Higher Tier Booklet

Year 11 Prompt Booklet Higher (A* – C grades) Name:

Grade C PROMPT sheet

C4 Round to one significant figure

C1 Understand & use proportionality  To increase a quantity by 5% Multiply the quantity by 1.05 (100+5 = 105)  To decrease a quantity by 5% Multiply the quantity by 0.95 (100–5) = 95

These all have ONE significant figure 300 80 2 0.7 0.05 0.003 C4 Estimate answers to calculations

C2 Calculate using proportional change To increase £240 by 15% (100+15 = 115) = 1.15 x £240 = £276 To decrease £240 by 15% (100-15 = 85) = 0.85 x £240 = £204

C2 Multiply & divide numbers 0-1  Multiply e.g. 0.2 x 0.4 Ignore decimal points & multiply numbers 2 x 4 = 8 Count the number of decimal places (2) The answer will have this many (2) 0.2 x 0.4 = 0.08 (2 decimal places)  Divide e.g. 8 ÷ 0.2 Multiply both by 10 80 ÷ 2 = 40

makes whole

C2 4 rules of fractions  Add & subtract Denominators must be the same  Multiply Multiply numerators; multiply denominators  Divide Invert fraction after ÷ Multiply numerators; multiply denominators

 Round each number to 1sf first e.g. 423 x 28 = 400 x 30 = 12000 = 20 568 600 600 e.g. 3.26 x 11.8 = 3 x 10 = 30 = 300 = 50 0. 58 0.6 0.6 6 e.g. 8.3 x 35.6 = 8 x 40 = 320 = 640 0.49 0.5 0.5 (÷0.5 = doubling the number being divided)

C5 Use a calculator efficiently Know your keys x2 x3 x

3

(-)

C6 Expand brackets and simplify Multiply everything inside the bracket by what is outside Then collect like terms together

3(x + 2) + 2(x – 5) =3x + 6 + 2x – 10 =5x - 4 Watch for the negative sign in front of the bracket It changes the sign inside the bracket

3(x + 2) - 2(x – 5) =3x + 6 - 2x + 10 =x + 16

C7 Draw a straight line graph 

To draw a graph of x + y = 7



Think of x and y coordinates that add to make 7

e.g. (4,3) (3,4) (2,5) (1,6) (0,7) (-1,8) ....

These are usually put into a table:



x y

-1 8

0 7

1 6

2 5

3 4

4 3



T

hen the

points are plotted and joined

y



To draw a graph of y = 2x -1

 

Some coordinates are usually given in a table You have to fill in the rest by following the rule of the equation ‘ whatever x is, multiply by 3 then -2’

x y

-3 -7

-2 -5

-1 -3

0 -1

1 1

2 3 2x2-1

2x0-1 2x-2-1 

Then the points are plotted and joined

3 5



To find the gradient of a line

 

The gradient of a line is its ‘slope’ It is measure by vertical ÷ horizontal

Gradient = 6 ÷ 4 = 1.5

C8 Solve inequalities in one variable a < b means a is less than b a ≤ b means a is less than or equal to b a > b means a is greater than b a ≥ b means a is greater than or equal to b

C9 Rearrange a formula

 e.g.

Inequalities can be treated like equations The solution can be shown on a number line e.g.1

l -1 e.g. 2

2x – 4 < 2 (+4 to each side) 2x < 6 (÷2 each side) x <3 l 0

l 1

l 3

l 4

l 0

l 1

l 2

e.g.

l 3

-7 ≤ 2x – 1 < 3 (+1 to each part) -6 ≤ 2x < 4 (÷2 each side) -3 ≤ x < 2

l l l -3 -2 -1

l 0

l 1

l 2

Once numbers have replaced letters:



Remember the order of operations BIDMAS Remember the rules for signs

- x -=+ - x +=-

3 7 11 15 19 23 .... +4 +4 +4 +4 +4

The term to term rule is +4 nth term = 4n -1 The nth term can be used to find the term in any position

e.g. 10th term means n=10 10th term = 4x10 – 1 = 39

C11 Plot quadratic functions 

Graphs of quadratic equations have x2 in and look like this:

l 3 4

C9 Substitute numbers into expressions



C10 Find the nth term of a linear sequence If the 1st difference is constant, it is linear

2x – 7 ≤ 5x + 2 (-2x each side) -7 ≤ 3x + 2 (-2 each side) -9 ≤ 3x (÷3 each side) -3 ≤ x (swap around) x ≥ -3 (swap inequality symbol)

l l l -3 -2 -1 e.g. 3

l 2

Use the same balancing steps as when you solve equations Make ‘t’ the new subject in: v = u + at (-u from each side) v–u= at (÷a each side) v–u= at a a t=v–u a

-- = + +- = -

or



To draw the graph of y = x2 + 4

 

Fill the table by following the rule Then join the points with a smooth curve

x y

-3 13

-2 8

(-2)2 + 4

-1 5

0 4

1 5

2 8 22 + 4

3 13

C12 Pythagoras Theorem

C14 Locus of point

For this right angled triangle:

LOCUS is the path or region a point covers as it moves according to a rule

hypotenuse a c



Fixed distance from a point – circle



Equal distance from two points perpendicular bisector



Equal distance from two intersecting lines – angle bisector



Perpendicular from a point to a line

b a = b2 + c2 2





If finding the hypotenuse ADD the squares of the other 2 sides Then square root If finding a shorter side SUBT the squares of the other 2 sides Then square root

C13 Find lengths, areas & volumes Formulae to learn: Area of rectangle = l x w

l w

Area of triangle = b x h 2

h b



P

Area of parallelogram = b x h h b Area of trapezium = ½(a + b)xh

a h

b Area of circle = π x r

2

r

Circumference = π x d d

Volume = Area of cross-section x length cross section length

C15 Bounds of measurement  

If 23cm is rounded to nearest whole cm 23 is between the whole numbers 22 and 24 l

l

l

22

23

24

22.5

23.5

lower bound

upper bound

C16 Compound Measures   

C18 Graphical representation

These triangles are useful Cover the quantity you are trying to find What is uncovered is the formula to use

D

S

Scatter diagrams – used to investigate correlation

e.g. Positive Correlation

M

T

D~Distance S~Speed T~Time

D

V

M~Mass D~Density V~Volume

Examples

Strong positive

Weak positive

If it shows correlation , draw a line of best fit on it Points which do not fit the trend are called OUTLIERS and should be ignored The line can be used to predict data Line of best fit

Speed = Distance Time

Time = Distance Speed

Distance = Speed x Time

C17 Plan a Statistical Enquiry         

Questions should be simple The answers need to be ‘yes or ‘no’ or a ‘number’ or from a choice of answers Tick boxes are useful Avoid responses open to interpretation Avoid leading questions Avoid open-ended questions Avoid biased questions Ensure the sample is large enough Ensure the sample will give valid results

Negative

No correlation

Frequency polygon – plot mid-points of bars & join

Histogram

Frequency polygon

C19 Estimate mean Time (t sec)

C22 Examine results of an enquiry Justify choice of presentation

x

f

fx

60 < t ≤ 70

65

12

780

70 < t ≤ 80

75

22

1650

80 < t ≤ 90

85

23

1955

90 < t ≤ 100

95

24

2280

100 < t ≤ 110

105

19

1995

∑f = 100 ∑fx = 8660

Mean = ∑fx = 8660 = 86.6sec ∑f

100

Modal class = 90 < t ≤ 100 (because this class interval has the largest frequency i.e. 24)

Median = ½ (100 + 1) th = 50.5th = 80 < t ≤ 90

C20 Compare distributions  

Compare an average using mean, median or mode.

Compare spread using the range (the higher the range, the bigger the spread of data)



Frequency polygons and stem & leaf diagrams are often used to compare 2 distributions on the same diagram

C21 Understand relative frequency This is the name given to an estimate of probability from an experiment or a survey Relative probability = No. times an outcome occurs Total number of trials

A scatter diagram would be used to find out if there is any correlation or relationship between two sets of data A frequency polygon would be used to compare two sets of data

Grade B PROMPT sheet B/1 Change recurring decimal to fraction If x = 0.4444444 10x = 4.4444444 9x = 4 x=4 9

If x = 0.54545 100x =54.545454 99x = 54 x = 54 99

B/2 Repeated percentage change To increase £12 by 5% per year for 4 yr = 1.054 x £12 To decrease £50 by 12% per year for 4 yr = 0.884 x £50 B/2 To find the original quantity ~If an amount has been increased by 5% Original amount = new amount ÷ 1.05 ~If an amount has been decreased by 12% Original amount = new amount ÷ 0.88 B/3 Standard Form ~ a x 10n a is between 1 & 10; n is an integer ~ When mult/div in standard form, work out number part separate from the power of 10 part e.g. 3x105 x 4x103 = 12 x108 = 1.2 x 109 ~ With a calculator use EXP or x10x B/4 Factorise a quadratic expression x2 – 3x – 4 = (x – 4)(x + 1) x2 – 25 = (x – 5)(x + 5) Difference of 2 squares

B/5 Expand 2 brackets  Use F O I L F L (x + 3)(x – 2) I O F O I L 2 x – 2x + 3x – 6 =x2 + x – 6

B/6 Change the subject of a formula  

Isolate the new subject Use balancing

Make c new subject Make x new subject f = 3c – 4 ax + bx = ay 3c – 4 = f (+4) x(a + b) = ay 3c = f + 4 (÷3) x = ay c =f+4 a+b 3

B/7 Evaluate algebraic formulae Rewrite the formula with numbers replacing letters 

WITH A CALCULATOR

Use fraction key

or

Use (-) key for negative numbers 

WITHOUT A CALCULATOR

Remember the rules for negative numbers

-+ = -- = +

-x+=-x-=+

B8 Solve simultaneous equations by an algebraic method  

  e.g.

Make the number of ys the same Add or subtract to eliminate the ys Same signs ~ subtract Different signs ~ add Find the value of x Substitute the value of x to find y 2x – 3y = 11 5x + 2y = 18

First plot the straight line.

Decide which side of the line to shade.

Leave the region required unshaded. e.g. x ≤3 y > –2 y < x

(x2) (x3)

4x – 6y = 22 15x + 6y = 54 Add the two equations to eliminate y 19x = 76 x = 4 Substitute x = 4 into one of the equations 5x + 2y = 18 5x4 + 2y = 18 20 + 2y = 18 2y = -2 y = -1

B8 Solve simultaneous equations graphically  

B/9 Represent inequalities graphically

Draw the graphs of the equations Find where they cross

R B/10 Identify graphs 

Learn the basic shapes of graphs

Linear graphs – straight line - equation in x Quadratic graph – parabola – equation in x2 Cubic graph – S—shape – equation in x3 Reciprocal graph – equation e.g y = 3 x

B/11 Effect of adding/multiplying by a constant on a graph Original graph y = x2 New equation y = x2 + 2 y = x2 - 2 y = 2x2 y= ½ x2

Solution is x = 2, y = 3

Change in graph Move up 2 Move down 2 Stretch from x-axis in ydirection – scale factor 2 Stretch from x-axis in ydirection – scale factor ½

B/12 Coordinates in 3D

B/14 Trigonometry

In 3D there are 3 axes, x, y and z The coordinates of a point are (x, y, z)

SOH

CAH

TOA

Example O S H

y 2

R

1 Q 1

O

1

A 2

3

x

2 z

4

C3

C

O H

T

A

EXAMPLES

S

P

A

B

sin x = 4 cos28o = x tan 28 = 5 5 5 x sin x = 0.8 x = 5xcos28 o x= 5 x = sin-1(0.8) x = 4.4 tan 28 x = 53.1o x = 9.4

On the grid the vertex P is (2, 1, 3) B/15 Difference between formulae for length, area and volume

B13 Similarity If one shape is an enlargement of the other, we say they are similar.  

Corresponding angles are equal Corresponding sides have proportional lengths

Example – these 2 triangles are similar

4.8m

6m x 8m

Scale factor = 6 ÷ 4.8 = 1.25 X = 8 ÷ 1.25 = 6.4cm N.B. Always draw the 2 triangles separately and the same way up – it is easier to spot the sides that correspond to each other

  

Numbers and π have no dimensions Length x length = area Length x length x length = volume

Example: 5abc + 3a2b (Ignore the numbers)  axbxc + axaxb  volume + volume  volume

B/16 Median, quartiles & interquartile range from cumulative frequency graph

B/18 Add or multiply two probabilities P(A or B) = p(A) + p(B)

200

P(A and B) = p(A) x p(B)

Cumulative frequency 160

Upper Quartile

120

If you get an answer to a probability question that is more than one, you have most certainly added instead of multiplied

Median

80

Lower quartile

B/19 Tree Diagrams

40

 When going along the branches.

MULTIPLY the probabilities

0 10

20

30

40

50

60

Mark

Median = 38 marks Upper quartile = 43 marks Lower quartile = 30 marks Interquartile range = 43 – 30 = 13 marks

 If more than one route is wanted, ADD the probabilities Example: The probability that Jane is late = 0.2 Day 1

B/16 Box plot LQ

M

Day 2 late – 0.2 x 0.2 = 0.04

UQ

0.2 late 0.8 0.2

B/17 Compare distributions.0000

  

Mean, median & mode compare size Range & interquartile range compare spread Distributions can be compared visually using a box plot

not – 0.2 x 0.8 = 0.16 late

0.8

late– 0.8 x 0.2 = 0.16 0.2 not late

0.8 not– 0.8x 0.8 = 0.64 late

To find the probability of late on only one day: day1 late

= =

&

day2 not late

0.16 0.32

OR

+

day1 & not late

0.16

day 2 late

Grade A/A* PROMPT sheet

 Rationalising the denominator This is the removing of a surd from the denominator of a fraction by multiplying both the numerator & the denominator by that surd

A/1 Use fractional & negative indices

In general:

a

Rules when working with indices:



ax x ay = a(x + y) a3 x a2 = a(3 + 2) = a5 23 x 22 = 2(5) = 32

ax ÷ ay = a(x - y) a7 ÷ a3 = a(7 - 3) = a4 37 ÷ 33 = 3(4) = 81

(ax)y = a (x y) (a3)2 = a6 (23)2 = 26 = 64

a0 = 1 y0 = 1 80 = 1

a-x = 1

ax/y = ( a )x

=

a2/5 = ( 5 a )2 322/5 = ( 5

1

A/2 Manipulate and simplify surds 25 is NOT a surd because it is exactly 5 3 is a surd because the answer is not exact A surd is an irrational number To simplify surds look for square number factors

25  3 = 5 3 Rules when working with surds:

b =

ab

3 x 15 =

45 =

9x5 = 9 x

5 =3 5

m a + n a = (m+n) a 2 5 + 3 5 = (2+3) 5 =5 5

a = b 72 = 20

x

12

(Multiply both top & bottom by √12)

4 3 2 3 = = 2 2

3

A/3 Upper & lower bounds

( a )x

a x

6

32 )2 =22

y



(Multiply both top & bottom by √b)

8

a-x/y =

75 =

b

12 12 6 12 12 = = = 12 2

a3



b a b = b

b

12

a

23

x

6

x

2-3 = 1 = 1

=

Example

y

a-3 = 1

=

b a

72 20

Square number =

36  2 4 5

=

6 2 2 5

=

Square number

e.g. if 1.8 is rounded to 1dp Upper bound = 1.8 + ½(0.1) = 1.85 Lower bound = 1.8 – ½(0.1) = 1.75

 Calculating using bounds Adding bounds Maximum = Upper + upper Minimum = Lower + lower Subtracting bounds Maximum = Upper - lower Minimum = Lower – upper Multiplying Maximum = Upper x upper Minimum = Lower x lower

a b

 If ‘a’ is rounded to nearest ‘x’ Upper bound = a + ½x Lower bound = a – ½x

3 2 5

Dividing Maximum = Upper ÷ lower Minimum = Lower ÷ upper

A/4 Direct and inverse proportion

A/6 Solve quadratic equations by formula

The symbol  means: ‘varies as’ or ‘is proportional to’

x = -b ± √b2 – 4ac 2a Example

Direct proportion If: y  x or y  x2 or y  x3 Formulae: y = kx or y = kx2 or y = kx3 Example 

y is directly proportional to x When y = 21, then x = 3

To solve:

3x2 + 4x – 2 = 0 a = 3 b = 4 c = -2

x = -b ± √b2 – 4ac 2a

(find value of k first by substituting these values)

y x

 y = kx

21= k x 3  k=7 y = 7x (Now this equation can be used to find y, given x)

Inverse proportion If: y  1 or y  1 or y  1 x x2 x3 Formulae: y = k or y = k or y = k 2 x x x3 Example 

a is inversely proportional to b When a = 12 and b = 4 a 1 a = k b b 12= k 4  k = 48  a = 48 b

A/5 Solve quadratic equation by factorising  Put equation in form ax2 + bx + c = 0 2x2 – 3x – 5=0  Factorise the left hand side (2x – 5)(x + 1) = 0  Equate each factor to zero 2x – 5 = 0 or x + 1 = 0 x = 2.5 or x = -1

x = -4 ±

√(-4)2 – 4(3)(-2) 2(3)

= -4± √16+24 6 = -4± √40 6 x = -4 +

√40 6

OR -4 - √40 6

x = 0.39(2dp) OR

-1.72 (2dp)

A/7 Solve quadratic equation by completing the square   

 

Make the coefficient of x2 a square 2x2 + 10x + 5 = 0 (mult by 2) 4x2 + 20x + 10 = 0 Add a number to both sides to make a perfect square 4x2 + 20x + 10 = 0 (Add 15) 4x2 + 20x +25 = 15 (2x + 5)2 = 15 Square root both sides 2x + 5 = ± √15 (-5 from both sides) 2x =-5 ± √15 x =-5 + √15 OR -5 - √15 2 2 x = -0.56 OR -4.44 (2dp)

A/8 Simplify algebraic fractions Adding & subtracting algebraic fractions Example 1

A/10 Solve simultaneous equations ~ one is a quadratic Rewrite the linear with one letter in terms of the other  Substitute the linear into the quadratic Example 

x+3+x–5 4 3

(common denominator is 12)

= 3(x + 3) + 4(x – 5) 12 12 = 3x + 9 + 4x – 20 12 = 7x – 11 12 Example 2 5 - 3 (common denominator is (x+1)(x+2) (x + 1) (x + 2) = 5(x + 2) – 3(x + 1) (x+1)(x+2) = 5x + 10 – 3x – 3 (x+1)(x+2) = 2x + 7 (x+1)(x+2)  Simplifying algebraic fractions Example 2x2 + 3x + 1 (factorise) x2 -3x – 4 = (2x + 1)(x + 1) (cancel) (x – 4)(x + 1) = (2x + 1) (x – 4)

x + y = 4 (find one letter in terms of the other  y=4-x 2 x + y2 = 40 (substitute y=4 –x) x2 + (4-x)2 = 40 (Expand (4-x)2) x2 + 16 – 8x + x2 = 40 2x2 – 8x + 16 = 40 (-40 from each side) 2x2 -8x – 24 = 0 (÷2 both sides) x2 – 4x – 12 = 0 (factorise) (x – 6)(x + 2) = 0 x = 6 or x = -2

A/10 Solve GRAPHICALLY simultaneous equations ~ one is a quadratic Draw the two graphs and find where they intersect Example y=2x2-4x-3 y=2x-1 

y 14 12 10 8 6

A/9 Solve equations with fractions

4 2

x + 4 = 1 Common denominator (2x-3)(x+1) 2x – 3 x+1 x(x+1)+ 4(2x-3) =1 (2x-3)(x+1) x2 + x + 8x - 12 =1 (2x-3)(x+1) x2 + 9x – 12 =1(2x-3)(x+1) x2 + 9x -12 = 2x2 –x -3 (-x2 from both sides) 2 9x -12 = x –x -3 (-9x from each side) 2 -12 = x –10x -3 (+12 to each side) 0 = x2 –10x + 9 (factorise) (x + 9)(x + 1) = 0 x = -9 or x = -1

–2

–1

0

1

2

3

4

x

–2 –4 –6

Solutions are x = -0.3 and x = 3.3 (points of intersection) 

Sometimes the equation has to be adapted~ rearrange the equation to solve so that the equation of the graph drawn is on the left.

On the right is the other equation to be drawn

A/11 Graph of Exponential function

A/13 Graphs of trigonometric functions

The graph of the exponential function is: y = ax Example y = 2x

LEARN THE SHAPES OF THE GRAPHS Graph of y=sin x

9 8 7 6 5 4 3 2 1 0 -4

-2

-1 ≤ sin x ≤ 1 0

2

4

Graph y = cos x

It has no maximum or minimum point It crosses the y-axis at (0,1) It never crosses the x-axis

A/12Graph of the circle The graph of a circle is of the form:

x2 + y2 = r2 where r is the radius and the centre is (0,0)

-1 ≤ cos x ≤ 1 Graph y = tan x

y 6 5 4 3 2 1

–6

–5

–4

–3

–2

–1

O

1

2

3

4

5

6

x

–1 –2 –3 –4 –5 –6

This a circle of radius 5 and a centre (0,0) The graph of this circle is



x2 + y2 = 52 x2 + y2 = 25

Tan x is undefined at 900, 2700 ....

Solutions to trigonometrical equations can be found on the calculator and by using the symmetry of these graphs Example: If sin x = 0.5 x = 300 , 1500 , (See the solutions on sin graph above or from calculator)

A/14 Transformation of functions f(x) means ‘a function of x’ e.g. f(x) = x2 – 4x + 1 f(3) means work out the value of f(x) when x = 3 e.g. f(3) = 32 – 4x3 + 1 = -2 In general for any graph y = f(x) these are the transformations

y = f(x) + a y = f(x+ a) y = -f(x) y = f(-x) y = af(x)

y = f(ax)

Translation 0 a Translation -a 0 Reflection in the x-axis Reflection in the y-axis Stretch from the x-axis Parallel to the y-axis Scale factor=a Stretch from the y-axis Parallel to the x-axis Scale factor= 1 a

A/15 Change the subject of a formula  The subject may only appear once Use balancing to isolate the new subject Example : To make ‘x’ the new subject A = k(x + 5) (multiply both sides by 3) 3  3A = k(x + 5) (Expand the bracket)

 The subject may appear twice Collect together all the terms containing the new subject & factorise to isolate it Example: to make ‘b’ the new subject a = 2 – 7b (multiply both sides by (b – 5) b-5 a(b – 5) = 2 – 7b (Expand the bracket) ab – 5a = 2 – 7b 7b + ab – 5a = 2

(+7b to both sides) (+5a to both sides) To leave terms in b together

7b + ab = 2 + 5a

(factorise the left side) To isolate b

b(7 + a) = 2 + 5a (÷(7 + a) both sides) (7 + a) (7 + a) b = 2 + 5a (7 + a)

A/16 Enlarge by a negative scale factor With a negative scale factor:  

The image is on the opposite side of the centre The image is also inverted

Example : Enlargement scale factor -2 about 0 y

 3A = kx + 5k (-5k from both sides) T

3A – 5k = kx (÷ k both sides) 3A – 5k = kx k k x = 3A – 5k k

O

x

A/17 Congruence 



Congruent shapes have the same size and shape, one will fit exactly over the other.

2 triangles are congruent if any of these 4 conditions are satisfied on each triangle

A/18 Similarity & enlargement 

For similar shapes when:

Length scale factor = k Area scale factor = k2 Volume scale factor = k3 Example

4cm

~The corresponding sides are equal ~ SSS

A

6cm

B

If height of A = 4cm & height of B = 6cm  Length scale factor = 6 ÷ 4 = 1.5 If surface area of A = 132cm2  Surface area of B = 132 x 1.52 = 297cm3 If volume of A = 120 cm3  Volume of B = 120 x 1.53 = 405cm3

~2 sides & the included angle are equal ~ SAS

A/19 Finding lengths & angles in 3D  

Identify the triangle in the 3D shape containing the unknown side/angle Use Pythagoras and trigonometry as appropriate

P ~2 angles & the corresponding side are equal ~ ASA

12 cm

4 cm 3 cm

Q

D

~Both triangles are right-angled, hypotenuses are equal and another pair of sides are equal ~ RHS

5 cm

B 10 cm

A 8 cm C

A/20 Sine Rule (non-right angled triangles) To find an angle use: sin A = sin B a b To find a side use: a b sin A = sin B

=

A/20 Cosine Rule (non-right angled triangles)

a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C

sin C c OR

c sin C

=

cos A = b2 + c2 – a2 2bc cos B = a2 + c2 – b2 2ac

Use SINE RULE when given:  two sides and a non-included angle  any two angles and one side Example: To find side b

cos C = a2 + b2 – c2 2ab

C

b

a=30cm

400

520

A

Example: To find angle C A

B b sin B b sin 520 b b

=

a sin C

=

30 sin 400

=

30 x sin 520 sin 400

Use COSINE RULE when given:  3 sides  2 sides and the included angle

c= 5cm

B

= 36.8 cm (1dp)

A 350 c= 7.1cm a= 9cm

sin C = sin A c a sin C = sin 350 7.1 9 sin C = sin 350 x 7.1 9 sin C = 0.4524.....

C = sin-1(0.4524.....)

C = 28.90(1dp)

a=8cm

C

cos C = a2 + b2 – c2 2ab 2 cos C = 8 + 62 – 52 2x8x6 cos C = 0.78125......

C = cos-1(0.78125......)

C = 38.60(1dp)

Example: To find angle C

B

b= 6cm

C

Example: To find side a A 1100 c= 4.2cm

B

a=?

b=3.7cm

C

a2 = b2 + c2 – 2bc cos A a2 = 3.72 + 4.22 – 2x3.7x4.2 cos1100 a2 = 41.96 a = 6.48(2dp)

A/20 Area of triangle –height not known Area = ½ ab sinC Area = ½ bc sinA Area = ½ ac sinB

VOLUME -FRUSTUM OF A CONE

H

H r

Example

R

A

R

CONE

=

0

38 a= 8cm

FRUSTUM OF CONE

Volume of frustum = Volume of whole cone – volume of cone removed

b=6cm

B

r

1 2 1 2 πR H πr h 3 3

C

A/21 Pyramid & Sphere – Surface Area

Area = ½ ab sinC = ½ x 8 x 6 x sin 380 = 14.8 cm2(1dp)

CURVED SURFACE AREA ~Curved surface area of a cylinder = 2πrh r

A/21 Pyramid & Sphere –Volume

h

VOLUME - PYRAMID Volume of Pyramid =

1 x area of cross-section x height 3

e.g. cone

~Curved surface of a cone = πrl

h h w r Volume =

1 x πr2h 3

l

Volume =

1 xlxwxh 3

h

l r

[NB To find ‘l’ use Pythagoras’ Theorem

VOLUME - SPHERE

l 2 = h 2 + r 2]

4 3 Volume of Sphere = πr 3

~Curved surface of a sphere = 4πr2 r r

A/22 Length of arc & area of sector

A/23 Circle properties

sector arc segment

r

The angle at the centre = 2 x the angle at the circumference

θ

The angle in a semi-circle is a right angle

Length of arc = θ x 2πr 3600

r θ

Angles in the same segment are equal

Opposite angles of a cyclic quadrilateral add up to 1800

Area of sector = θ x πr2 3600

A/22 Area of segment radius of circle=7cm

640

The perpendicular from the centre to a chord bisects the chord

Tangents from a point to a circle are equal

7cm Area of segment = area of sector – area of triangle = θ x πr2 - ½ ab sinθ 0 360 = 640 x π x 72 - ½ 7x7x sin640 3600 = 5.35cm2 (1dp)

The angle between a tangent and a chord is equal to the angle in the alternate segment

A/24 Vectors 

 Vector addition Adding graphically, the vectors go nose to tail

Vector notation

a=

This vector can be written as 3 2 B

or a or AB

3 2

b=

-2 2

C

b

a

a+b

B

A A vector has magnitude(length) & direction(shown by an arrow) Magnitude can be found by Pythagoras Theorem

a



AB = √32 + 22 = √13 =3.6

A The combination of these two vectors:

AB + BC = AC = a + b 

3 + -2 2 2

A parallel vector with same magnitude but opposite direction B

=

a

1 4

P

 Vector subtraction Adding graphically, the vectors go nose to tail

A Q

a=

Vector PQ is equal in length to AB but opposite in direction so we say:

3 2

b=

-2 2

PQ = -a

B 

-b

A parallel vector with same direction but different magnitude B

D

a

a A

C

a-b

The combination of these two vectors:

A

AB - BC = AC = a - b

C

3 - -2 2 2

Vector CD is twice (scalar 2) the magnitude but same direction so we say:

CD = 2a

A negative scalar would reverse the direction

=

5 0

AC is called the RESULTANT vector



The sum of vectors M B

P

Example

N

A

AB = AP + PM + MB The vector AB is equal to the sum of these vectors or it could be a different route: go via N

AB = AN + NB

start point

end point

An inspector wants to look at the work of a stratified sample of 70 of these students.

Language

Number of students

Greek

145

Spanish

121

German

198

French

186

Total

650

No. from Greek = 145 x 70 ≈ 16 650 No. from Spanish = 121 x 70 ≈ 13 650 No. from German = 198 x 70 ≈ 21 650

A/25 Sampling The sample is:  a small group of the population.  an adequate size  representative of the population Simple random sampling Everyone has an equal chance e.g. pick out names from a hat Systematic sampling Arranged in some sort of order e.g. pick out every 10th one on the list Stratified sampling Sample is divided into groups according to criteria These groups are called strata A simple random sample is taken from each group in proportion to its size using this formula: No from each group = Stratum size x Sample size Population

No. from French = 186 x 70 ≈ 20 650 This only tells us ‘how many’ to take - now take a random sample of this many from each language

A/26 Histograms

A/27 Probability – the ‘and’ ‘or’ rule

Class intervals are not equal Vertical axis is the frequency density The area of each bar not the height is the frequency Frequency = class width x frequency density Frequency density = frequency ÷ class width   

P(A or B) = p(A) + p(B) Use this addition rule to find the probability of either of two mutually exclusive events occurring e.g. p(a 3 on a dice or a 4 on a dice) =

To draw a histogram Calculate the frequency density Example

Age (x years)

Class width

f

Frequency density

0 < x  20

20

28

28÷20 = 1.4

20 < x  35

15

36

36÷15 = 2.4

35 < x  45

10

20

20÷10 = 2

45 < x  65

20

30

30÷20 = 1.5

1 1 2 + = 6 6 6

P(A and B) = p(A) x p(B) Use this multiplication rule to find the probability of either of both of two independent events occurring e.g. p(Head on a coin and a 6 on a dice) =

1 1 1 x = 2 6 12

Scale the frequency density axis up to 2.4 Draw in the bars to relevant heights & widths

A/28 Probability – Tree diagram for successive dependent events

To interpret a histogram

When events are dependent, the probability of the second event is called a conditional event because it is conditional on the outcome of the first event

20 Frequency density 16 (number of books 12 per £)

Example 2 milk and 8 dark chocolates in a box Kate chooses one and eats it. (ONLY 9 left now) She chooses a second one This can be shown on a tree diagram

8 4 0 0

5

10

15 20 25 30 Price (P) in pounds (£)

35

40

1st

NOTE: On the vertical axis each small square = 0.8

f = width x height

0<P5

5 x 8 = 40

5 < P  10

5 x 12 = 60

10 < P  20

10 x 5.6 = 56

20 < P  40

20 x 1.6 = 32

M

1 2

Price (P) in pounds (£)

2nd

2milk 8dark

M

10

MM= 2 x 1= 2 10 9 90

9 8 9

8 10

2

D

D

M

9 7 9

MD= 2 x 8 = 16 10 9 90

DM= 8 x 2 =16 10 9

D

90

DD= 8 x 7 = 56 10 9

90