downwards

iv x = 0, 1

b y = (x − 3)2 + 4

d i

ii (0, 6)

iii

9 a i (−3, 0) and (3, 0)

9 8 7 6 5 4 3 2 1

ii (0, 13.5) iii y = −1.5 (x − 3) (x + 3) b i (−1, 0) and (5, 0) ii (0, −1.67) iii 10 a i (1, −4)

−4 −3 −2 −1−1 −2 −3 −4

ii (0, −3)

iii

x 1

y 2

iv x = −3, 2

1 −2

−1

x 1

e i (2.5, 12.25) iii

−2 −3 −4

iv x = −1, 3 b i (−3, −8)

ii (0, 10)

ii (0, 6)

13 y 12 11 10 9 8 7 6 5 4 3 2 1 −3 −2 −1−1

x 1 2 3 4 5 6 7

iv x = −1, 6

Answers mathspace.co

755

6.01 Cost, profit, revenue

f i iii

Understanding

y 3

1 B

2 1 −1

−1

x 1

2 a True

b False

c True

d True

3 Find the point of intersection of the lines for revenue and cost, or solve the equation: Revenue = Cost.

−2 −3

Fluency

−4

4 a Proﬁt of $5340

b Proﬁt of $730

c Loss of $500

d Loss of $480

5 a $326.40

11 a y = ±2 b x = ±15 c x = 15, x = 1

6 a R = 26x

b R = 1890x

7 a $2250

b $2610

8 a C = 170x + 220

b C = 520x + 440

9 a $400

b $300

f g m = 0, −6

c Profit of $100

d 10

e $100

12 a b = 0, 22

b t = 9, 5

c y = 6, 4

e x = −4, 2

10 a The cost is $10 000 and revenue is $0 b The cost is $30 000 and revenue is $40 000 c 4 11 a

13 The error is in Step 2. The correct factoring should be (x + 1) (x + 7). The numbers in the factors must add to 6 and multiply to 7, not the other way around. 14 a k = 7

15 a x = 3 m

b 36 m

Dollars 900 800 700 600

c 12 m

16 a The y-intercept is the height that the ball was thrown from. The ball was thrown from 260 m. b The x-intercept is the time it takes for the ball to reach the ground. The ball reaches the ground after approximately 5.4 seconds. c 332.25 m

C = 40x + 380

500

R = 120x

400 300

x 2 b $120

c $380

d $600

12 a i The revenue per item is $150

17 a b x = 0, x = 30

ii The ﬁxed costs are $400 iii For 3 items a loss of $70 iv For 10 items a proﬁt of $700

756

b $108

c $1260

Mathspace Victorian Curriculum – Year 9 mathspace.co

e $860

b i The revenue per item is $25.5

20 a

Dollars

ii The ﬁxed costs are $1250

700

iii For 3 items a loss of $1203.5

600

iv For 10 items a loss of $1095 c i The revenue per item is $3500 ii The ﬁxed costs are $10 000

400

iii For 3 items a loss of $3250

300

iv For 10 items a proﬁt of $12 500

200

13 a $1800

100

b $560

14 a $200

C = 9x + 600

500

c $1240

R = 8x

b $220

c Loss

d Loss of $20

15 a Proﬁt of $500

b Loss of $250

16 a Proﬁt of $400

b Loss of $100

c 4 17 a C = 25x + 200, R = 50x b 8

x 14

b The weekly cost is always greater than the weekly revenue. This is because the revenue per smoothie is less than the cost per smoothie. c She needs to increase the revenue per smoothie, or decrease the cost per smoothie, so that the revenue per smoothie is greater than the cost per smoothie.

c Loss of $100 Problem-solving Reasoning

21 a C = 15x + 8000

18 a From the original break-even point where 10x + 120 = 12x and x = 60, if the cost increases to 11 this means that 11x + 120 = 12x and x = 120. So the number of units sold would have to double just to break-even, from 60 to 120.

22 a $420 23 a

12x = 10.5x + 120

600

x = 100 This means that it’s harder for a business to break-even when the ﬁxed price increases from $120 to $200, as they need to sell 100 items instead of 80 to cover their costs. 19 a i N o, the business will never make a loss because the revenue is always greater than the cost. ii P = 15x b i T he cost will be a horizontal line at C = 500, where C is the y-axis. ii P = 20x − 500

c $700

Dollars 800

x = 80

b $300

900

b Calculating the number of items if the ﬁxed price rises from $120 to $200 or if the cost per item rises from $10 to $10.50,

12x = 10x + 200

b 800 meals

c $25 per meal

C = 50x + 390

700

500

R = 110x

400 300

x 2

b $210

c 19.1%

d P = 60x − 390

6.02 Income Understanding 1 Wage is a rate of money earned per time period, so the total amount earned is based on how much time you end up working. Salary on the

Answers mathspace.co

757

other hand means that you earn a set amount independent of the number of hours worked.

predicted income under Offer A is 10% of $4000 plus $500, which is only $900.

3 C

Offer B is also the safer offer because she will earn more even if she has overestimated the value of sales she will make.

Fluency

24 a 20a + 40b, or 20(a + 2b)

2 a False

b True

c True

b 40a

4 $17 264

c 100b

5 $294

25 Answers vary, but may include:

6 $8373.50

9 a $912

b $1824

c $47 424 d $3952

Job A has an annual value of $34200, so total income is greater per year with Job B. Job A only requires 40 weeks of work, but Job B is working 48 weeks. So maybe Job A is preferable to work less for still a reasonable amount of money.

10 a $6333

b $2923

c $1462

Problem-solving

7 $2696.08 8 a $54 840 b $53 040 c Offer 1

d $38

11 $272

26 a $27.63

b $7800

12 a y = 52x

b 41.5 hours

27 a 22.5 hours

13 a y = 36x

b $64 800

c 7 hours

14 a $28.50

b $741

28 At least 36 hours

15 a $49 400

b $53 300

29 $240

16 a 65 hours

b $24.65

30 2%

c $1725.50

31 a Profit = Line C Cost = Line A Revenue = Line B

17 $5902 18 $59 489.55 19 a y = 0.02x + 1 700

b $6678

c $175 000 20 a y = 0.005x + 800

c $3000

b 15 hours

b 10 units

6.03 Simple interest Understanding

b $1225

c $110 000 21 a $2500

c 18.1%

b 10% d $15 000

Reasoning 22 No, Mbacke is incorrect.

1 a $5400

b $1080

2 a Divide by 12

b Multiply by 12

3 a True

c True

b False

Fluency 4 a $342

b $7942

The conversion 1 month ≈ 4 weeks is not very accurate. To calculate the monthly pay for the manager job, he should multiply by 52 then divide by 12 which gives $2946.67.

5 $10 080

23 Elna should choose Offer B.

8 a 1.5 years

Her predicted income under Offer B is 5% of $4000 plus $1000 which is $1200. Her

9 a 7.56%

b 10.95%

c 7.84%

10 a 1.5%

b 0.5%

c 0.12%

758

Mathspace Victorian Curriculum – Year 9 mathspace.co

6 a $2500

b 5 years

7 1.82 years b 6 quarters

11 a $693.00

b $692.75

c $2164.80

d $657.99

12 a $4996.53

b $10 329.30

c $1470.75

d $8835.44

e $10 509.60

f $4980.36

13 Option 2

29 $2857.14 30 Option 1

6.04 Re-investing with simple interest Understanding

14 Option 1

1 Starting with the opening balance, add the interest and subtract any repayments.

15 $1634

2 a True

16 a 8%

b 9%

b False

c True

d False

e True

c 4%

Fluency

17 $80

18 $78 585 19 Fortnightly Reasoning 20 Yarel has multiplied an annual interest rate by the number of months, so the units of time in his calculation are not the same. He must either convert 22 months to

years or convert the

interest rate of 8% p.a. to

per month.

4 a

Month

Opening Balance

Interest

Closing Balance 73 073

73 000

73 073

73 146

73 219

73 292

73 365

73 438

Quarter

Opening Balance

Interest

Closing Balance

44 000

880

44 880

880

45 760

880

46 640

880

47 520

21 The interest from Option 1 will be 0.5 × 2 × 12 = 12% of the principal, while the interest from Option 2 will be of the principal, so Option 1 will earn Ariela the most interest. 22 When calculating interest we multiply by the loan term, so a shorter loan term will result in less interest paid. When we calculate repayments, we divide by the number of repayments. If making monthly repayments, for example, the number of repayments is 12 × T. A smaller T value means a smaller number of monthly repayments, which means that the repayment amount may be greater.

b 3 quarters or 9 months c $3520 5 a

Month

Opening Closing Interest Repayment Balance Balance

2500

700

1875

700

1250

700

625

700

b 4 months

Problem-solving

c $300

23 a $960

b $290

24 a $2400

b 6%

25 a $2784

b $5490

27 800

417

3200

25 017

d $2490

25 017

417

3200

22 234

417

3200

19 451

417

3200

16 668

c $10 764 26 a $1000 27 5 years

b $200

c 2%

6 a

Quarter

Opening Closing Interest Repayment Balance Balance

b $12 800

28 25.36% p.a.

Answers mathspace.co

759

7 a 15 years

b 0.5% per month

c $598 500

made their normal repayment, they would be paying back more than they owe.

d $283 500

8 a $1600

b 10 weeks

c 1% per fortnight

d $8400

e $400 9 a

Month

Opening Balance

Interest

Closing Balance

9400

9447

47.24

9494.24

47.47

9541.71

47.71

9589.42

b $189.42 c Under simple interest Page would have earned $188 interest, so she has earned $1.42 more than with simple interest. 10 a $8754.88

Month

15 When being charged simple interest, making higher repayments can decrease the number of months taken to pay off the loan, which decreases the total interest paid. When interest is calculated as a percentage of the total amount, making a higher monthly repayment also decreases the interest charged on the next month, so the effect is even greater. Problem-solving

b 0.4%

16 a Year 1: $10

11 a 0.5% b

14 The equivalent monthly rate for Option A is 0.75% simple interest. Since they have the same rate, Option B would earn Ibryam the most interest because the interest is calculated as a percentage of the total amount (principal plus interest previously earned) instead of just the principal.

Opening Closing Interest Repayment Balance Balance

868.48

70 71 72

Year 2: $10 Year 3: $10 Year 4: $10

4.34

250

622.82

3.11

250

375.93

b Year 1: $10

375.93

1.88

250

127.81

Year 2: $11

127.81

0.64

128.45

Year 3: $12.10

c 6 years

Year 4: $13.31

d $17 878.45

e $2878.45

Interest ($) Interest on total amount

12 a 0.75% b

Month

Opening Closing Interest Repayment Balance Balance

146 26 452.84 198.40

5000

21 651.24

147

21 651.24 162.38

5000

16 813.62

148

16 813.62 126.10

5000

11 939.72

149

11 939.72

89.55

5000

7029.27

150

7029.27

52.72

5000

2081.99

151

2081.99

15.61

2097.6

c 12 years and 7 months

12 10 Simple interest 8 6 4

d $752 097.60 e $450 000

2 Year

Reasoning 13 If the opening balance plus interest is less than the normal repayment amount, the borrower only needs to pay back that amount. If they

760

Mathspace Victorian Curriculum – Year 9 mathspace.co

d Simple interest points lie in a horizontal line because the amount of simple interest stays the same each year. The interest calculated

as a percentage of the total amount increases by a greater amount each year, so the points curve upward, rather than lie in a straight line.

Opening Balance

Month

Closing Balance

Interest

85 000

127.5

85 127.5

17 a i 6 months to repay the loan and total interest is $45.84

85 127.5

127.69

85 255.19

127.88

85 383.07

ii 4 months to repay the loan and total interest is $36.38

85 383.07

128.07

85 511.14

128.27

85 639.41

128.46

85 767.87

Month

Opening Balance

Interest

Closing Balance 30 600

b $9.46 18 a i $112 b

38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2

ii $97.59

8 a

Amount owing on loan ($) Option B

Option A

Time (Months) 2

1 a Proﬁt of $6 090

b Proﬁt of $780

c Loss of $500

d Loss of $480 b $270 d Loss of $70 b $5 200

c 13.51%

4 a The weekly retainer is $600 and the commission is 5% b $57 200 c $20 000 5 a $476.92 c $4 482.00 6 a $850

612

31 212

624.24

31 836.24

636.72

32 472.96

c Simple interest rate of 2.06% per month

Month

Opening Repay- Closing Interest Balance ment Balance

945.60

4.73

280

670.33

3.35

280

393.68

1.97

280

115.65

0.58

116.23

c 5 years 10 a i C

d $16 636.23

e $2 636.23

ii A

iii B

b 80 units 11 a C = 5x + 10 000

b 1 000 cups

c $15 per cup

Chapter 6 review

3 a $29.62

600

30 600

b $2 472.96

c Ashly should choose Option B since the amount will decrease over time.

c Loss

30 000

9 a 0.5%

2 a $200

1 2

b $787.50 d $795.43 b $9 350

12 No, Aria is not correct. The conversion 1 month ≈ 4 weeks is not very accurate. To calculate the monthly pay for the tutoring job, they should multiply by 52 then divide by 12 which gives $3 250. 13 Leo should choose the Executive package. His predicted income under the Executive package is 8% of $5 000 plus $1 800 which is $2 200. His predicted income under the Strategist package is 15% of $5 000 plus $1 200, which is $1 950. The Executive package is also the better offer because he will earn more even if he has overestimated the value of the projects he will complete.

Answers mathspace.co

761

14 J aden has multiplied an annual interest rate by the number of months, so the units of time in his calculation are not the same. He must either convert 18 months to

years, or convert the

interest rate of 6% p.a. to

per month.

7.01 Similar right-angled triangles Understanding 1 a i ∠Q

ii ∠R

iii ∠P

b QP c RP

15 a Year 1: $12 Year 2: $12

2 a Each pair of corresponding angles must be equal.

Year 3: $12

b Each pair of corresponding sides must be proportional.

Year 4: $12 b Year 1: $12 Year 2: $12.96

3 a False

b True

Year 3: $14.00

e False

f True

c False

d False

Year 4: $15.12 c 18

Fluency

Interest ($) Interest on total amount

16 14

Simple interest

10 8 6 4 2

Year 1

16 a i $162.50

ii $132.02

Amount owing on loan ($) 45

Option 1

35 30 25 20

b y = 5 mm

c y=6m

d y = 9 cm

e y = 7.6 m

f y = 12.025 m

7 a r = 55, s = 35

b m = 3.4, n = 15

c x = 10, y = 45

d y = 10, x = 6

,y=6

8 a

Reasoning 9 a Yes. Since a pair of corresponding sides on the right triangles is in the same ratio as the two hypotenuses, the remaining sides will also be proportional by Pythagoras’ theorem.

10 All pairs of angles in the two triangles are equal. It is possible to ﬁnd the missing angle since the sum of the angles in a triangle are equal to 180°. The side lengths given of both triangles are in the same proportion. 11 All the triangles

5 3

c Option 2

Mathspace Victorian Curriculum – Year 9 mathspace.co

H x°

J x°D

Time (Months) 2

d Yes

6 a y = 17.5 cm

c Yes

b Yes. Since a pair of corresponding angles in the right triangles has the same measure, all pairs of corresponding angles are equal.

Option 2

762

b Yes

d The scenario from part (b) returned the larger investment by approximately $6.08

b A and B

c A and B 5 a No

4 a A and D

x° CA

x° EA

16 cm 4 cm 12 cm

closet

Since each triangle has a right angle and shares angle x, the remaining angle in all the triangles will be equal to (90 − x) °. This shows that all the angles are equal, so all the triangles are similar.

2.4 cm

Problem-solving

18.4 cm

12 564 cm

Reasoning

13 14.7 m 14 43.75 m 15 The height of the roof is about 4.33 m.

7.02 Scale factors Understanding 1 a NL

b NM

2 a RS

b ∠RUT

3 a True

b False

c False

e True

f True

g True

d True

12 She is incorrect. For a quadrilateral to be similar, the AAA similarity test does not hold. We also need to show that all corresponding sides are in the same ratio. But some sides have a scale factor of 6 while others have a scale factor of 2, so these are not similar shapes. 13 Dorothy is correct. If the picture wasn’t distorted, it would be in the ratio of 1 : 1, not 1 : 3. This makes it look taller and narrower that in reality. If he had enlarged it by a scale factor of 3, it would be 300 × 300 px. 14 The scale factor of

enlarged length would be 3x, so if we use a scale

Fluency

factor of

we get

5 a x = 28, y = 89

b a = 64, b = 85

c p = 71, q = 38

d j = 112, k = 68

Problem-solving

e r = 80, s = 118

f x = 150, y = 117

15 a 30 cm

b c = 15, d = 30

16 3

6 a f = 9, g = 5 c j = 24, k = 22

d s = 4, t = 12.5

e a = 11, b = 26

f e = 27, f = 11

g a = 6, b = 25

h x = 16.5, y = 7

would reduce it back to its

original size. For example, if a side had length x the

4 Smaller

7 a 3

b 3 d

, the original length.

b 60°

17 L = 5.3 18 a 2 cm

b 1 cm

19 a i Display correctly ii Distorted or cropped iii Distorted or cropped

8 a i 2

ii x = 8

iv Display correctly

b i

ii a = 32

v Distorted or cropped

c i

ii x = 12

d i

ii m = 4

11 a

vi Display correctly b i 100 cm2

ii Taller

7.03 Scale and area

9 4 10 a 2

8 cm

b 12.7 cm

Understanding 1 a True 2 a 60 cm

b False

c True b 1 cm

3 0.04

Answers mathspace.co

763

7.04 Scale and volume

Fluency 4 a x = 20

b x = 567

c x = 30

Understanding 1 a False

5 72 cm2

b False

2 a 8 cm

6 3136 cm2

b 64 cm 3

c 512 cm

7 84 cm

c True 3

d Multiplies it by 8

3 1.5 = 2.25 more times

8 a 5 b The larger triangle’s perimeter is 5 times the smaller triangle’s perimeter.

Fluency 4 a x = 1280

b x=4

c x = 36 000

9 324 cm

5 a 4

b 64

c 54 528 cm3

10 2 m

6 a 4

b 64 3

7 1080 cm

Reasoning 11 The enlarged rectangle has an area that is k2 times the area of the original rectangle.

8 5 9 x = 8.2

12 The circle has been dilated with a scale factor of 2. This means that its area will increase by a factor of 22 = 4, and 4 × 78.5 = 314 cm2.

13 a 4

Reasoning

b 16

c 24 L

d The amount of paint needed is not simply four times the amount for the ﬁrst ceiling because the paint covers area, not length. The area scale factor is the square of the length scale factor, so the amount of paint needed is sixteen times the amount for the ﬁrst ceiling. Problem-solving 14 18 m2 2

15 334.89 m

b 10 800 cm2

16 a 12 17 a 76 cm2

b 6 cm, 12 cm and 15 cm c 684 cm2 d 9 e k2 b 3584 cm2

18 a 4 19 a 2.5 cm

b 2 519 424 L 2

20 48 400 km 2

21 a 4x cm

12 a The surface area will increase by factor of 32 = 9. b The volume will increase by factor of 33 = 27. 13 Capacity is the amount that an object can hold, given in litres in this case. The required volume (which is the space the esky takes up) of the new esky is eight times the old esky. This means that the required dimensions of the new esky are times the old dimensions. If Basili buys an esky with dimensions 8 times his old one he will end up with an esky with a capacity of 20 × 83 = 10 240 L, which is much larger than he needs. 14 a Since all sides of a cube are the same length, the side lengths of two different cubes are always in the same ratio. All the faces are also squares, and so all adjacent sides are perpendicular to each other. This means any cube is similar to another cube. b

b x cm

c 1 cm

22 k = 12

764

Mathspace Victorian Curriculum – Year 9 mathspace.co

10 4

Problem-solving 15 a 1.4

b 1.96

16 a

c 2.744

d 0.343

11 a 21 cm

b 13.6 cm

12 a y = 50

b y = 812.5

17 a 1 : 4

b 1 : 64

c m3 : n3

13 128 cm2

18 a 64

b 4

c h = 80

14 3.5 m

19 a 2

b 4

c 8

20 a 1 : 2

b 1:8

c 8

21 a 20 c 0.6429 cm3 22 a 1.5

d 16

b 7.6 cm

15 The area of the dilated triangle is p2 times the area of the original triangle. 16 0.75 m2

d 5143 cm3

17 a x = 116

b 2.25 times greater

18 a 4

b x = 72 c 16 000 cm3

b 64

19 x = 3 cm

Chapter 7 review 1 a Yes

b Yes

2 a y = 19.5

b y=7

3 a r = 62, s = 28

b a = 65, b = 7.5

4 a One triangle is the one with the woman as the ﬁrst side, the side length of 1.8 m as the second side, and the line from the woman’s head to point S as the hypotenuse. The second triangle is the one with the tree as the ﬁrst side, the side length of 5.4 m as the second side, and the line from the top of the tree to point S as the hypotenuse. These two triangles are similar because the share angle S as a common angle, and they both have right angles. This means they are similar by the AAA similarity test. b 4.8 m

20 Capacity refers to the volume an object can hold (expressed in litres or similar units). The desired volume of the new ﬁsh tank is 8 times smaller than that of the old tank. This implies that the required dimensions of the new tank are

= 0.5, which is half the dimensions of

the old one. This means the manager’s estimation is correct. 21 a Yes because all corresponding sides are in the same ratio:

b 1.6

c 2.56

d 4.096

e 0.512

8.01 Pythagoras’ theorem

5 27 m 6 a i

ii z = 36

Understanding

b i

ii z = 6

1 a True

7 a f = 83, g = 68

b a = 117, b = 63

8 a r = 15, s = 30

b c = 16, d = 6

9 Jay is incorrect. For two quadrilaterals to be similar, all corresponding sides must be in the same ratio. In this case, Jay only dilated the rectangle vertically, so only two pairs corresponding sides are in the same ratio. Two sides have a scale factor of 2, while the others have a scale factor of 1. So, the rectangles are not similar.

b False

c True

d True

2 132 + x2 = 192 3 a i △ ABC, △ ABD, △ BCD ii AC

iii BC

iv BD

v AB

b i No

ii No

iii Yes

iv Yes

Fluency 4 a c = 25 c c = 29 m

b c = 15 d c = 50 cm

Answers mathspace.co

765

5 a c=

b c=

Problem-solving

c c=

d c=

16 a d = 105

e c=

f c=

g c=

h c=

17 x = 3.9

6 a y = 12

b x = 16

c b = 21

d a = 18

7 a a=

b b=

c k=

d m=

e q=

f r=

b 14 595 m2

18 a 11.2 cm

b 12.8 cm

19 a x = 5.39 m

b y = 5.39 m

c 21 m

d $777

20 a 6

b x = 6.8

21 a x = 10 cm

b 96 cm

8 h = 10

22 1379 mm

9 165 cm

23 a 1178 m

10 25 cm

8.02 Trigonometric ratios

11 x = 70

1 a Z

12 a She made a mistake on 4 as she simplified extracting

c − b =to ca− b. This is incorrect

because the expression further simplified.

c2 − b2 =cannot a2 be

b Since we are solving for a length, we know that it must be positive. If she did show the negative square root, it would need to be discounted in the final answer. 1

c2 = a2 + b2

c tan

4 a True

b True

c True

5 a Adjacent: BC

Opposite: AC

c Adjacent: YZ

Opposite: XZ

d Adjacent: YZ

Opposite: XZ

c2 − b2 = a a2

c 20 cm

15 The new side lengths will be 6, 8 and 10 units. 62 + 82 = 36 + 64 = 100 102 = 100 So Pythagoras’ theorem still holds for the enlarged triangle.

Mathspace Victorian Curriculum – Year 9 mathspace.co

Opposite: AC

b Adjacent: BC

6 a 52 cm

14 35.78 cm. Since opposite sides of a rectangle are equal, then using Pythagoras’ theorem to ﬁnd each diagonal will gave us the same answer of 35.78 cm.

d True

Fluency

c2 − b2 = a2

Since a2 + b2 = c2 then it is a right-angled triangle.

d c

b ∠A b A

b 7.52 + 102 = 56.25 + 100 = 156.25 and 12.52 = 156.25

c X

3 a H

c2 − b2 = a2

Since a2 + b2 ≠ c2 then it can not be a right-angled triangle.

b X

2 a ∠C

13 a 52 + 72 = 25 + 49 = 74 and 92 = 81

766

b 14 minutes

Understanding

Reasoning

c 3.6

b 48 cm d

7 a Sine

b Cosine

e Sine

f Cosine

c Tangent d Tangent

8 a i

iii

b i

iii

c i

iii

d i

iii

e i

iii

f i

iii

g i

iii

h i

iii

Problem-solving

9 a i x = 12

b i x=4

c i x = 15

17 a ∠BCA

d i x = 20

18 a ∠BCA

10 a

19 a

20 tan θ =

tan θ =

b i Tangent

a tan θ = QS

c i Cosine

a tan θ =

d i Sine

ii sin(25°) =

2a tan θ = QR

12 a i

QR = 2a tan θ

b i

11 a i Sine

21 a 19 : 21

b 19 : 21

c 21 : 28.32

d 57 : 84.96

Reasoning

13 The opposite side is doubled in length, the adjacent side is unchanged and the hypotenuse will increase to compensate but not by double.

is between 3.5 and 5 m. And then using Pythagora’s theorem,

So the sine ratio ratio

will increase, the cosine

will decrease and the tangent ratio will exactly double.

14 a

is the same, which is .

15 From the diagram we have the following

as required.

This means d comes out between 25.25 and 25.49 m.

1 Adjacent and Opposite 2 a Yes

d Yes, this is true for all right-angled triangles. This is because in any right-angled triangle, the opposite side in reference to one angle is the adjacent side in reference to the other angle, and the hypotenuse remains the same for both angles. So, sin θ = cos α for all right-angled triangles.

min d2 = 252 + 3.52 and max d2 = 252 + 52

Understanding

c Yes, the value of the ratio for sin θ and cos α

, cos θ =

will ensure that the roof height

8.03 Side lengths

relationships: sin θ =

≤ tan x ≤

tan θ =

b No

c Yes

d Yes

3 a Use the tangent ratio: tan 45° = b Use the sine ratio: sin 30° = c Use the cosine ratio: cos 60° = Fluency

which is equal to tan θ

4 a 0.990

b 0.883

c 9.514

d 0.342

e 1.072

f 0.629

g 0.588

h 0.087

5 a 8.019

b 11.911

c 4.254

d 4.096

e 2.669

f 1.641

g 1.751

h 0.016

Answers mathspace.co

767

6 a 0.135

b 0.106

c 0.183

d 0.093

Problem-solving

e 4.877

f 0.313

g 0.083

h 0.208

16 a a = 6.26

b b = 3.77

17 a AD = 5.20

b BD = 12.85 c AB = 18.05

c d=5

7 a f = 16.308

b y = 5.032

c f = 5.129

d y = 6.359

e x = 10.037

f x = 10.941

8 a g = 9.465

b h = 15.243

c b = 8.260

d x = 81.144

20 a i y = 90.24 ii w = 65.67

iii x = 24.6

e x = 15.475

f f = 25.812

b i y = 77.80 ii w = 26.74

iii x = 51.1

19 a r = 11.3 cm

b 12.7 cm

21 29 m

9 AC = 15.7 10 a = 12 cm 11 a h = 21.03

18 l = 20.6 m

8.04 Angles b f = 21.53

12 d = 3303.31 m

Understanding 1 a AC

b BC

c BC

d AC

e ∠ACB

Reasoning 13 Since we are given the angle, the opposite side, and the adjacent side, we will use the tangent ratio to solve for b. Opposite Opposite Opposite tan ==θθOpposite == tanθtan θtan Use the tangent ratio Adjacent Adjacent Adjacent Adjacent Substitute tan θ = , 33 3b3b bb == == 10 5510 10 Opposite = b, Adjacent = 10 55 10 33 33 bb bb 10 ==××1010×× 10 10 Multiply both sides by 10 ××1010× =×=10 10 55 55 1010 10 Evaluate using a calculator 66==b66 b == bb

This shows b = 6 m 14 Ethan has only checked for the cosine ratio assuming that 10 cm is the adjacent side to the 30° angle. Instead Ethan can use the sine ratio with opposite side to ﬁnd that:

2 a i

iii

b i

iii

c i

iii

d i

iii

e i

iii

f i

iii

3 Inverse trigonometric functions (sin−1, cos−1, tan−1) help you ﬁnd the unknown angles in a right-angled triangle when you know the side lengths. On the other hand, regular trigonometric functions (sin, cos, tan) are used to work out the ratios of the side lengths of a right-angled triangle when one of the angles is already known. Fluency

So we end up with the hypotenuse being 20 cm. This shows that it is possible to have a rightangled triangle with an angle of 30° and side lengths of 10 cm and 20 cm. 15 a i 0.98 and 0.98

ii 0.92 and 0.92

iii 0.71 and 0.71

iv 0.53 and 0.53

b Sine of an angle is equal to cosine of the complementary angle. c The angles are complementary angles. That is, θ + α = 90°.

768

Mathspace Victorian Curriculum – Year 9 mathspace.co

4 a 45°

b 64°

c 27°

d 16°

e 50°

f 83°

g 55°

h 45°

5 a 0°

b 0°

c 12°

d 60°

e 90°

f 45°

g 73°

h 41°

i 0°

j 60°

k 26°

l 66°

m 66°

n 54°

o 24°

6 a 59.46°

b 61.15°

c 83.63°

7 a x = 74°

b x = 69°

c x = 74°

d x = 74°

e θ = 36°

f x = 62°

g x = 67°

h y = 47°

i θ = 67°

j x = 67°

8 θ = 54°47′

Problem-solving

9 tan 33°

19 x = 26.57°, y = 50.19°, z = 13.24°

10 a 0.2

b 0.6

c 1.2

d 0.4

e 0.8

f 1.7

11 a 27°

b 45°

c 63°

d 76°

12 a

b θ = 29.05°

20 x = 21.9°, y = 48.8° 21 a

22 a x = 29.74°

b y = 150.26°

23 59°

13 θ = 47.73°

24 θ = 21.80°

14 θ = 25.02°

25 a QR = 9.75 cm

Reasoning

8.05 Applications of trigonometry

15 a A = 66.4° b To create a right-angled triangle from the isosceles triangle given in the question, we must add a vertical line which divides the isosceles triangle into two halves (the line of symmetry). Then we know that the hypotenuse is 10 cm and the adjacent side to angle A is 4 cm. The cosine function is the most appropriate choice as it uses the hypotenuse and adjacent sides to an angle.

b θ = 33.02°

Understanding 1 α 2 a σ

b θ and σ

3 Tangent 4 a

1.36 m

16 sin θ = cos θ when θ = 45°. When the sine and the cosine are equal,

44°

This means that the opposite and adjacent sides must be equal, which occurs in a right isosceles triangle with angle θ = 45°. 17 If the sine ratio of an angle is greater than the cosine ratio of the same angle, it means that the side opposite the angle is longer than the side adjacent to the angle. This occurs when the angle is greater than 45°. Angles in a rightangled triangle (which aren’t the right angle) must be less than 90°. Therefore, the angle is in the range (45°, 90°). 18 False. When the gradient is doubled, it does not mean that the angle between the line and the positive x-axis is also doubled. The relationship between the gradient and the angle is not linear. Instead, the gradient is related to the angle through the tangent function: m = tan θ. So, when the gradient is doubled, the angle between the line and the positive x-axis will not necessarily double, but the tangent of the angle will be doubled.

b Sine Fluency 5 h = 11.15 6 d = 285 m 7 θ = 41.81° 8 θ = 43.79° 9 y = 58.02 10 a

2.79 m 2.25 m

b θ = 53.75°

Answers mathspace.co

769

11 a 74.85 m

b 28.31°

12 a x°

3.46 m

1.9 m

24 m

Finally, we can set up equations for the angle based on each length and evaluate:

b θ = 16.26° 13 θ = 23.9 14 d = 814 m 15 d = 1857.11 Reasoning 16 Luca is correct, because the two sides given are the opposite side and the adjacent side. 17 a A diagram to represent the situation is shown.

26° 48’

The shorter side is opposite the angle made by the diagonal and the longer side, and the diagonal is the hypotenuse. The sine ratio is used when we have the opposite side and the hypotenuse.

The only distance which gives an angle between 10° and 20° is 86 cm, so the ladder is safest to climb when it is 86 cm from the wall. 19 Vern is correct. As the ship drifts, the angle decreases. The angle, x, is approximately 38°. Problem-solving 20 a 73 500 m

b h = 41 101

21 a

b y = 4.51

65°

18 Since the distance from the base of the ladder to the wall is given in centimetres, and the ladder is given in metres, we ﬁrst convert the distances to metres by dividing by 100:

20 m

b 44.49 m

190 cm = 1.9 m

22 a 63.3 m

42 cm = 0.42 m

23 Angle SPR = 33.61°

86 cm = 0.86 m

24 36.7 m

Drawing a diagram helps organise the known information and determine which trigonometric ratio is needed. Using the ﬁrst length given, a diagram can be drawn as shown.

1.6 m

b θ = 36.68

Chapter 8 review 1 a

2 a

3 Pythagoras’ theorem can only be used to ﬁnd missing side lengths in right-angled triangles. Since the angles in a triangle sum to 180°, we

770

Mathspace Victorian Curriculum – Year 9 mathspace.co

can ﬁnd the missing angle by subtracting the other two angles from 180°. 180 − 28 − 63 = 180 − 91

b 5.14

c 0.97

d 6.78

e 3.09

f 0.85

g 2.23

h 0.33

12 a v = 21.87

= 89

11 a 0.64

This show that the triangle does not have a right angle, so Pyhtagoras’ theorem cannot be used. 4 d = 1844 m 5 1 020 cm

b s = 8.15

c y = 74.67

d g = 32.91

e t = 1410.93

f h = 40.45

13 a 30°

b 45°

c 30°

d 40°

e 37°

f 36°

g 75°

h 55°

14 a z = 70°

6 d = 21.3

b w = 50°

c m = 36°

7 a Cosine

b Sine

d j = 28°

15 71.6°

8 a i sin θ =

ii cos θ =

iii tan θ =

16 θ = 49.40°

b i sin θ =

ii cos θ =

iii tan θ =

17 a θ = 30°

b x = 2.60

18 H = 43 m

9 a i y = 21

ii tan θ =

b i y = 30

ii tan θ =

10 a Since sin θ =

, the hypotenuse has a

19 Ezekiel is correct. As Alena moves closer, the angle increases. At the beginning of Alena’s hike, the angle from the ground to the peak is about tan−1

length of , and the opposite side from θ also has a length of 1. Using Pyhtaogras’ theorem, we ﬁnd the third side to have a length of 1.

≈ 15°. After moving toward the

mountain, the angle is about tan−1

≈ 18°.

This shows the angle is increasing as Alena moves toward the mountain. 20 θ = 37.7°

9.01 Area Understanding

1 a False

This means the triangle is an isosceles right-angled triangle. b Any triangle with side lengths in the form x, x, . For example: 1 1

b True

2 a i Incorrect

ii Approximate ii Approximate

d i Correct

ii Exact b 0.7 cm

c 270 000 m

4 a 36 cm2

b 145 mm2

d 54 m2

e 121 cm

f 1539 cm2

g 30 m2

h 420 cm2

d 5 ha

Fluency

c 144 cm

ii Exact

c i Incorrect

d False

b i Correct

3 a 300 cm

c True

e True

j 81 cm2

i 90 m

k 276 mm

l 34 m2

m 76 mm2

n 75 mm2

Answers mathspace.co

771

Base

Height

Area

5 mm

2 mm

10 mm2

15 cm

7 cm

105 cm2

10 cm

5 cm

50 cm2

7.5 m

37.5 m2

c The area of a kite with diagonals of x and y is half of the area of a rectangle with length and width of x and y.

a=5m h=8m

b=9m 2

A = 56 m

7 h = 16 cm

d The area of a rhombus with diagonals of x and y is half of the area of a rectangle with length and width of x and y.

b = 18 cm

A = 288 cm2 2

8 a 20 cm

b 3.3 cm

c 27 cm2

d 80.0 cm2 y

9 130 mm2 10 a 169π cm2 c 121π cm2 11 a 283.5 cm2 c 804.2 cm2

b 484π mm2 289 d π m2 4 b 452.4 mm2

13 a 4 cm c 28.5 cm

d 153.9 m2

b 17 mm

16 Nadine is correct. The new radius will be .

d 3.16 mm

If we substitute this into the area formula it becomes area.

Reasoning 14 a The area of a triangle with base b and height h is half of the area of a rectangle with length and width b and h. h

h b

15 No. Doubling the side lengths does not double the area. The current area is 54 cm2. Doubling the side lengths would result in an area of 216 cm2 which is more than double the area.

12 314.16 cm2

b The area of a parallelogram with base b and height h is the same as the area of a rectangle with length and width b and h.

Mathspace Victorian Curriculum – Year 9 mathspace.co

. This is a quarter of the original

For example: r = 6 cm; A = 36π cm2 6 36π r = = 3 cm; A = 9π cm2 = cm2 2 4 17 The circumference of a circle is C = 2 × π × r, so we can divide the circumference by 2 × π to .

ﬁnd the radius,

Now, we can put the radius into the area formula A = π r2 to get A = π × 82 = 64π. We can leave this as an exact value or round to get the approximate area. 18 a a + b

772

b h(a + b)

5 a 96 cm2

Problem-solving 19 a 150 000 m2 20 a

e 194.85 units

Width (m)

Area (m2)

200

100

20 000

350

7350

100

2400

400

16 000

100

10 000

b 120 cm2

d 63 cm2

7 a i One semicircle and one triangle ii 98.87 cm2 b i One rectangle and four semicircles ii 233.30 mm2 8 a 636.2 cm2 2

c Farm 3 and Farm 2 d Farm 5

9 a 40.8 cm2

e 11 150 m2

b 242.9 cm2 2

21 a 4 m

b 5 plants

22 42 m2 2

23 a 84 cm

b 276.3 m2 d 70.7 cm2

c 44.6 cm

f 513 cm2

6 a 144 cm2 c 136 cm

b Farm 1 and Farm 4

c 242.9 cm

d 17 398.5 mm2

10 a 245.46 m2

b 461.81 mm2

11 a 414 m2

b 318 m2

b 24 tiles Reasoning

24 150.92 m

12 a A = 2w2 − s2

25 1884.96 m2

9.02 Area of composite shapes

Remember that the area of a triangle is given by Area =

Understanding b False

c True

b No

13 There are two triangles with base 18 cm and height 8 cm, and one smaller triangle with base and height each 8 cm.

26 14.1 cm2

1 a True

d 180 cm2

c 463 mm

b 15 ha

Length (m)

Farm

b 143 mm2 2

× base × height.

d True

2 a Rectangle and triangle b To calculate the total area of a composite shape composed of a rectangle and a triangle, you would ﬁnd the area of the rectangle and the area of the triangle separately, using their respective formulas, and then add the two areas together. 3 a Square and triangle b To calculate the total area of a composite shape composed of a square and a triangle, you would ﬁnd the area of the square and the area of the triangle separately, using their respective formulas, and then subtract the area of the triangle from the area of the square. Fluency

Evaluate each product ﬁrst Evaluate the addition

14 Joshua’s approach is correct. In calculating the area of composite shapes, it’s important to break down the composite shape into its simpler components, ﬁnd the area of each component, and then add these areas together. The composite shape in this case is made up of a rectangle and a semicircle, so it’s appropriate to calculate the area of the rectangle and the semicircle separately, and then add these areas together. Joshua correctly calculates the area of the rectangle as 10 cm × 6 cm = 60 cm2 and the area of the semicircle as . Adding these

4 a 52 m

b 24 m

c 76 m

together gives the total area of the composite shape as approximately 74.13 cm2.

Answers mathspace.co

773

On the other hand, Lila’s approach is incorrect because she calculates the area of a full circle instead of a semicircle. As a result, she overestimates the total area of the composite shape.

Fluency

15 Ava is correct in this situation. The problem describes a composite shape made up of a square and a quarter-circle. The diameter of the circle is equal to the side length of the square, not the radius. Hence, the radius of the quarter-circle is half the side length of the square. Therefore Ava’s calculation of the areas of the square and quarter-circle are correct, while Max’s calculation incorrectly uses the side length of the square as the radius of the quarter-circle.

6 A rectangular prism is made up of 6 faces. Option A has 3 faces and option D has 5 faces, so they are not possible. Option B shows the incorrect measurements, so option C is the correct net.

Problem-solving 16 144.62 cm2

4 a Triangular prism

b Rectangular prism

c Hexagonal prism

d Trapezoidal prism

5 B

7 a Triangular prism

b Pentagonal prism

c Octagonal prism

d Trapezoidal prism

8 a

17 96 pieces of the 15 cm tile, 126 pieces of the 30 cm tile 18 a Example answer:

50 m

3.2 m 30 m

19 a 18 m

e ii

f viii

g v

h vii

Reasoning

c 365.37 kg

b 7.13 m2

20 a 28.5 m

d 35.37 m2

c $427.56 e $3537 21 a 4025 cm2 2

c 6507 cm

b 3867.9 cm2 d $1440

Understanding b False

c True

11 The net of a rectangular prism shows that it has 6 faces. By finding the area of each face of the net it is possible to find the surface area of the rectangular prism. 12 If you’re given a net with six congruent square faces, you can form a cube from it. You can prove this by showing that all faces, edges, and angles are the same, which is the definition of a cube.

9.03 Nets of prisms

d False

2 6 faces 3 2

774

d i

b $864 2

1 a True

b vi

c iv

10 Possible answers:

b 1461.49 m2

9 a iii

Problem-solving 13 $46.83 14 a $47.635

Mathspace Victorian Curriculum – Year 9 mathspace.co

b $49.89

c Box 1

9.04 Surface area of prisms

(1 × 1 × 27 = 27 cubic units) but its surface area is 2 × (1 × 1 + 1 × 27 + 1 × 27) = 110 square units.

Understanding 1 a 6

b 5

c 6

d 8

2 a False

b True

c True

d False

16 Vlad is incorrect. Even though they have the same base area and the same height, the surface areas will be different because the faces of the prisms are different.

3 C Fluency 4 a

15 John is correct. The surface area of the rectangular prism is 2(lw + lh + 2wh) = 2 × (8 × 5 + 8 × 3 + 3 × 8) = 158 square units.

6 cm 6 cm

Substituting into the area of a trapezium Substituting into the area of a rectangle Factorising out the common term, l

Sum of all 6 sides

b 216 cm

5 a 54 cm2

b 864 cm2

c 222 m2

d 180 cm2

e 232 m2

f 484 mm2 2

6 a 756 cm

b 487 cm

c 434 cm2

d 264 cm2

7 a 306 cm2

b 510 cm2

8 a d = 18.0 cm

b 492 cm2

Simplify

Problem-solving

18 a 172 cm2

b 13.11 cm

19 11 cm 20 14 m 21 474 cm2

9 708 cm2

22 22b2 cm2

10 21 108 cm

11 152.62 m2 2

12 a x = 10 cm and surface area is 432 cm

b x = 5 cm and surface area is 1220 cm2 13 a 295.99 cm2

Substitute

b 153.30 m2

Reasoning 14 No, a cube and a rectangular prism with the same volume do not necessarily have the same surface area. This is because the surface area depends on the dimensions (length, width, height) of the prism, not just the volume. A cube has equal dimensions, while a rectangular prism can have different dimensions, leading to different surface areas. For example, a cube with sides of length 3 units has a volume of 3 × 3 × 3 = 27 cubic units, and a surface area of 6 × (3 × 3) = 54 square units. A rectangular prism with dimensions 1 unit by 1 unit by 27 units has the same volume

23 a 52x2 m2

b 156x2 m2

24 a 3240 m2

b 3175.2 m2

25 318.8 cm2

9.05 Volume of prisms Understanding 1 a Square, 21 cm

b Triangle, 8 cm

c Rectangle, 3 cm 2

Volume

d Trapezium, 3 cm

Capacity

1 cm3

1 mL

50 cm3

50 mL

750 cm3

750 mL

1 m3

1 kL

1 m3

1000 L

1 m3

1 000 000 mL

3 a True

b False

c False

d True

Answers mathspace.co

775

Fluency 3

b 156 m

b 2250 cm3

4 a 126 cm

5 a 336 cm

c 560 mm3 3

e 52.5 m 3

6 a 35 m

c 63 cm3

d 240 cm3 f 240 cm3 b 5 824 mm3 d 360 cm3

7 16 m3 8 120 000 cm3

consider Prism B with a larger base area of 15 square units but a shorter height of 2 units. Its volume would be 15 × 2 = 30 cubic units. Despite Prism B having a larger base area, its volume is equal to Prism A due to its shorter height. However, if the height of Prism B was even less (for instance, 1 unit), its volume would be 15 × 1 = 15 cubic units, which is less than the volume of Prism A. Thus, a prism can have a larger base area but a smaller volume if its height is signiﬁcantly shorter.

9 h = 6 cm

Problem-solving

10 y = 9

21 1.245 cm

11 l = 30 cm

22 486 L

12 a 450 mL

b 0.45 L

24 4112.4 mm3

13 80 L 14 a 40 mL

23 784 cm3

b 87 880 000 mm3

c 2197 containers 15 b = 18 m Reasoning 16 No it is not possible, because the volume can be found by multiplying base by height.

25 a 7605 cm3

b 9

26 There are many possible answers. Prism A: 2 by 8 by 9 and Prism B: 4 by 3 by 6. Another possible solution is Prism A: 8 by 6 by 1 and Prism B: 2 by 3 by 4. 27 a 240 m2

b 4800 L

c 100 mm

17 Locker A contains a volume of 192 000 cm3 and Locker B contains a volume of 195 750 cm3

28 a y = 80 − 2x

b 80x2 − 2x3

29 a 3.52 m3

b 8.8 minutes

Stuart should choose locker B as it contains a larger volume.

9.06 Accuracy of measurement

18 If the length, width, and height of a rectangular prism are all doubled, the volume will increase by a factor of 23 = 8, so the volume will be 8 times the original volume.

Understanding

19 If two prisms have the same volume, it doesn’t necessarily mean they have the same base area or height. It is possible for one prism to have a larger base area but a shorter height, or vice versa. An example is a prism with base area of 6 cm2 and a height of 4 cm. Its volume will be 24 cm3, same with a prism with base area of 8 cm2 and a height of 3 cm. 20 Yes, it’s possible. The volume of a prism is a product of its base area and height. For example, consider Prism A with a base area of 10 square units and a height of 3 units. Its volume would be 10 × 3 = 30 cubic units. Now,

776

Mathspace Victorian Curriculum – Year 9 mathspace.co

1 a 22 mm

b 33 mm

c 46 mm

d 65 mm

2 a iii

b iv

c ii

3 a True

b True

c False

d i

Fluency 4 a 0.6 kg 5 a 0.5 kg

b 2.6% b 0.5 g

6 a 0.05 mL

c $5 b 0.5 mm

c 0.05 g

d 0.5 s

e 0.05 °F

f 0.005 kg

g 0.05 m

h 0.005 °C

7 a 1 cm d i No

b 90 cm

c 0.5 cm

ii No

iii No

iv Yes

8 a i 6.45 m

ii 6.35 m

b i 5.5 km

ii 4.5 km

c i 14.005 kg

ii 13.995 kg

Fluency

d i 4884.5 m

ii 4883.5 m

4 a 219.91 m2

b 5 kg

5 872.9 m2

9 a 10 kg

c Lower bound: 65 kg, upper bound: 75 kg

3 a Yes

b No

b 169.65 m2

6 a i 75.40 cm2

ii 100.53 cm2

b i 75.40 m2

ii 131.95 m2

c i 201.06 m

ii 301.59 m2

ii 353.43 m2

13 19.995 m

8 a 565.49 cm2

b 188.50 m2

14 12.05 m

c 314.16 cm2

10 a 50 km/h 11

b 5 km/h

c 55 km/h

d i 226.19 m

12 0.4%

7 144π cm2

15 a Maximum: 175, minimum: 165 b Maximum: 3050, minimum: 2950 2

16 a 94.4025 m

e 9424.78 cm

f 66 853.09 cm2

9 a radius = 15 m

b height = 92 m

10 height = 73 m

b 92.2625 m2 17 a 1188 cm3

b 1026.375 cm3

c 1365.625 cm3

d 177.625 cm3

e 15.0% Reasoning 18 We don’t know what the number has been rounded to (eg. nearest 5, 10 or 50). 19 a 12.57 m2

b 12.56 m2

c 0.01 m2

d In real-world situations, using rounded or approximated values can lead to error in calculations. In this case, it could mean not having enough paint for the circle. Problem-solving 20 a 26.25

b 0.6

21 a 1 mm

b 79.5 mm

22 a Upper bound: 272.6925 m2, lower bound: 269.3925 m2 b Upper bound: 66.2 m, lower bound: 65.8 m 2

23 31.5725 m

24 a 7.26 m/s

b 7.13 m/s

11 height = 51 cm Reasoning 12 Matea is incorrect. Doubling the height of a cylinder will not double its surface area. The area of the curved part of the cylinder will be doubled but the circular bases will not be doubled. The overall surface area will increase but not be doubled. For example, the surface area of a cylinder with radius 3 cm and a height of 5 cm is 48π cm2. While the cylinder with radius 3 cm and a height of 10 cm will have a surface area of 72π cm2. 13 If the surface area is ﬁxed and you increase the radius of the cylinder, its height must decrease. This is because the surface area of a cylinder is directly proportional to both its radius and its height. So if one increases and the total surface area is to remain the same, the other must decrease. 14 Yes. By increasing the total amount of paper, you will increase the diameter which will affect the total surface area of the toilet paper roll, although the height will remain the same.

9.07 Surface area of cylinders

Problem-solving

Understanding

15 593.76 cm2

1 a C = 2π x 2 a True

d 15 896.46 cm2 2

b False

b A = π x2

16 125.66 cm2

c False

17 742.11 cm2

Answers mathspace.co

777

Reasoning

18 5.2756π m2 19 a 326.73 cm2

13 The volume of the smaller can is 1152π cm3 and the volume of the larger can is 9216π cm3. The volume of the bigger can compared to the smaller one is 8 times larger. Lyn is incorrect

b 353.43 cm2

c Vincent’s cylinder has larger surface area by 26.7 cm². 20 452.6 m2

14 a 1130.97 cm3

21 101.8 m

c 360π cm3

e To go from cm to mm3, we multiply by 1000. This means that any rounding that we make can be ampliﬁed. Since in part (a) we only rounded to two decimal places, that means we lost one decimal place of accuracy as when we multiply by 1000, the decimal places “moves” three positions. There is a difference of 3 mm3 due to rounding. Converting from an exact value instead of a rounded value gives a more accurate volume reading for mm3.

22 1376.02 cm 23 138.98 m2

24 a TSA = 4π r2 + 6π rh b 4.95 cm 25 1.81 m2

9.08 Volume of cylinders Understanding 1 Circle 2 a

3 a 201.06 cm2

b 153.94 cm2

4 a 256π

b 231π

5 a True

d 1 130 973 mm3 3

b False

c False

Depending on the context this may or may not matter. In a scientiﬁc experiment, this could make a big different, especially if this inaccurate measurement is being used repeatedly. If this was a measuring cup for cake batter, this would not make a signﬁcant difference as this is less than 1 mL off.

d False

Fluency 6 a 452.4 cm3

15 Bob should buy the second can because it has a larger volume. The volume of a cylinder is given by the formula V = π r2 h. Even though the height of the second can is less than the ﬁrst, the increase in the radius (from the increase in diameter) has a greater effect on the volume because the radius is squared in the formula.

b 5598.3 m3

c 5674.5 cm3

d 19 088.3 mm3 or 19.1 cm3

7 a 350π mm3

b 896π cm3

c 3610π km

d 307.2π m3

8 a 294.5 cm3

b 10 995.6 cm3

d 69 115.0 cm2

c 402.1 m

e 23 605.9 cm 3

9 a 5780π cm 3

c 2873π m

Problem-solving

f 2413.7 cm3 or 2 413 748.5 mm3

16 2.8 cm

b 1408π mm d

10 a 137.4 cm3

17 16 286.02 L 3

π km

18 275.08π cm3

b 527.8 mm3

c 150.8 m

19 2 387.61 cm3 3

d 7.6 cm or 7634.1 mm 3

11 a 1837.83 m

c 1633.63 m3

b 829.38 cm

d 7037.17 cm3 or 0.01 m3

12 81π mm3

778

b 1 130 970 mm3

Mathspace Victorian Curriculum – Year 9 mathspace.co

20 72π cm3 21 a 29.69 cm3 b 169.2219 g costs $84.61

18 574.6 L

Chapter 9 review 2

1 a 150 mm

2 a 12.25π m2

b 107.5 mm

19 a 0.05 mL

b 38.48 m2

c 0.005 kg

b 0.5 s d 0.05 °C

20 a 280 cm3

3 19 mm

b 219.375 cm3 3

4 a 180 m

b 24 pieces 2

e 25.2% 2

5 a 187.5 mm

b 563.5 mm

6 a 229 cm

b 108 cm2

7 a 197.92 cm2

b 196.77 mm2

8 54 stones are needed. 900 ÷ 60 = 15 stones to cover the length of the pool, so 30 stones are need for both lengths. 600 ÷ 60 = 10 stones are needed to cover the width of the pool, so 20 stones are needed to cover both widths. However, this leaves the four corners without any stone, so we need to add 4 more stones. This gives a total of 54 stones. 9 a Square prism or rectangular prism

21 a 7.3 m

b Minimum: 7.25 m Maximum: 7.35 m

22 a 527.79 mm2

b 5183.63 cm2

23 a Radius = 12 m

b height = 42 m

24 a 417.83 cm

b 537.21 cm2

25 a 2309.1 mm3

b 5654.9 m3

26 a 981.7 mm3

b 175.9 cm3

27 219.91 m3 28 Radius= 2.19 cm

10.01 Experimental probability

b Triangular prism or right-angled triangular prism 10 a

d 70.625 cm3

c 350.625 cm

Understanding

1 C 2 In situations which involve experiments such as having a “loaded die” or historical events such as the winners of a contest. 3 True Fluency

11 a iii

b i

12 a 304 cm2 c 412 cm2

c ii

b 478.22 cm2 d 148 m2 b 2032 m2

13 a x = 8 m

6 2

14 The original surface area is 2 (3x + 4x + 12x ) = 2 (19x2) = 38x2. The new dimensions would be 6x × 2x × 8x, so the new surface area is 2 (12x2 + 16x2 + 48x2) = 2 (76x2) = 152x2. This means the surface area increased by 152x2 − 38x2 = 114x2 square metres. 3

15 a 480 cm

c 114 mm3 16 h = 7 cm 17 39.375 mL

b 138

5 a

7 a 5 times

8 a 12 times

9 2% 10 a 59%

b 41%

b 112 m

d 10 200 mm3

11 12 a 13 a 240 desserts

c b 45%

Answers mathspace.co

779

14 a

15 a

16 a

17 a

the long run, with a large number of trials, the experimental probability should approach the theoretical probability more closely. However, in a relatively small number of trials (like the 19 tosses in this case), there can be variations that cause the experimental probability to differ from the theoretical probability.

d 1

c c

18 a 102 times b i

iii

19 a 50 times

20 a

21 a

22 a

Problem-solving 31 a Summing the results over the 3 days results in 120 total spins and 32 total “Yes”. Thus, a possible spinner could look like this

NO YES

b 58.90 cm2

23 60 commuters 24 57 times

b 72 people

32 a

c 14 months

25 a 48 000 people

b 480 people

26 a

b 160 tablets

27 a 400 deliveries

10.02 Venn diagrams and two-way tables

d 150 deliveries

Understanding

c Bicycle 28 a 71%

b 73%

c 69%

33 Player A

1 a iv: A ∪ B, the union of events A and B b i: Sample space

Reasoning

c iii: A ∩ B, the intersection of events A and B

29 a 0.52

d ii: A′, the complement of event A

b The coin seems like it might be fair, but more tosses should be done to check.

b 2 plants

c 0.504

3 a True

d The more ﬂips that are performed the closer the coin’s experimental probability gets to the theoretical probability. It is probably a fair coin.

4 a 13 students

30 a

c The experimental probability and theoretical probability are not the same. The experimental Probability is greater than the Theoretical Probability. This difference might be due to chance or randomness in the experimental results. In

780

2 a 6 plants

Mathspace Victorian Curriculum – Year 9 mathspace.co

b True

c 5 plants

c False

d False

e False b 25 students

c 30 students Fluency 5 a

6 a

7 a 28 players

b 1

c 8 players

Walks regularly

Does not walk regularly

Total

Dogs

Cats

Total

108

9 a 60 10 a

16 a Month

Took off on time Delayed 135

Total

180

June

140

200

Total

275

105

380

b 380 ﬂights

c 30.0%

d 35.5%

Total

Non-ﬁction

Total

c 50% Laptop owners Tablet owners

17 a

11 a

Borrowed Available Fiction

May

Freelancer

Physical book

Likes smooth Likes crunchy Total peanut butter peanut butter Tea

105

Coffee

Total

180

b 180 individuals 12 a

d Yes

Enjoy

Don’t enjoy

Total

Reasoning

Teenagers

Seniors

Total

18 Joanna is correct. 38 people speak French. Trent forgot to include the people who could speak French and Spanish. 19 a 6 people

b 13 a

Likes pizza

Doesn’t like pizza

Total

Likes sushi

Doesn’t Like sushi

b 45 people 14 a Children

15 a

Not vegetarian

Vegetarian

Total

c Goran is correct. 14 people play only 1 sport. Goran added the people that played both tennis and football, then took away the players that played both sports twice as people that played in both sports are counted in football and in tennis. 13 + 11 − 5 − 5 = 14 d The new probability will be lower than the probability calculated in part (b).

Adults

20 a Samuel added the number of students that joined the soccer and basketball clubs 75 + 78 = 153. This is 11 more than the total number of students that were asked to join. This means that 11 students are found in both groups. b Yes, because more students joined the basketball club.

Answers mathspace.co

781

Problem-solving 21 a 27 movies d i

27 a b 43 movies

c 15 movies

iii

22 a 70 plants

c Sketching

Not Total rectangular

Emerald

Not emerald

Total

b i

23 a

Rectangular

28 a

iii

Restaurant Bowling alley

Pottery

5 15

29 25 4

Theme park

b 12 people c i

b 128 people ii

iii

c d i

24 a i

iii

29 a

iii

b Example answers: 2

i Not a primary student ii Not a primary student that catches the bus to school

2 M

iii Either a senior student that does not catch a bus or a primary students that catches a bus 25 a i

iii

b Example answers: i Not a non-musician

A 1

26 a 7 employees

1 Project management

c $45.80

iii A child musician

Leadership development

ii A musician who is not a child

3 1

1 0

Probability that a plant at random has aphids but not mites is c

782

Mathspace Victorian Curriculum – Year 9 mathspace.co

30 a

Detected Detected Total Pesticides as Organic

7 a

Heads

H,1

H,2

H,3

H,4

H,5

H,6

Tails

T,1

T,2

T,3

T,4

T,5

T,6

Fruit has pesticides

Fruit is organic

140

b 12

Total

120

200

b 60%

c 66.67%

8 a

Understanding 1 a True

b False

c True

d False

2 a Equal probability b Not equal probabilities c Equal probability 3 Sample space = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

10.03 Arrays and simple tree diagrams

9 a

Fluency 4 a P (9) = b P (less than 5) = c P (more than 6) =

1, A

1, B

1, C

2, A 2, B 2, C

3, A 3, B 3, C

b 9

10 a

d P (7 or higher) =

b P (more than 6) = c P (6 or less) + P (more than 6) = 1 d (a) and (b) are complementary events, so they cover all possible outcomes and have a sum of 1. 6 a 8

Cyan (C)

iii

Pink (P ) White (W )

C, B

P, B

W, B

Grey (G)

C, G

P, G

W, G

Red (R)

C, R

P, R

W, R

Yellow (Y )

C, Y

P, Y

W, Y

b 12

11 a 8 b {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} c i

b i

Black (B)

e P (an even number) = 5 a P (6 or less) =

iii

v 12 a

Answers mathspace.co

783

13 a

Red

Problem Solving

Black Red

17 a

Black

18 a

Red

Red 19 a

20 a i

iii

vii

viii

ii False

iii True

iv False

d 1

Red Red

Black Red

Black Black b

Black Sally

Peter

Alvin Peter Sally

Sally

Alvin

Peter Alvin

Peter

Sally

Alvin Peter

Alvin

Sally

14 An array would be more efficient as there are 36 possible outcomes. Drawing up that many possibilities on a tree diagram would take a long time. However, filling in a 6x6 grid would not take as long. An array presents all possible outcomes at once, making it easy to see every combination. This is particularly beneficial when you need to calculate probabilities for specific events or sums of the dice. 15 The probability the next flip is a heads is . This is because each flip is independent of the others, and for any given flip, there are only two possible outcomes: heads or tails. Since there are no specific factors or conditions mentioned that would affect the flip, the probability is the same. 16 The probabilities are not the same. For both dice, there is only one way this can happen, so the probability is

If we made an array for Myah, we could see that the sample space has 36 outcomes. While for Frankie the sample space has 36 outcomes. So and

These are different, so the chance of rolling a 6 and then a 2, is not the same for Frankie and Myah.

784

21 a e 0

22 a The 12-sided and 20-sided dice b The 4-sided and 6-sided dice c Any combination of dice

Reasoning

the probabilities would be

b i True

Mathspace Victorian Curriculum – Year 9 mathspace.co

10.04 With and without replacement Understanding 1 a With replacement

b Without replacement

c With replacement

d Without replacement

2 The sample space changes between experiments with and without replacement. With replacement, the sample space remains constant, while without replacement, it decreases after each draw. 3 a True

b False

c False

d True

4 {23, 24, 32, 34, 42, 43} Fluency 5 a B

B B W G

Sample Space BB BB BW BG

B B W G

BB BB BW BG

B B W G

WB WB WW WG

B B W G

GB GB GW GG

Sample Space B

W G B W G B B

G B B

WG GB GB GW

W 6 a Red

Blue

Green

Yellow

b Red

Blue

Green

Olivia

Sarah

BG BB BW BG WB

David Micheal Sarah

Olivia

David Micheal Sarah Olivia Micheal Sarah

David

Olivia David

Micheal

9 a Red Blue Green Yellow Red Blue Green Yellow Red Blue Green Yellow Red Blue Green Yellow

Sample Space 8 78

7 8

b {78, 79, 87, 89, 97, 98} c 10 a

Sample Space G G G B G G G B G G G B G G G B

Blue Green Yellow Red

Green Yellow Red Blue

GG GG GG GB GG GG GG GB GG GG GG GB BG BG BG BB

b 11 a

Blue

Green

Yellow Red Yellow

7 a Without replacement b {BG, BY, GB, GY, YB, YG} b

1 2

5 5

Answers mathspace.co

785

c i

iii 0

iv 1

b i

Red

Blue Green

15 a

b 0

16 a

Red

Blue

17 a {MO, MN, ME, MY, OM, ON, OE, OY, NM, NO, NE, NY, EM, EO, EN, EY, YM, YO, YN, YE}

Green Red Green

Blue

b i

iii

Sample Space

Reasoning

9 5

69 65

18 When picking items with replacement, the number of choices (branches) and the chance of each outcome stay the same throughout. Without replacement, the number of choices decreases for each level, and the chance of picking each item changes.

5 6

35 36

13 a

3 b i 0

For example, imagine a bowl with 2 balls (A and B). If you pick balls with replacement, there are always 2 choices at each level, and you can have 4 different results (AA, AB, BA, BB). But if you pick balls without replacement, there are 2 choices in the beginning, and then only 1 choice afterward. So, you can only have 2 results (AB, BA).

ii 1

14 a C

786

iii

v 12 a

C C L L C C L L C C L L

Sample Space CC CC CL CL CC CC CL CL CC CC CL CL

C C C L C C C L

LC LC LC LL LC LC LC LL

Mathspace Victorian Curriculum – Year 9 mathspace.co

19 a The marbles are drawn with replacement. You can tell because after a marble is drawn in the tree diagram, it does appear as an option in the next level of branches. b The labels R1, R2, and R3 are used instead of just R to show that there are three different red marbles. Even though the marbles are drawn with replacement, using these labels helps us see that there are multiple red marbles in the bag, and each one can be drawn separately. 20 a Yes, getting YY is possible when you select two letters one after the other without putting them back from the word YUMMY. There are two Ys in the word. The chance of selecting

YY is

24 a

because you can select one Y ﬁrst

and then the other Y next. In fractions, that’s . b MY. First, we calculate the chance of getting UM. That’s

. Now, considering the

word YUMMY, we notice there are two Ys and two Ms. This gives us a bigger chance of ending up with MY. We can work out this probability as

second selection remains . So the overall .

Problem Solving b

An event with a probability of 0 would be drawing two counters with the same number, for example, drawing both counters with the number 4, which is impossible as each number appears only once in the bag. b Answers may vary. An event with a probability of 1 would be drawing two counters with a sum greater than 8, as any pair of counters taken from the bag would have a sum larger than 8 (like 4 + 5 = 9 and the sums only increase from there).

3 4

b i

10.05 Multi-stage probability Understanding 1 a True

b False

c True

d False

e False 2 a

3 a

1 6 H 1 2

An event with a probability of less than 0.5 would be drawing a smaller number like 4 as the outcome. The possible draws for this outcome are (4, 5), (4, 6), (4, 7) and (4, 8),

1 6 1 6

1 2

or 0.2,

T 1 6

considering all the 20 possible pairs of counters.

1 3 1 3

R B Y G R B Y G

b No

23 a Without replacement ii

42 42

ii There would be 20 × 19 = 380 ﬁlled cells, as each card can only be paired with different cards, and the array would be symmetrical along the diagonal (eliminating duplicate combinations).

c Answers may vary.

b i

25 a 1

22 a Answers may vary.

which has a probability of

c i T here would be 20 × 20 = 400 ﬁlled cells, as every combination of the 20 cards can occur.

would be , and since we put M back, the

21 a

these probabilities, we see that getting MY is more likely than UM.

probability of getting MY would be

. If we compare

c If we select with replacement, the probability of getting MY increases. The ﬁrst selection

iii

4 a 9

b 5

c 4

Answers mathspace.co

787

Fluency 5 a B

7 12

10 a

11 a

12 a

13 a

c 0.48

14 a

15 a

7 12

5 12

7 12

5 12

b 0.6

7 a

1 2

4 7

3 7

1 2 2 3

1 3

b 6 a 0.64

16 a

17 a

e c Reasoning

8 a B 1 9 1 3

3 8

5 8

1 8

5 9 R

18 a

B 2 8

iii

v 0

vii

5 8

1 8

B 3 8

4 5

1 5

4 5 Miss

788

G P

9 a 1 5

4 5

19 Making a selection without replacement means that the number of remaining items will decrease by one. If non-simpliﬁed fractions are used, the denominator of the fractions for the probabilities will decrease by one at the next stage. Depending on what type of item was selected, the numerator of the fraction may also decrease by one, or may stay the same.

4 8

b i

b No, the probabilities would change because there would be fewer black cards, and fewer cards in total, for the next draw.

1 5

Hit 4 5 Miss

4 5

Hit 4 5 Miss

1 5

Hit Miss

20 They are both correct. The sum of the probability on any set of branches must be 1, so adding up the P(GB) + P(GG) is the same as doing P(Gardener large prize) × 1. We can see this if we do: P (Gardener large prize)

Hit

Multiply across, add down

Miss 1 5 1 5

4 5

Mathspace Victorian Curriculum – Year 9 mathspace.co

Hit Miss Hit Miss

Common factor of Add fractions

b Section A area is 14.29 m2 while Section B area is 85.71 m2

21 There are 4 numbers on a die less than 5. The probability of rolling a number less than 5 on any die is .

Chapter 10 review

Probability=

1 a 10 times

c The experimental probability says that rolling a 4 is twice as likely as expected with

Problem Solving 22 a

theoretical probability, . We expected to see a four 10 times, but it only appeared 4 times. The data so far suggests that the dice is not fair. It is also possible that this high number of 4s is random and the die is fair.

23 a b i 38% 24 a 0.05

ii 9%

iii 28%

b 0.6

c 0.15

25 a

–1 2

H –1 2

Relative frequency 43% 40%

This is close to:

Theoretical probability

Makes

b i 0.6 0.4

0.8

Makes

0.2

Misses This means she may have labelled the 12 sides with ﬁve 1s, ﬁve 2s, one 3, and one 4.

c 0.08

3 4

Misses

b 0.4 27 a

ii 22.17%

3 a Adding the results of all of her friends. The relative frequencies are:

–1 8

3 – 8

26 a

b i 12%

–1 2 T

2 a 600

3 – 8

–1 2

d 0.2

–1 8

b 1249.2 mm2

4 a 18

b 100%

c 3 b

5 a

Has been Has not been Total out of state out of state

28 33% 29 a A x Garden 1−x B

3 5

Rose

2 5

Daisy

1 4

Rose

3 4

Daisy

The probability of drawing a ﬂower from Section A is represented by x, and from Section B is 1 − x.

Pet

No pet

Total

b i 6

Australian resident

iii

Likes yeast spread

3 16

Answers mathspace.co

789

11 a

Likes yeast spread

Does not like yeast spread

Total

Australian resident

Not Australian resident

Total

7 a i P (1) =

Sample Space 5 35 6 36 8 38

ii P (less than 2) =

iii P (more than 1) = b P (less than 2) + P (more than 1) = These two events are complementary, because less than 2 includes 0 and 1, while more than 1 includes 2 and 3. This makes up the whole sample space, so would have probability of 100%. 8 a

c i 0

4 10 B

790

Mathspace Victorian Curriculum – Year 9 mathspace.co

63 65 68

3 5 6

83 85 86

6 10

Sample Space C CC

4 10

6 10

4 10

2 10

6 10 3 10

b 13 a

3 11

2 11

2 10

1 10

6 10 3 10

6 11

c True. The ﬁrst step would have 6 branches for the ﬁrst die. The second step would split each of the 6 branches into 6 more branches, making 6 × 6 = 36 branches.

{SUT, STU, UST, UTS, TSU, TUS}

6 10

means P (Even) = and P (Odd) = , so an even product is more likely.

b Let S represent the star sticker, U represent the unicorn sticker, and T represent the tractor sticker.

3 5 8

12 a

b True. We can only make an odd product by multiplying two odd numbers. An even product can result from the product of two evens or an even and an odd. There are 27 even products and 9 odd products. This

10 a Without replacement

53 56 58

b {35, 36, 38, 53, 56, 58, 63, 65, 68, 83, 85, 86}

9 a False. There are 10 outcomes less than or equal to 5 and 26 outcomes greater 5. Since these are not the same, their probabilities would be different, so are not equally likely.

3 6 8

2 10

5 10

b i

iii

B W

Sample Space BB BW

14 a Hit 1 6

Miss Hit

1 6 Miss

1 6 1 6

5 6

Hit

1 6

Miss

Miss Hit

Substitute

Miss

Add

1 6

5 6

So P (one four) =

5 6

Miss Hit

Simplify

17 a

–1 5

4 – 5

1−x

A –1 3

Miss

15 a i

iii

2 – 3 B

c $25.33

b i

iii

11.01 Types of data Understanding

16 Less than two 4s, means one or none.

1 a False

2 A census involves collecting data from an entire population, while taking a sample means collecting data from a group or subset from within the population.

Probability of not rolling a four is Multiply

To find P (one four), we can find the probability of the one being in each possible spot, first, second, third, fourth, or fifth and add them all together. Probability of one four is Multiply

b True

c True

d False

3 Answers vary, but may include: • Use random sampling: Ensure that the sample is representative of the entire population by selecting participants randomly. • Design neutral survey questions: Avoid leading or loaded questions that suggest a particular answer. Questions should be worded neutrally to avoid influencing the respondent’s answers.

• Ensure anonymity and confidentiality: Allow respondents to remain anonymous and Probability assure them that their responses will be of one four confidential. is Multiply

This probability can be repeated 3 more times, for a four in the third, fourth and ﬁfth spots. The probability of each of these will also be

Fluency 4 a Numerical

b Categorical

c Categorical

d Categorical

e Categorical

f Categorical

Answers mathspace.co

791

g Numerical

h Numerical

i Categorical

j Categorical

k Numerical

l Categorical

5 a Nominal

b Nominal

c Ordinal

d Nominal

e Ordinal

f Ordinal

g Ordinal

h Nominal

6 a Discrete

b Continuous d Continuous

e Discrete

f Discrete

7 a Continuous

d The question uses emotive or leading language as it suggests social media is time wasting.

b Discrete numerical c Categorical nominal

Question should be: “How much time do you spend on social media each day?”

d Categorical nominal

e The question makes a false assumption that a person uses shoe polish.

e Discrete numerical f Continuous numerical g Continuous numerical c Biased

d Biased

9 a Leading Question

b Loaded Question

c Leading Question

d Leading Question

10 a Not biased, because the sample size is small and the sample is not representative of the population. b Biased, because the group of people being asked are already users of public transport and so will beneﬁt if the government spends more money on the public transportation. c Biased, because the question is leading and contains negative information about shark nets. 11 a Yes

b Yes

c No

12 a No

b Yes

c Yes

13 a Emotional language b Question is unclear: how large is “large”? c This question is too personal. Individuals might not be comfortable discussing this openly. Reasoning 14 a The question uses emotive or leading language by including the Prime Minister’s beliefs.

792

c The question makes a false assumption that the person has a sibling and that the sibling can drive. Question should be: “Do you have a sibling? If yes, how often do they drive you to school?”

g Continuous

b Fair

b The question asks more than one question and makes a false assumption that a person has one of the accounts. Question should be: “Do you have a Snapchat account?” and “Do you have a Twitter account?”

c Continuous

8 a Fair

Question should be: “Do you think taxes are too high, too low, or just right?”

Mathspace Victorian Curriculum – Year 9 mathspace.co

Question should be: “Do you use shoe polish? If yes, what brand do you use most often?” 15 Collecting both categorical and numerical data in a survey about dietary habits can provide a more comprehensive understanding of the habits. Categorical data can help identify types of food or meals preferred, while numerical data can indicate quantities or frequencies, providing insight into the extent of these habits. 16 To understand commuting habits, they should collect both categorical and numerical data. Categorical data might include types of transport used (bus, train, car, bicycle, etc.), while numerical data could include distance travelled, time taken, and frequency of travel. 17 The reliability of this study is questionable due to the small sample size and the sample not being representative of all students (it only includes students from a single school). To make a claim about all students, a larger, more diverse sample would be needed. Problem-solving 18 a Country b No, because the question uses emotive or leading language and suggests that country music is cool. It also asks two questions

rather than one, as it’s possible to like both types of music (or neither). 19 a I would conduct the survey during lunch breaks in common areas like the cafeteria, ensuring a diverse sample by including students from different grades and activities. The questionnaire would list various after-school activities for students to select their favourites, providing insights into the most popular options. b The data collected would be categorical (nominal) as it involves the classiﬁcation of activities into categories. c To ensure that your data collection methods are unbiased and accurate, you could use stratiﬁed sampling by selecting a representative sample of students from each grade level, ensuring that the survey is anonymous, and encouraging honest responses.

11.02 Data displays Understanding 1 A frequency table would help when dealing with a large data set. It would help to reduce the time taken to add all the scores as well as reduce the possibility of errors due to the amount of scores. 2 Stem-and-leaf plots are best used for smaller sets of data that are integers or decimal numbers. For example, student test scores, ages of people, etc. 3 By counting the number of dots above a particular score Fluency 4 a

20 a To study the relationship between hours of sleep and academic performance, you would need to collect numerical data on the number of hours of sleep per night for each student and their academic performance (e.g., exam scores, GPA). b You could collect this data by using selfreport questionnaires for hours of sleep and obtaining academic performance data from the school records.

21 a The clothing company should collect data on customer preferences, purchasing habits, frequency of visits to their store or website, and average spending per visit. They could also collect demographic data such as age, gender, and location. b This data could be used to tailor marketing campaigns, optimise inventory and product selection, and create targeted promotions to improve sales. 22 a b This case might lead to bias because the survey is not truly representative of the entire city’s population.

Score (x)

Frequency ( f )

Cumulative frequency

32 40

Total

Score (x)

Frequency ( f )

Cumulative frequency

105

48 156

125

Total

Score (x)

Frequency ( f )

120 445

Cumulative frequency

119

4046

125

244

4500

160

404

6080

143

547

5720

185

732

7770

137

869

6028

Total

869

34 144

Answers mathspace.co

793

Score (x)

Frequency ( f )

Cumulative frequency

150

1800

160

2400

170

1870

180

1800

190

3040

200

2800

Total

5 a 8

Frequency ii 3

iii 1

Frequency

b i

20 18 16 14 12 10 8 6 4 2 0

iv 1.09

v 1

2 0

0 1 2 3 Number of siblings

7 a

Score Frequency

58 61

81 93

b 21

c 5 students

d 4 students

e 73.8 b 193

c 164 beats per minute 10 a

1 2 3 Number of pets 3

iii 4

Number of students

b 178 beats per minute

iv 3.85

b The data is bi-modal has modes of 49 and 52 and a range of 5. The least frequent score is 50. The median is 49 and the mean is 49.625

v 4

11 a

8 6

Number of websites visited daily(x)

Frequency Cumulative ( f ) frequency

ii 3

9 a 179 beats per minute

c i

8 a 174

Number of students

c 250 and 280

Number of bedrooms

ii 2

794

13 710

b 1500

6 a i Number of pets

vi Number of students

3 4 5 Number of bedrooms

Number of siblings

Number of students

iii 1

iv 1.6

v 2

Mathspace Victorian Curriculum – Year 9 mathspace.co

b 3.36

Total

94 c 4

12 a

Score

Frequency

b 20 13 a 201

c 17

d 10

b 10.58

c 9

of 15, 15 will become the new mode, as it now occurs more frequently than any other value.

e 27.84

• The median is the middle value when the data is ordered. Adding three new values of 15 will shift the middle of the dataset, likely bringing the median closer to 15. The exact median will depend on how the numbers are arranged, but it will shift towards the middle of the new dataset.

14 a Percussion

b 23 people

Problem-solving

c 3 people

d 6 people

19 a 126 ng/L

Reasoning 15 A frequency table would help to see the most common number of hours spent on homework (mode). You could quickly calculate the mean and median with less chance of making an error. 16 When calculating the “average”, Alison found the mean, while Sarah found the median. Both answers are correct as measures of centre, however the average is usually taken to be the mean, so this would mean that Alison is correct. 17 a False, a dot plot only shows the results from one variable not two. b True c False, while the mode can be easily determined from the highest peak in the dot plot, the mean and median can’t be directly observed but can be estimated by looking at the general distribution of the data.

c 105 ng/L

ii 37.5%

20 a

Responses

b i 5

ii 17

iii 18

c 6 d The range will increase while the mode and median remains the same. The new range is 28, new mode is unchanged and is still 17 and new median is also unchanged at 18. 21 a 21

b 4

c 10

22 a 5 Total number Number of of paintings artists completed

d False, the range can be determined by identifying the maximum and minimum values on the dot plot. e False, dot plots can be useful for both small and large data sets. 18 • The range is the difference between the highest and lowest values. Since the new values (15, 15, 15) are within the existing range (10 to 28), the range will remain the same. • The mode is the value that appears most frequently. Before adding the new values, the mode was 11. After adding three occurrences

b 17 places

d i 15

b 5

Cumulative frequency

Total

107 c 4.28

23 1

Answers mathspace.co

795

11.03 Back-to-back stem and leaf

b Both groups have the same median, 24 books.

Understanding

c Both groups

1 2

10 a Yes c Neither

2 18, 34, 40

11 a The lowest score for Hotel A is 3 desserts.

3 Answers vary, but may include:

b Hotel B

• Displaying distributions • Comparing central tendencies and variability • Drawing conclusions

History

b $36 d Teachers

13 Visually, Class A clearly has higher test scores overall.

Math 7

Key: 5|7 = 57

Oliver

Valentina 0

14 • Adal’s results: 14, 15, 17, 17, 18, 21, 24, 26, 27, 28 Mean: 20.7 • Dado’s results: 7, 10 15, 18, 21, 22, 23, 26, 29, 30 Mean: 20.1

Key: 4|1 = 14

Hence, Adal appear to be the best performer because his mean result is higher.

Problem Solving 15 a i 45

Key: 1|2 = 12

6 a 59

b 67

c Wildcats

7 a Alina

b Layla

c Yes

8 a 70.3 kg

The mean for Class A is approximately 78.1. The mean for Class B is approximately 58.8. The median for Class A is 80. The median for Class B is 60.5. When you compare the mean and median for both classes, we notice that the measures of centre are higher for Class A, than for Class B.

Key: 5|5 = 55

b i 45.5

ii 40.5 ii 43.8

c i No, as it will be their scores’ outlier if added. d Yes

b 62.9 kg

c • Under 30 years of age: 53, 56, 57, 61, 61, 63, 66, 68, 71, 73 Mean: 62.9 • Under 30 years and over: 62, 63, 63, 68, 69, 71, 72, 74, 78, 83 Mean: 70.3 Hence, the group “30 years and over” appear to be heavier because their mean weight is higher. 9 a The lowest score for University students is 7 books.

796

c Neither

Reasoning

12 a Teachers c $42

Fluency 4

b Dylan’s

Mathspace Victorian Curriculum – Year 9 mathspace.co

ii Since there are 11 scores now, the new median will move at the 6th position which is 42. The new mean will be which makes the new Sri Lanka’s mean higher now than India’s mean (45.5). iii First median is 40.5 while the new median is 42 which is still less than India’s median. First mean is 43.8 while the new mean is 48.45 which is now higher than India’s mean. So the measure of centre that was least affected by adding the outlier is the median.

16 a 87

b 95

c 73.86

b 50%

c 25%

d 79.13

8 a

e 76.67 17 a $109

Score (x)

. The probability of selecting a player

2 1

50–59

Score (x)

Frequency ( f )

30–39

40–49

The coach’s team, Team Amazing has a higher probability of selecting a player with a score that is greater than 70. 19 a 50%

b 44%

c Summer challenge

Understanding Frequency

30–39

40–49

2 8

70–79

80–89

Score (x)

11.04 Grouped histograms

Class

50–59 60–69

9 a 20–29, 30–39, 40–49, 50–59 and 60–69

d 46 books

e 39 books

3 4

30–39

with a score of greater than 70 from Team Amazing is

10–19 20–29 40–49

18 a b The probability of selecting a player with a score of greater than 70 from Team Amazing

Frequency ( f )

20–29

30–39

40–49

50–59

60–69

10 a 10–19, 20–29, 30–39, 40–49, 50–59, 60–69 and 70–79

50–59

60–69

70–79

10–19

80–89

20–29

2 a Ungrouped

b Grouped

c Ungrouped

d Grouped

3 a False

b False

c True

Score (x)

d True

Fluency

Frequency ( f )

30–39

40–49

50–59

60–69

70–79

11 a Number of hours Class

Class centre

1–9

10–18

19–27

28–36

37–45

15 10 5 0

b 23 5 30–39 6 a 9.3

b 1–5

7 a 9.9

b 1≤x<6

b 5–9

spent on studying

20 Frequency

4 a

0 5 10 15 20 25 Hours spent on studying

Answers mathspace.co

797

b 0

12 16 4 8 Number of faults

13 a 7

b 4

14 19 < height ≤ 28 15 a 140.9 cents c

b 11 days

7 6

Frequency

3 2 1 120.9 125.9 130.9 135.9 140.9 Price (in cents per litre)

d 140.9 < p ≤ 145.9 16 a 25 students

b 13 students

c 16%

Frequency

17 a

20 18 16 14 12 10 8 6 4 2 0

5 10 15 20 25 Number of books

b i Y es, the highest frequency is in the range 5 ≤ x < 10, indicating that the majority of readers fall in this category. ii Yes, the sum of frequencies in the ranges 0 ≤ x < 5 and 5 ≤ x < 10 is greater than the sum of frequencies in the ranges 10 ≤ x < 15 and 15 ≤ x < 20, indicating that most readers read less than 15 books.

20 a Salvia’s is more useful, as the class intervals are smaller so we can describe the shape of the data with more accuracy and identify trends in more detail. b No, Lee is not correct as the histogram shows that 51 patients lived between 21–30 years. In fact, by using Salvia’s histogram, we can see that 0 patients lived 30 years after being diagnosed.

21 a

0 20 40 60 80 100 120 Price ($)

20 18 16 14 12 10 8 6 4 2 0 20 40 60 80 100 Commute time (minutes)

Reasoning b 60 mm ≤ x < 80 mm

c Yes d No. The table shows ranges of lengths and not individual leaves. We do not know if any of the leaves in the range 0 ≤ x < 20 are less than 5 mm.

798

Problem-solving

b 20–39 and 80–99

18 a 35

Number of books read

iii Yes, the frequency in the range 20 ≤ x < 25 is only 2, indicating that a small number of readers read 20 or more books.

19 a

Frequency

20 18 16 14 12 10 8 6 4 2 0

Frequency

12 a Faulty machinery

Mathspace Victorian Curriculum – Year 9 mathspace.co

b i No

ii Yes

iii No

iv No

22 a

Speed (km/h)

Fluency

Frequency

110 ≤ x < 120

120 ≤ x < 130

c Symmetrical

d Positively skewed

6 a Negatively skewed

130 ≤ x < 140

e Positively skewed

f Negatively skewed

140 ≤ x < 150

150 ≤ x < 160

g Positively skewed

h Symmetrical

Total

b 51.6% 23 a

b Positively skewed

Battery life (minutes)

Frequency

400 ≤ x < 500

500 ≤ x < 600

600 ≤ x < 700

700 ≤ x < 800

7 a Unimodal

b Bimodal

c Unimodal

d Unimodal

8 a i 21 years

ii 11 years

iii 30 years

iv 23.32 years

b Positively skewed 9 a No

b 26 and 36

10 a Yes, 30s − 40s

b 48

c Positively skewed

800 ≤ x < 900

900 ≤ x < 1000

11 a Multimodal but not bimodal

1000 ≤ x < 1100

b 4% − 5% and 8% − 9%

1100 ≤ x < 1200

1200 ≤ x < 1300

Total

117

b 89.7%

c 9.4%

12 The data is clustered around 80-89 and around 100-109. 13 a Yes 14 a Yes

11.05 Shape of data Understanding 1 Possible answers:

b 1 to 4

c 3

d Positively skewed b 9 − 10

d Negatively skewed 15 a Yes, 45 − 49

b 47

c Negatively skewed

• The data is evenly distributed around the mean. It means that the left side of the distribution (less than the mean) mirrors the right side (greater than the mean).

16 a Grouped data

• The left side of the distribution (less than the mean) mirrors the right side (greater than the mean).

18 a No

• The mean, median and mode are very similar. 2 Positive skew refers to a distribution where the tail is on the right side of the distribution. Negative skew refers to a distribution where the tail is on the left side of the distribution. 3 Clusters within a data set can be identiﬁed by looking at areas where data points are concentrated. These are typically groups of data points that are relatively close together compared to the rest of the data set. 4 Bimodal

c 9

b 151.5 cm

c Negatively skewed 17 Positively skewed b 15 and 21

c Symmetrical

Reasoning 19 In a positively skewed distribution, the mean is typically greater than the median, which is greater than the mode. This is because the mean is affected by the long tail of high values, pulling it further to the right. 20 C lusters in the age data collected across the 5 suburbs might form due to common characteristics or attributes shared by people within each suburb. For example, one suburb might have a cluster of younger people due to the presence of schools and playgrounds, while another suburb might have a cluster of seniors

5 Greater than 60

Answers mathspace.co

799

due to retirement communities or more affordable housing for older adults.

c 90, which was obtained by Alice. d 72, which was obtained by Bob.

Problem-solving

9 a Team X

21 The heavy right skew in the sales data suggests that most salespeople have low sales numbers, while a few have very high sales numbers.

b i 12.3

ii 12.1

10 a i 72

ii 81

22 The teacher might consider that the symmetrical distribution indicates a more consistent performance across students, while the positively skewed distribution suggests that most students scored lower, with a few high achievers. The teacher should also consider using measures of centre and spread to compare the two classes more accurately.

11.06 Compare data sets Understanding 1 We can compare data sets using: • Measures of central tendency (mean, median and mode) • Measure of spread (range) • Shape of the data 2 Yes

ii 73

b City B 12 a 10.3 cm b 10.1 cm e Plant A

c 9 cm

d 3 cm

f Plant B

13 a i 36

ii 30

b i 45.5

ii 43.8

c Team A is quicker if we use the average running time to compare the two teams. This can be supported by the answer in part b. 14 a i 34

ii 38

b i 27.7

ii 30.1

c Adelaide is warmer if we use the average temperature to compare the two cities. This can be supported by the answer in part b. 15 a Yes

b Learning Tree

c Yes

16 a 4

b 3

c 5

d 6

c 28

d 27

17 a 77.69

4 Answers may vary. • Reveals patterns: Graphs show relationships between variables, enhancing comparisons. • Simplify complex data: Graphs break down large data sets into a more understandable format, aiding comparisons. • Quick processing: Graphs help visualise data trends quickly, making comparisons easier. Fluency 5 a 21 minutes

b 25.7 minutes

c Primary students 6 a i False

ii False

b Yes

c Yes

iii True

Reasoning 18 Looking at the data, although the ranges are identical (both are 7 kg). The mean for Group 1 is 58.3 kg. The mean for Group 2 is 63.4 kg. The mean values are quite different. Group 2 has a noticeably higher average weight than Group 1. This difference in mean weights might suggest that these two groups of turtles are not of the same species. However, it’s important to note that the determination of species is based on multiple factors, not just weight. Additional data and biological information would be needed to conclusively determine species. 19 a

b 17

c i 3.4

ii 3.2

8 a i 86.25

ii 78

b 75.69

e Group 1

Here are some of the reasons:

b 82.125

800

11 a i 120

e Team X f Team Y

3 Graph 1

7 a 7

b Class B

Mathspace Victorian Curriculum – Year 9 mathspace.co

Mean

Median

Mode

Range

Class 1

2.7

Class 2

8.2

for Team Y. Team Y has a more consistent performance than Team X. Team X has a slightly more diverse range of successful free throws for each player.

b Class 2, because their mean, median and mode were higher. 20 Company Alpha, because the mean is higher than the median, which means the higher salaries must be very large. 21 a

Chapter 11 review

Department A

Department B

1 a Continuous numerical

Mean

35.71

51.43

b Discrete numerical

Median

c Discrete numerical

Mode

35, 40

45, 50, 55

d Continuous numerical

Range

b Department B seems to have older employees as the mean, median and mode ages are all higher for Department B. Problem-solving 22 a Positively skewed

b Location A

2 a No

b Product B 24 a Positively skewed

ii 2.5 b Barry’s Burgers d Range

e Frank’s Fries

f Median

25 a Person A

b Range

c Person A

d Mean

26 a Symmetrical

b 20.17%

c 13.2

d 23.8

e Pens because the daily mean is higher.

Goals scored

Frequency

27 a Team X lowest: 6

Team X highest: 10

Team Y lowest: 7 Team Y highest: 10 b Team X: 7.7 Team Y: 8.4 c Team X: 7.5 Team Y: 8 d

= 0.5

= 0.8

f T he mean and median is consistently higher for Team Y. The probablity that a randomly selected player will have made 8 or more successful free throws, is signiﬁcantly higher

d No

b Answers may vary. One possible answer: The clinic could test the new method with a mix of different patients, make sure they compare it properly to the old methods, and double-check all the details they write down for mistakes. It’s also important to see if anything else might be making the patients feel better.

c Product A

c Frank’s Fries

c Yes

3 a Answers may vary. One possible answer: The clinic should note down how long (in days) it takes for patients to feel better using the new method compared to the old ways. They might also ask patients to rate how they felt during the treatment.

c Location B 23 a i 3

b No

b Mean: 2.62 goals, Median: 3 goals, Mode: 1 and 4 goals, Range: 5 goals c The mean represents the average number of goals scored and is typically where the dots in the dot plot are most densely clustered. The median is the middle value, so on the dot plot, it would be where the number of dots on the left and right are equal. The mode represents the most frequent score. On the dot plot, it appears as the column with the most dots or the highest peak.

Answers mathspace.co

801

5 a

Score (x)

Frequency ( f)

Cumulative frequency (cf )

20–24

25–29

30–34

35–39

40–44

45–49

50–54

11 a No

b Yes, 1–4

c 3

d Positive skew/Right skew

12 a Section 1

b Section 2

c They have the same range. d The mean is equal to the median. 13 a 24.47 years

b 35 years

c Positively skewed

b 96

14 a 115

c Positively skewed

b i 30

ii 37

d 34

c i 65.6

ii 106.1

d The range of pulse rates and the mean pulse rate increase after exercise.

6 a Ace Rockets: 45.2 minutes Thunders: 43 minutes Ace Rockets

8 a

b Sirin’s

400 500 600 700 800 900 1000 1100 1200 1300 Battery Life (mins)

c Neither b 89.7%

Frequency

110 ≤ s < 120

4 12

130 ≤ s < 140

140 ≤ s < 150

150 ≤ s < 160

Total:

c 9.4%

16 The symmetric distribution for Labradors suggests consistent sales throughout the year, indicating steady popularity. The positive skew for Beagles suggests that they were less commonly sold, with occasional periods (or perhaps promotions) where they sold in higher numbers. 17 a Victoria: 3935.6, South Australia: 1296.8

b 51.6% Number of Calls

15 10

120 ≤ s < 130

10 9 8 7 6 5 4 3 2 1 0

25 20

Key: 6|2 = 62

Speed (km/h)

9 a

35 30

Key: 1|6 = 61 7 a Yes

15 a

Thunders

Frequency

b Victoria: 204 651.2, South Australia: 67 433.6 c Yes d 211 533.75 units

b 196 minutes

802

10 Symmetrical and bimodal

Mathspace Victorian Curriculum – Year 9 mathspace.co