MATH 8 by YOUG JI INTERNATIONAL SCHOOL

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Chapter 1 Special Products and factoring Lesson 1.1 Polynomials A polynomial is an algebraic expression that represents a sum of one or more terms containing whole number exponents on the variables. A polynomial with one term is called a monomial. A polynomial with two terms is called a binomial A polynomial with three terms is called a trinomial We can also classify polynomials using their degree. The degree of a monomial in one variable is the number of times the variable occurs as a factor in the monomial. For example: -5ab2c3 is of degree 1 in a , 2 in b and 3 in c. The degree of a monomial is the total number of times its variables occur as factors. Hence, the degree of -5ab2c3 is 1 + 2 + 3 or 6. The degree of nonzero constant monomial is 0. The degree of polynomial is the greatest of the degree of its terms. For example, in the polynomial a2b + a2b2 – ab a2b has degree 3, a2b2 has degree 4, and ab has degree 2 thus, the degree of a2b + a2b2 – ab is 4. Degree 1 2 3

Example 2x + 3y 2 3x + 2x + 1 4x3 – 2x2 + x – 5

Classification linear polynomial quadratic polynomial cubic polynomial

The degree of a term that has only one variable is the exponent of that variable. The degree of a polynomial that has only one variable is the highest power appearing in any of the terms. The degree of a term that has more than one variable is the sum of the exponents of the variables The degree of a polynomial in more than one variable is the highest sum of the exponents of the variables in any of the terms. The conventional way of writing a polynomial in one variable is to write the terms in order of descending (or increasing) degrees. We then say that the polynomial is written in descending order. When a polynomial is written with decreasing exponents, the coefficient of the term is called the leading coefficient.

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In some instances, it is convenient to write the polynomial in ascending (or increasing) order. In this case, we begin with the term with the lowest exponent. The terms that follow must have increasing degrees. Example: Descending order: 4x5 + 3x3 + 2x + 9 Ascending order: 9 + 2x + 3x3 + 4x5 In a polynomial, terms that are exactly alike or that are alike except for their numerical coefficients are called like terms. The following are examples of like terms: a2b, 7a2b, 9a2b The following examples are not like terms: ab, a2b, 3ab2 A polynomial containing two or more like terms can be simplified by adding these terms. This process is combining like terms. A polynomial is simplified when all like terms have been combined and only unlike terms remain. Like terms 4x and 7x -9 and 7 4a2b and -5a2b

unlike terms 4x and 3y -9 and 7b 4a2b and -5ab2

Combining like terms, such as 3x and 2x , involves an important property of numbers, the distributive property. For all numbers a,b and c, a(b + c) = ab + ac and (b+c)a = ba + ca The distributive property can be used to simplify an expression that contains like terms. 3x + 2x = (3 + 2)x = 5x Practice Exercise 1.1 A. Simplify by combining like terms 1. 5x + 9 – (6x – 7 ) 2. –(2x -3) -3 -2x 3. –(3a-2b) – 3(b – a) 4. 4a + 5b – 6a + 7b 5. 5x + 6y – 11 + 8x – 12y + 14 B. Give the degree of each polynomial 1. -5x + 10x3 2. 7 – 4x3 3. 2x5 + 5x3 – 7x4 4. -3x2 + 8 -4x 5. 8 – 6x + 2x3 C. Tell whether each expression is a polynomial. If it is a polynomial, identify it as Monomial, binomial, or trinomial. 1. -73 2. 2 + 5x2 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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3. x2 + 4xy + y3 4. -25a2b7 + 8c 5. 11 + 2y

Lesson 1.2 Product with Monomials When multiplying monomials, you will always often use the following laws of exponents: Laws of Exponents 1. Product Rule for Exponents: To multiply powers having the same base, keep the base and add the exponents. 2. Power Rule of Exponents: = To find the power of a power of a base, keep the base and multiply the exponents. 3. Power of the Product Rule: ( To find the power of a product, find the power of each factor then multiply the resulting powers. Example: a. b. c.

To multiply polynomial and monomial, multiply each term of the polynomial by the monomial. Example: Volume = length x width x height Length= 7x + 6 Width= 6x Height = 2x Volume =(7x+6) (6x)(2x) = (7x+6) 12x2 = 84x3 + 72 x2 Answer the following: 1. 2. 3. 4. 5.

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Lesson 1.3 Product of Two or More Polynomials FOIL method in finding the product of binomials. Example: (x2 – 5)(x2 + 7) =(F) first term = x2 x2 = x4 (I) Inner term =( -5)(x2) = -5x2 (O) Outer term =( x2 )(7) = 7x2 (L) last term = (-5)(7) = -35 = x4 + 2x – 35 The outer and inner products in the FOIL method are often like terms. Answer the following: 1. (m – 10n)(3m + 5n) 2. (12c – 9)(12c – 9) 3. ( )(

)

4. 5. (0.5 + 5y)(2x – 0.5)

Lesson 1.4 Division of Monomials Quotient Rule for exponents For every positive integer m and n, and x = when m> n; = when m<n; Example: Rule 1: if the exponent of the numerator is greater than that of the denominator = when m> n (x Rule 2: If the exponent of the numerator is less than that of the denominator = when m<n (x Rule 3: if the exponents of the numerator and the denominator are equal, the quotient is 1 = Practice:

when m = n

1) 2) 3) 4) 5) The area of a parallelogram is 84x 6 square feet. Represent the height of the parallelogram if its base is 14x3 feet. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Lesson5 Division of Polynomials by another Polynomials Rule: 1. Arrange the terms of both the dividend and the divisor in either ascending or descending order. 2. Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. 3. Multiply the entire divisor by the number obtained in step 2 and subtract the product from the dividend. The remainder will be the new dividend. 4. Divide the new dividend by the first term of the divisor as before, and continue to divide in this way until the dividend is exact or until the remainder is of a lower degree than the divisor. 5. If there is a remainder, write it over the divisor and add the fraction to the part of the quotient previously obtained. Example The area of a rectangular garden is x 2 + 3x – 40 square feet. If the length is x + 8 feet, find a binomial that represents the width. Solution The area of a rectangle is the product of its length and width A = lw W= = Use long division to solve for the width. X–5 X+8 x2 + 3x – 40 -(x2 + 8x) - 5x – 40 -(-5x – 40) 0 The width of the garden is x – 5 feet.

Practice 1. X2 + 8x + 7 ÷ x + 7 2. X2 – 4x – 45 ÷ x + 5 3. X2 – 2x – 48 ÷ x + 6

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Lesson 1.6 Special Products Perfect square binomial pattern (x + y)2 = x2 + 2xy + y2 ( x – y) = x2 – 2xy + y2 In words, the square of a Binomial is the square of the first term plus (or minus) twice the product of the two terms plus the square of the last term. When you use these special product patterns, remember that x and y can be numbers, variables, or even algebraic expressions. Trinomials of the form x2 + 2xy + y2 or x2 – 2xy + y2 are called perfect square trinomials because each is the result after squaring a binomial. Example Find the square of each binomial (4x + 3y)2 = (4x)2 + 2(4x)(3y) + (3y)2 = 16x2 + 24xy + 9y2 Cube of a binomial Pattern For all numbers x and y (x + y)3 = x3 + 3x2y + 3xy2 + y3 Example Find the volume of the cube shown at the right V = e3 V = (2 + 3x)3 V = 23 + 3(2)2(3x) + 3(2)(3x)2 + (3x)3 V = 8 + 36x + 54x2 + 27x3 Product of sum and difference pattern The sum of the two numbers x and y represented by x + y is multiplied by their difference represented by (x + y) ( x – y) = x2 – y2 In words, the product of the sum and difference of two numbers is the difference of the squares of those two numbers. Example (4t + 7)(4t – 7) = 16t2 – 49 (51)(49) = (50 + 1)( 50 – 1) = 502 – 12 = 2500 – 1 = 2499 Practice Multiply and simplify 1. (a + 4)3 2. (b - 2)3 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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3. (x + 4) ( x – 4) 4. ( 5x + 1)(5x – 1 ) 5. ( a + 6 )2

Lesson 1.7 Factoring Polynomials Recall that you can apply the distributive property to simplify expressions. Thus, 3(4x + 5) = 3(4x) + 3(5) = 12x + 15 Each term in the parentheses is multiplied by 3 to find the product. You can reverse the process to find the factors of the related product. The reverse process is called factoring. The greatest common factor or GCF of x2 and 2x is x. for the polynomial x2 + 2x, the factored form is x(x+2). Factoring and GCF Factoring is finding two or more factors (GCF) of two or more monomials is the common factor having the greatest numerical factor and with variables having the least degree. Thus, the term axn is the GCF of a polynomial if 1. a is the greatest integer that divides each of the coefficients of the polynomials, and 2. n is the smallest exponent of x in all the terms of the polynomial. Example: factor completely a. 5x + 10 = find the factor of each term of 5x + 10 Factors of 5x = (5)(x) Factors of 10 = (5)(2) The GCF is 5. Use the GCF to rewrite the polynomial 5x + 10 = 5(x) + 5(2) = 5 (x+2) use distributive property So, 5x + 10 = 5 (x+2) Practice: Factor each expression completely a) 25 + 45x b) 24a + 48b c) 4m2 12m – 15 d) 3a3b4 + 9ab3 + 12ab4 e) ab2c + a2b3c2 + a3bc3

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Lesson 1.8 Difference of Two Squares Difference of two squares If x and y are real numbers, variables, or algebraic expressions, then X2 – y2 = ( x + y ) ( x – y) In words, the difference of the squares of two terms I the product of the sum and difference of those terms. Example X2 – 9 = x2 - 32 (x + 3 )( x – 3) X2 – y2 ( x + y)( x – y) If the terms of a binomial have a common factor, first factor out the common factor. Then continue factoring . Example 28x3 – 7x = 7x(4x2 – 1 ) factor out 7x, the GCF 2 2 = 7x [(2x) – 1 ] express the second factor as the difference of two squares = 7x(2x + 1)(2x – 1) the factors are the sum and difference of the squared term Practice Factor each completely 1. a4 – b6 2. c2 – 81 3. 4h2 – 49 4. 1 – 25q4 5. 16r4 – 121

Lesson 1.9 Sum and difference of two cubes Sum and difference of two cubes Let x and y be real numbers, variables, or algebraic expressions. Factoring sum of two cubes x3 + y3 = ( x + y ) ( x2 – xy + y2) Factoring Difference of two cubes x3 – y3 = (x – y)(x2 + xy + y2) The sum of two cubes as the cube of a first quantity plus the cube of a Last quantity, we have the formula F3 + L3 = (F + L)(F2 – FL + L2) The difference of two cubes as the cubes as the cube of the first quantity minus the cube of the last quantity we have the formula F3 – L3 = ( F – L )( F2 – FL + L2) Example Factor completely 64 – p6 = 82 – (p3)2 = (8 – p3)(8 – p3) = ( 2 – p)(4 + 2p + p2)(2 + p)(4 -2p + p2) Show that 16 + 4p2 + p4 = (4 + 2p + p2) ( 4 – 2p + p2) YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Practice 1. m3 – 64 2. 27 + 8p3 3. 125 + 8q3 Lesson 1.10 Factoring Perfect Square Trinomials Rule Factoring perfect square trinomials X2 + 2xy + y2 = ( x + y)2 X2 – 2xy + y2 = ( x – y)2 Example Factor each completely X2 + 18x + 81 = x2 + 2(9x) + 92 = (x + 9)2 Practice Factor completely 1. a2 + 6a + 9 2. 1 + 8c + 16c2 3. 4d2 + 20d – 25

Lesson 1.11 Factoring Quadratic Trinomials x2 + bx + c Procedure in factoring quadratic trinomials 1. List all pairs of integers whose product is c. 2. Choose a pair, m and n, sum is b, that is m + n = b 3. The factorization of x2 + bx + c is x2 + bx + c = ( x + m)(x + n) 4. If there are no such integers m and n such that m + n = b, the trinomial cannot be factored and is called prime. Example X2 + 3x – 10 Solution Find two numbers whose product is -10 and whose sum is 3. A table is helpful when factoring a trinomial of the form x 2 + bx + c Factors of -10 sum 1,-10 -9 -1, 10 9 2, -5 -3 -2, 5 3 the numbers we need are -2 and 5 X2 + 3x – 10 = (x -2)(x +5) Practice Complete the factorization 1. X2 + 11x + 28 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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2. X2 + 12x + 35 3. X2 + 15x + 56

Lesson 1.12 Factoring Trinomials Points to remember in factoring ax2 + bx + c 1. If the terms of a trinomial do not have a common factor, then the terms of a binomial factor cannot have a common factor. 2. If the constant terms of a trinomial is a. Positive, the constant terms of the binomials have the same signs as the coefficient of x in the trinomial. b. Negative, the constant terms of the binomials have the opposite signs. Example 2x2 – 13x + 15 Solution The terms have no common factors. Because c is a positive and b is a negative, only the negative fctors of 15 need to be tried Factors of 2 factors of 15 1,2 -1,-15 -3, -5 Write the trial factors. Use FUIL to check the middle term. Trial factors middle term (x -1)(2x – 15) -15x – 2x = -17x (x -3)(2x – 5) -5x – 6x = -11x (2x -1)(x – 15) -30x – x = -31x (2x -3)(x – 5) -10x – 3x = -13x Thus, 2x2 – 13x + 15 = (2x – 3)(x – 5) Factoring ax2 + bx + c ( a 1) by grouping 1. Find ac 2. Find the factors of ac whose sum is b 3. Rewrite the middle term (bx) as a sum or difference using the factors I n step 2. 4. Factor by grouping Example Factor using the ac-product or factoring by grouping method 2x2 + 15x + 28 Solution: 1. Find ac: ac = (2)(28) = 56 2. Find the factors of 56 whose sum is the numerical coefficient of the middle term, 15 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Factors of 56 sum 1, 56 57 2, 28 30 4, 14 18 7, 8 15 this is the desired sum 2 3. Express the middle term of 2x + 15x + 28 as the sum of the factors in step 2. 2x2 + 15x + 28 = 2x2 + (7x + 8x) + 28 2x2 + 15x + 28 = 2x2 + 7x + 8x + 28 4. Factoring by grouping 2x2 + 7x + 8x + 28 = ( 2x2 + 7x) + ( 8x + 28) = x ( 2x + 7) + 4 (2x + 7) = (2x + 7) (x + 4) Thus, 2x2 + 15 + 28 = (2x + 7) (x + 4) Factoring by grouping can be used extensively for polynomial expressions containing four or more terms Rules 1. Group the terms that have a common monomial factor. There will usually be two terms. Sometimes the terms must be rearranged. 2. Factor out the common monomial factor from each group. 3. Factor out the remaining binomial factor, if there exists. Factoring a polynomial over the integers 1. Is there a common factor? If so, factor out the GCF 2. Is the Polynomial a binomial? If so, can it be factored by any one of the following forms? Difference of two squares: x2 – y2 = ( x + y ) ( x – y) Sum of two cubes : x3 + y3 = ( x + y) ( x2 – xy + y2) Differences of two cubes: x3 – y3 = (x – y)(x2 + xy + y2) 3. Is the polynomial a trinomial? If it is a perfect square trinomial , use any of the following forms: X2 + 2xy + y2 = (x + y)2 X2 – 2xy + y2 = ( x –y )2 If it is not a perfect square trinomial, use trial and error or by grouping Does the polynomial contain four or more terms? If so, try to factor by grouping. Example 3x3 + 15x2 + x + 5 Group the first two terms and then the last two terms 3x3 + 15x2 + x + 5 = (3x3 + 15x2) + ( x + 5) = 3x2(x + 5) + (x + 5) = ( x + 5)( 3x2 + 1)

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Practice 1. 2x2 – 7x – 4 2. 5x2 + x – 18 3. 4x2 + 4x – 15 4. 3x2 – 5x – 2 5. 9x2 – 47x + 44

Chapter 2 Rational expression Lesson 2.1 Rational Expressions

A rational expression in one variable is an expression that can be written in the form where P and Q are polynomials in one variable and q Example: Evaluate the expression Solution: = =

when x = 0 x = 1

or =

Practice Evaluate each rational expression 1)

when x = 4

when c = 4 and d = 2

when g = -2 and h = 2

Lesson 2.1 Division of Polynomials

To divide a polynomial by a polynomial with more than one term, use a procedure similar to a long division in arithmetic. For example: (3x2 – 14x – 12) divided by x – 2 X–2 3x2 – 14x – 12 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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X–2

3x 3x – 14x – 12 3x2 – 6x 8x – 12 2

3x – 8 3x2 – 14x – 12 3x2 – 6x – 8x – 12 – 8x + 16 28

Steps: 1. Write the problem in long division format. 2. Divide the leading term in the dividend, 3x2 by the leading term in the divisor, x. So, 3x2 ÷ x = 3x. This is the first term in the quotient. 3. Multiply the result 3x by the divisor x – 2 and write the result under the dividend. 3x(x – 2) = 3x2 – 6x 4. Subtract the quantity 3x 2 – 6x. To do this, add its opposite. Then bring down the next term of the dividend, 12. 5. Divide the first of the result, 8x by the leading term in the divisor, x. 8x ÷ x = 8. This is the second term. 6. Multiply the result, 8 by the divisor x – 2. 8(x – 2) = 7. Write the result under the dividend. 8. Then subtract. The remainder is – 28. Do not continue because the degree of the remainder is less than the degree of the divisor. The quotient is 3x – 8 The remainder is – 28 The solution to a long division can be written in the form: Quotient +

Hence: (3x – 14x – 12 ) ÷ (x – 2) = 3x – 8 + or 3x – 8 . Division of polynomials can be checked in the same way as division of real numbers. Dividend = (divisor)(quotient) + remainder 3x2 – 14x – 12 = (x – 2)(3x – 8)+ (-28) = 3x2 – 14x + 16 + (- 28) = 3x2 – 14x – 12 Note: when using long division to divide polynomials, you continue the division operation until the remainder is 0 or the degree of the remainder is less than the degree of the divisor. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Practice: 1. 9x2 – 4 ÷ 3x – 2 2. 5x2 + 12x + 4 ÷ 5x + 2 3. X2 + 9x + 14 4. 8c3 + 5c2 + 2c4 – 4c – 4 ÷ c2 + 3 5. 2x3 + 7x2y – 4xy2 – 15y3 x2 + 2xy – 5y2

Lesson 2.2 Simplifying rational expressions Property of equivalent fraction If a, b, and c are nonzero real numbers Then, Simplifying Rational Expressions 1. Factor the numerator and denominator: 2. Write a product of two rational expressions, one factor containing the GCF of the numerator and denominator, and the other containing the remaining factors. 3. Rewrite the factor containing the GCF as 1. 4. Multiply the remaining factors by 1. Example: Simplify the rational expression = When we divide out common factors, we are actually using a shortcut method known as cancellation. Cancellation is a method of dividing a factor in the numerator by the same factor in the denominator. Other example = Division of negatives The quotient of any nonzero expression and its negative is -1. Practice: 1. 2. 3. 4.

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Lesson 2.3 Multiplying and Dividing Rational Expressions To multiply fractions, we multiply the numerators together, multiply the denominators together, and then simplify. If necessary, we may use the rule where b we can rewrite this rule so that it implies to multiply rational expressions. Also, when we multiply fractions, we can simplify before or after multiplying, by dividing out factors common to both the numerator and denominator. For example,

Multiplying rational expressions 1. Write each numerator and denominator in factor form. 2. Divide out any numerator factor with any matching denominator factor. 3. Multiply the numerators and also the denominators. 4. Simplify if possible. Example: 1.

2. To divide fractions, we may use the rule

The procedure for dividing rational expressions 1. Write the equaivalent multiplication statement using the reciprocal of the divisor. 2. Factor the numerator and denominator. 3. Divide out any numerator factor with any matching denominator factor. 4. Multiply the numerators and denominators 5. Simplify, if possible. Example: Divide = Practice Find the quotient 1. 2. 3. 4. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Lesson 2.4 Adding and Subtracting Rational Numbers Rule Addition/Subtraction of Rational Expression If and and any two rational expressions, then â&#x20AC;&#x201C;

+ = and â&#x20AC;&#x201C; = , where C When adding or subtracting rational expressions, each numerator must be treated as if it were enclosed in parentheses. For some subtraction problems, it may be necessary to insert parentheses within each numerator. When subtracting numerators, apply distributive property when removing the parentheses Example Subtract = To add or subtract rational expressions with different denominators 1. Find the least common denominator (LCD) 2. Write the equivalent expression of each rational expression 3. Add or subtract the numerators and keep the LCD 4. Simplify the result, if possible. Least Common Denominator (LCD) The least common denominator (LCD) of a set of fractions is the least number that can be divided by the denominator of each fraction exactly. Finding the LCD 1. List the different denominators that appear in the rational expression. 2. Factor each denominator completely. 3. For each unique factor, compare the number of times it appears in each factorization. Write a factored form that includes each factor for the greatest number of times it appears in the denominator factorization. Example Add or subtract =

= =

Practice Perform the operation 1. 2. 3. 4.

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Lesson 2.5 Complex Fraction A complex fraction is a fraction that has a rational expression in the numerator, the denominator, or both. The following are examples of complex fractions: , ,

and

Simplifying Complex Fractions Method 1: Write the numerator as one fraction and the denominator as one fraction. Then, invert the denominator and multiply. Method 2: Multiply the numerator and the denominator of the complex fraction by the LCD of the fractions within the numerator and the denominator. Then simplify. Example 1: simplify

Example 2: simplify Example 3: Simplify

= =

= -6 =

Practice: Write each complex fraction as a single fraction in reduced form. 1.

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Lesson 2.6 Equations Involving Rational Expressions A rational equation is an equation that contains one or more rational expressions. Solving rational equations 1. Eliminate the rational expressions in the equation by multiplying both sides of the equation by the LCD. 2. Solve the equation 3. Check your solution Example Solve for x.

Solution: The LCD of the denominators is 20. 20( ) ( ) 4x + 5 = 10x 5 = 6x =x If zeros appear in the denominators, the fractions are undefined. An extraneous solution is an apparent solution that does not solve the equation Example Number problem Find two consecutive integers such that the sum of one-third of the first and one-fourth of the second is 9. Solution Let n = the first integer n + 1 = the second integer Equation: n + = 9 One-third of the first and one fourth of the second is 9 Motion problem An express train travels 150 km in the same time that a freight train travels 100 km. if the express train goes 20 km per hour faster than the freight train, find the rate of each train. Solution Let r = rate of the freight train r + 20 = the rate of the express train

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We first construct a table d r Freight train 100 r

Express train 150 r + 20

Equation The fact that traveled time is the same leads to the equation The equation is a proportion, hence we cross multiply 100(r + 20) = 150(r) 100r + 2000 = 150r -50r = -2000 r = 40 r + 20 = 40 Therefore the freight trains rate is 40 kph and the express trainâ&#x20AC;&#x2122;s rate is 60 kph Practice Solve the rational equation 1. 2. 3. 3

4. 6 + Chapter 3 Integral exponents Lesson 3.1 Bases and exponents Natural Number Exponents If n is a natural number, then n are factors of x In the expression xn, x is called the base and n is called the exponent. The whole expression is called the nth power of x. An exponent which is a natural number tells how many times its base is to be used as a factor. An exponent of 1 means that its base is to be used one time as a factor, an exponent of 2 means that its base is to be used two times as factor, and so on. Thus, 41 = 4 = (-3p)(-3p) When 2 is used as an exponent, the base is squared When 3 is used as an exponent, the base is cubed

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Example 1 Write each product using exponents a. â&#x20AC;&#x201C;(9 Solution: find the repeated factor in each expression. This is the base. Then count the number of times it is used as a factor. This is the exponent. The number 9 is used as a factor 7 times. â&#x20AC;&#x201C;(9 = -(97) = -97 b. Use parentheses to show that both a and b are used as factors 5 times.

Example 2 Identify each base. Next, write the expression as a product of factors then multiply. 1)

( )

( ) = 25 ( ) ( ) ( ) ( ) ( ) 2) the base -5; (-5)3 = (-5)(-5)(-5) = -125 Example 3 Show that -54 and (-5)4 have different values -54 = -(5)4 = -(5 = - 625 4 (-5) = (-5)(-5)(-5)(-5) = 625 Since -625 Exercises: Identify each statement as true or false. 1. 9 is the exponent of 2. is the written in exponent form. 3. is written in exponent form. 4. 5 is the base of 5. is read as b to the power of a. 6. = 10 7. =axb 8. =1 9. =3x3x3x3 10. =8x8x8x8

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Lesson3.2 Multiplication Property of Exponents Rule: Product rule for exponents If m are natural numbers, then Example 1 Simplify each expression a. b. c. Solution : a. keep the base and add the exponents. = 3+5=8 b. apply commutative and associate properties. multiply the coefficients. Keep the base and add the exponent. 3+6=9 c. ( ) apply commutative and associative properties. multiply the coefficients. Keep the base and add the exponents. 4 + 6 = 10 Example 2 The number of ants in a colony after 7 weeks is what does the expression represent in this situation? Rewrite the expression using a single exponent. Solution: The expression represent the number of ants in colony after 11 weeks.

Hence,

can be written as

Example 3 Write down the value of each expression. a. b. c. example 4 write each expression using one exponent. a. solution: a. b.

c. ( ) keep the base and multiply the exponents. 2 . 5 = 10 keep the base and multiply the exponents. 7 . 8 = 56

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keep the base and multiply the exponents. 3.2=6

Example 5 Write each expression using a single exponent. a. b. c. solution: a. apply the product rule for exponents. apply the first power rule for exponent. b. apply the product rule for exponents. apply the first power rule for exponents. c. apply the product rule for exponents. apply the first power rule for exponents.

Example 6 Evaluate each expression. a. b. c. d. example 7 write each expression without using parentheses. Assume that there are no division by zero. a. solution: a. c.

d. b. d.

Example 8 Make chains his dog to a stake in an open circular area of his backyard to exercise without running. a. find the area in which the dog may exercise if the chain is m meters long. b. if mike replaces the chain with another that is twice the original length, find the new extended exercise area. c. if the original chain is 4 meter, find the original area and the extended area. Solution: a. let m = original length of the chain in meters. The area of a circle: substitute m in place of r. The area of a circle of radius m meters in square meters. b. let 2m = length of the new chain in meters. The area:

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square meter c. if m = 4, then the original area is square meters. The extended area is square meters. Exercise Fill in the blank to make a true statement. 1. To multiply exponential expression with the same base, retain the _____ and _____ the exponents. 2. To raise an exponential expression to a power, retain the _____ and ______ the exponents. 3. To raise a product to a power, raise each that power ______ to that power. 4. To raise a quotient to a power, raise the _____ and the ______ to that power.

Lesson 3.3 Division Property of Exponents Rule: quotient rule for exponents If m and n are natural number, and m > n and x 0, then In words, to divide exponential expression with the same base, keep the base and subtract the exponents. Proof:

Addition of zero. Law 2:

Example 1 Simplify and express your answers in exponential form. a. Solution:

Quotient rule for exponents. Subtract the exponents.

Product rule for exponents Add the exponents Quotient rule for exponents Subtract the exponents.

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Product rule for exponents Add the exponents. Quotient rule for exponents and reduce to

lowest terms. Example 2 Simplify and express your answer in exponent form. a. Solution:

c. Second power rule for exponents Simplify. Second power rule for exponents Simplify.

Second power rule for exponents Simplify. Product rule for exponents Simplify.

Second power rule for exponents First power rule for exponents Simplify.

Example 3 The distance that light travels in one year is kilometers. a. how far does light travel in one second? b. how long does it take light to travel from the sun to the earth? (The distance from the earth to the sun is about kilometers). Solution: a.

So, light travels approximately b. distance = rate time so time =

km in one second.

time = it takes approximately 500 seconds, or 8 minutes and 20 seconds for light to travel from the sun to the earth. Exponential Equations YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Property of equivalent exponents For b > 0 and b , if = , then m = n We can use this property to solve an exponential equation. If we can write each side of an exponential equation with the same base, then we can equate the exponents. Example: = 64 = 82 X=2 9x = 243 32x = 35 2x = 5 X= 32(x-5) = 93x -1 32(x-5) = 93x -1 = (32)3x – 1 2x – 10 3 = 36x- 2 2x – 10 = 6x – 2 -4x = 8 X = -2 Practice: A. Simplify each. Example: ( ) = (b3)3 = b9 1. 2. 3.

(

)

(

)

4. B. Solve for x 1. = 16 2. 3. 4.

= 256 + = 106 = 283 Lesson 3.4 Zero and negative exponents

Rule in zero exponent X0 = 1, where x Any number, excluding zero, raised to the zero power is equal to 1 Example =1 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Practice 1. 2. 7p0 3. (7p)0 Negative exponents Rule x-n = and xn = , where x Example = 72 â&#x20AC;&#x201C; 5 = 7-3 1. = = Example 5-3 = Practice Simplify 1. 7-1 2. 53-1 3. 9-2 4. 4-4 5. 01 + 70 Lesson 3.5 Exponential Equations Property of Equivalent Exponent For b > 0 and b , if bm = bn, then m = n Example Solve for exponential equation 8x = 64 8x = 82 X=2 Practice Solve for x 1. 2x = 16 2. 5x = 125 3. 4x = 32 4. 3x = 81 Chapter 4 Linear equations in Two Variables Lesson 4.1 Rectangular Coordinate System Procedure in plotting a point 1. Begin at the origin (0,0), move to the right or left along the x axis using the number indicated by the absolute value of the first coordinate. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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2. From the position on the x-axis, move up or down using the number indicated by the absolute value of the second coordinate. 3. Draw a dot to represent the point described by the coordinates. Procedure in identifying quadrants To determine the quadrant for a given ordered pair, consider the signs of the coordinates (+,+) means the point is in the first quadrant (-, +) means the point is in the second quadrant (-,-) means the point is in the third quadrant (+, -) means the point is in the fourth quadrant Procedure in identifying the coordinate of the plane To determine the coordinates of a given point in the rectangular system 1. Draw a vertical line passing through the point and the x-axis. The number associated to the point on the x-axis is the first coordinate 2. Draw a horizontal line passing through the point and the y-axis. The number associated to the point on the y-axis is the second coordinate. Example Use the coordinate plane to find the coordinates of M,A,T,H Solution M(7,2) A(-5,8) T(4,-6) H(-4,0) Practice Plot the following points on the coordinate plane 1. A(6,4) 2. B(3,-3) 3. C(-10, -3) 4. D(3,0) 5. E(-2,3) Lesson 4.2 Linear Equations in two variables Linear equation in two variables is an equation that can be written in the standard form Ax + By = C Where A, B, and C are real numbers, and A and B are both nonzero. Example Determine whether or not each equation is a linear equation in two variables 5x = 10 + 3y Solution Write the equation in standard form 5x = 10 + 3y 5x - 3y = 10 The equation is now in the form Ax + By = C. thus, A = 5, B = -3, and C = 10 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Solutions of a linear equations are ordered pairs that make the equation true Example The solutions of the equation 15x + 15 = 75, namely: (0, 15), (1,12), (2,9), (3,6), (4,3),… are plotted on the coordinate plane shown at the right. A line can pass through all these points. To find a solution to any linear equation in two variables, select a particular value for x and substitute it for x in the equation, then solve the resulting equation for y. it is also possible to first select a value for y and substitute it for y in the equation and then solve resulting equation for x Example Find three solutions of 4x + y = 12 Solution We find three ordered pairs that satisfy the given equation If x = 0, then if x = 1, then if x = 3, then 4x + y = 12 4x + y = 12 4x + y = 12 4(0) + y = 12 4(1) + y = 12 4(3) + y = 12 0 + y = 12 4 + y = 12 12 + y = 12 Y = 12 y=8 y=0 Therefore,(0,12)is a solution therefore, (1,8) is a solution therefore,(3,0) is a solution Instead of listing the ordered pair solutions as (0,12), (1,8) and (2,4), you can also list the solutions in a table of values. 4x + y = 12 x y 0 12 1 8 2 4 Practice Solve for y using the given equation. Then complete each table of values. 1. Y = 4x – 1 x y 0 -1 -2 1 2 2. 3x – y + 2 = 0 x y 0 -1 -2 1 3

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3. X â&#x20AC;&#x201C; 2y = -6 x y 0 -1 -2 1 2 Lesson 4.3 Relations and Functions A relation is a set of ordered pairs. The domain of a relation is the set of first coordinates. The range is the set of second coordinates. A function is a relation in which each element of the domain corresponds to exactly one element of the range Example: determine whether the following relations are function or not. a. {(1,2),(2,5),(3,10),(4,17)} b. {(1,2),(2,3),(2,0),(3,5)} Solution: a. Each element in the domain, {1,2,3,4} is assigned no more than one value in the range;1 is assigned to only 2;2 is assigned only to 5; 3 is assigned only to 10 and 4 is assigned only to 17. Therefore it is a function. b. Since the domain element 2 is assigned to two different values in the range, 3 and 0, it is not a function. If we are given a set of ordered pairs, we can easily determine whether the relation is a function or not by simply looking if each first element is used only once in a given set. The following characteristics of a function will help us decide when we test for a function when two sets of numbers are given. 1. Each element in domain X must be matched with exactly on element in range Y. 2. Some elements in Y may not be matched with any element in X. 3. Two or more elements in X may be matched with same element in Y. Here are ways to describe the function 1. Mapping diagram Input x output y 1 2 2 5 3 10 4 17 2. Table of values Input, x 1 2 3 4 Output, y 2 5 10 17 3. Graph â&#x20AC;&#x201C; the graph represents a function if an only if no vertical line intersects the graph in more than one point. 4. Rule or correspondence YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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f:x x2 + 1, x = 1,2,3,4 notice that the ordered pairs of numbers, the mapping diagram, the table of values, and the graph clearly shows that each value of y is obtained by adding 1 to the square of x. Hence, this is the rule or correspondence expressed in words for the said relation. 5. Equation: The rule or correspondence can be described by the equation y=x 2 + 1 Example: Represent the rule â&#x20AC;&#x153;Add one to each x-coordinateâ&#x20AC;? with a table of values, an equation, a graph, or a diagram. Solution: a. Table of values Input, x Output,y 9 10 -45 -44 8.28 9.28 x x+1 b. Equation: y = x + 1 c. Graph x-y coordinate d. diagram Domain Range -45 -44 8.28 9.28 9 10 Vertical line test A graph represents a function if and only if no vertical line intersects the graph in more than one point Example: use the vertical line test to identify graphs in which y is a function of x. a. b. c. d.

a. b. c. d.

y is not a function of x y is not a function of x. Two values of y correspond to an x-value. y is not a function of x. Two values of y correspond to an x-value. y is a function of x

Activity 1 A. Determine whether or not each relation is a function. Give the domain and range of each relation. 1. {(2,3),(4,5),(6,6)} 2. {(4,5),(4,6),(5,5),(5,6)} YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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B. State whether or not each relation is a function 1. Input, D output, R 3 4 9 -3 2. x 4 4 4 4

y 4 8 12 16

C. Determine whether or not each equation defines y as a function of x. if it is a function, find the domain and range. 1. x+y=9

2. x + y3 = 8 D. find the domain and range of each relation 1. f(x) = 4x2 + 3x + 1 2. g(x) =

Functions, like numbers, can be added and subtracted. Functions can also be multiplied or divided. Because functions usually given in an equation form, we perform these operations by performing operations on algebraic expressions that appear on the right side of the equations. For example, we can combine the following two functions using addition: F(x) = 3x + 1 and g(x) = –3 Add the terms on the right side of the equal sign of f(x) to the terms on the right side of the equal sign of g(x). Thus, (f + g) x = f(x) + g(x) = (3x + 1) + (x2 – 3) = 3x – 2 + x2 = x2 + 3x – 2 Sum, difference, product and quotient of functions Let f and g be any two functions. The sum f + g, difference f-g, product fg, and the quotient

are functions whose

domains are the set of all real numbers common to the domain of f and g, and defined as follows: 1. Sum: (f + g) (x) = f(x) + g(x) 2. Difference: (f-g)(x) = f(x) – g(x) 3. Product: (fg)(x) = f(x). g(x) 4. Quotient: (x) = where g(x) 0 Example: Let f(x) = x2 – 5 and g(x) = 5x + 4 a. (f + g) (x) = f(x) + g(x) = (x2 – 5) + (5x + 4) = x2 + 5x – 1 b. If f(x) = 3x – 2 and g(x) = x2 + 2x – 3 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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(x) =

Activity 3 A. Find f + g, f – g, fg, and

. Determine the domain of each function.

1. F(x) = 3x + 4, g(x) = 2x – 1 2. F(x) = 2x – 5, g (x) = 4x2 3. F(x) = x – 4, g(x) = √ 4. F(x) = , g(x) = Inverse of a function Let f and g be two functions such that f (g(x)) = x for every x in the domain of g and g(f(x))= x for every x in the domain of f. The function g is the inverse of the function f, and is denoted by f-1 ( read “ f-inverse”.) Thus, f ( f-1(x)) = x and f-1 (f(x)) = x The domain of f is equal to the range of f -1, and vice versa. Example: show that each function is an inverse of the other. a. F(g(x)) = x and g(f(x)) = x F(x) = 6x g(x) = F(g(x)) = 6 (g(x))

g(f(x)) = (

)

=6( ) = =x =x Finding the inverse of a function f can be found as follows: 1. Replace f(x) with y in the equation for f(x). 2. Interchange x and y. 3. Solve for the new y from the equation in step 2. If this equation does not define y as a function of x, the function f does not have an inverse function and tis procedure ends. If this equation does define y as a function of x, the function of x, the function f has an inverse function. 4. If f has an inverse function, replace y in step 3 by f -1(x). The result can be verified by showing that f(f-1(x)) =x and f-1(f(x)) = x. Example: find the inverse of the relation described by the set of ordered pairs { } Solution: the inverse of a relation is the set of ordered pairs obtained by switching the coordinates of each ordered pair in the relation Switch the coordinates of each ordered pair. { } Original relation { } Inverse relation Find the inverse of f(x) = x3 + 3 Step 1: replace f(x) with y : y = x3 + 3 Step 2: interchange x and y: x = y3 + 3 Step 3: solve for y: x – 3 = y3 =√ √ =y √ -1 -1 Step 4: replace y with f (x): f (x) = √ -1 Verify this result by showing that f(f (x)) = x and f-1(f(x)) = x YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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f-1(x) = √ f(x) = x3 + 3 f-1(f(x)) = √ f(f-1(x)) = (f-1(x))3 + 3 =√ = √ +3 =√ =x–3+3 =x =x Since f(f-1(x)) = f-1(f(x)) = x, then f-1 is the inverse of f. The graph of a function can tell us if it represents a function with an inverse. The horizontal line test for inverse functions A function f has an inverse that is a function, f -1, if there is no horizontal line that intersects the graph of the function f at more than one point. Example: which of the following graphs represent functions that have inverse functions? a. . b.

Solution a. b. c. d.

has an inverse function no inverse function no inverse function has an inverse function

Activity 4 I. Find (f (x) in f(x) = 2x, g(x) = x + 5 II. Find the inverse of a function and state whether f -1 is a function or not. } a. f = { b. f(x) = 2x2 – 3 III. Find the f(g(x)) and g(f(x)) and determine whether each pair of functions f and g are inverses of each other a. f(x) = 2x, g(x) = b. f(x) = - x, g(x) = -x IV. Find the equation for f-1(x). then verify if your equation is correct by showing that f(f-1(x)) = x and f-1 (f(x)) = x in f(x) = PRACTICE Evaluate each function at the indicated values 1. a. f(-1) b.(-3 + x ) 2. a. g(2) b. (a + b) 3. a. h(3) b. h (½)

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Lesson 4.4 Linear Equation Linear function is a function that can be written in the form f(x) = ax + b Where a and b are real numbers with a and f(x) are not both equal to zero. f(x) = 4x + 5 is a linear function because it is written in the form f(x) = ax + b, with a = 4 and b = 5 The graph of a linear function f(x) = ax + b is exactly the same as the graph of the linear equation y =ax + b. If a = 0, then we get f(x) = b, which is called a constant function. If a = 1 and b = 0, then we get the function f (x) = x, which is called the identity function. When we graph a function given in function notation, we usually label the vertical axis as f(x) rather than y. The slope The slope m of a line is the ratio of the change in the y-coordinates to the corresponding change in the x-coordinates. Determining slope If the coordinates of two points on a line are (x 1,y1) and (x2,y2), the slope m can be found as follows. m= , where x1 x2 y1 is read â&#x20AC;&#x153; y sub 1â&#x20AC;?. The number 1 is called a subscript. Example: Find the slope of the line containing each pair of points. (-2,1) and (4,6) Solution: Let (-2,1) = (x1,y1)and (4,6) = (x2,y2). m= m= m= The slope of the line is It does not matter which ordered pairs is selected as (x1,y1) Intercepts of a line 1. The x-intercept of the line is the value of x when y = 0. To find the x-intercept, let y = 0 and solve for x in Ax = c 2. The y-intercept of the line is the value of y when x = m0. To find the y-intercept, let x = 0 and solve for y in By = C x-intercept: the point where the graph intersects the x-axis y-intercept: the point where the graph intersects the y-axis slope-intercept form of a line The linear equation y = mx + b is in slope-intercept form. The slope of the line is m. its y-intercept is b Important ideas on slopes YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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1. A line with positive slope rises from left to right. The greater the slope the steeper the line rises. 2. Lines with negative slopes fall from left to right. 3. A horizontal line has 0 slope. 4. A vertical line has undefined slope. 5. Parallel lines have the same slope but different y-intercepts 6. The slopes of two perpendicular lines are negative reciprocals of each other. PRACTICE Find the slope of a line containing each pair of points 1. (4,5),(11,9) 2. (4,8),(1,5) 3. (-2,-2),(7,1) 4. (-3,6),(-3,1) Lesson 4.5 Writing Equations of lines The equation of a line with a slope, m and y-intercept, b is given by Y = mx + b Example Write an equation of a line given a slope and containing the given y-intercept. Express the equation in the form y = mx + b m = - 54, b = 7 Solution Substitute -4 in place of m and 7 min place of b into the slope intercept form y = -4x + 7 The point slope form of a line The point slope of a line passing through (x1, y1) with slope m is given by y – y1 = m(x – x1) Example Find the equation of the line that passes through (5,8) and with slope 4. Then write the equation in slope-intercept form. Solution First, write the point-slope form of the line y – y1 = m(x – x1) y -8 = 4(x – 5) substitute 8 for y1, 5 for x1 and 4 for m Solve for y and write the equation in slope-intercept for y = mx + b Y – 8 = 4x – 20 Y = 4x – 12 Alternate solution Replace m by 4, x by 5 and y by 8 in the slope-intercept form Y = mx + b 8 = 4(5) + b 8 = 20 + b B = -12 Since b = -12, we can write y = 4x – 12 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Formula for two-point form of a line The equation of the line passing through (x1,y1) and (x2,y2) is given by y – y1 = This equation is called the two-point form of a line Example Find the equation of the line that passes through the points (6,2) and (3,7) Solution Substitute the coordinates of the points (6,2) and (3,7) for (x 1, y1) and (x2,y2), respectively into the two-point form y – y1 = y–2= y–2= -3y +6 = 5x – 30 5x + 3y = 36 The intercepts form of a line where a and b are the x-and y-intercepts, respectively of the line. Example: find the equation of the line in standard form that passes through (6,0) and (0,3) Solution: The x-and y-intercepts of the line are 6 and 3, respectively. Using the intercepts form of the line we obtain or x + 2y = 6 Thus the equation of the line in standard form is x + 2y = 6 Practice A. Find the equation of the line having the specified slope and containing the indicated point. 1. m=2;(5,-2) 2. m = ; (-1,2) B. Use the given pairs of points to find the equation of the line in two-point form. 1. (3,4) and (8,1) 2. (-4,-2) and (5,3) 3. (8, -2) and (4, -2) Chapter 5 System of Linear Equations and Inequalities Lesson 5.1 Systems of Linear Equations and their Solutions A system of linear equations is a set of two or more linear equations that have variables in common. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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A p-air of equations of the form a1x + b1y = c1 and (a1, b1) a2x + b2y = c2 (a2, b2) is called a system of linear equations in two variables Example x+y=5 2x – 3y = 12 A solution of a system of equations is an ordered set of numbers that makes all equations in the system true. Checking solutions to a system of linear Equations 1. Replace each variable in each equation with its corresponding value and 2. Verify if each equation is true Example Determine whether (2,-1) is a solution of the system of equations 4x – 3y = 11 and 5x – y = 11 Solution Substitute (2,-1) in each equation 4x – 3y = 11 5x – y = 11 4(2) – 3(-1) = 11 5(2) – (-1) = 11 8 + 3 = 11 10 + 1 = 11 11 = 11 11 = 11 Practice Determine whether the ordered pair is a solution to the given system of equations 1. (5,3) x–y=2 X+y=8 2. (1,2)

3x – y = 1 2x + 3y = 8

3. (-1, 7)

4x + y = 3 2x – 3 = y

Lesson 5.2 Solving Linear Systems by Graphing Definition Consistent: A system of linear equations that has at least one solution. Inconsistent: A system of linear equations that does not have a solution Dependent: A system that has an infinite number of solutions Independent: A system that has an infinite number of solutions Classifying systems of Linear Equations To classify a system of linear equations, write the equations in slope-intercept form and compare their slopes and y-intercepts. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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1. If their slopes are different, then the system is consistent with independent equations and has a single solution. 2. If their slopes are equal and the y-intercepts are also equal, then the system is consistent with dependent equations and has an infinite number of solutions. 3. If their slopes are equal with different y-intercepts, then the system is inconsistent and has no solution. Or Any system of linear equations in standard form a1x + b1y = c1 a2x + b2y = c2 1. A single solution when 2. An infinite number of solutions when

and

3. No solution when Example Classify the system as consistent and independent, inconsistent and Independent, or consistent and Dependent 2x = 8 – 3y 2x + y = 6 Solution Write the equations in slope-intercept form, them compare their slopes and y-intercepts. 2x= 8 – 3y 2x + y = 6 3y = -2x + 8 y = -2x + 6 y= - + The slopes are different. Hence, the system in consistent and the equation is independent Practice Determine whether each linear system has one solution, no solution, or an infinite number of solution. 1. X + 2y = 4 2x + 4y = 8 2. 2x – 5y = -7 4x + 3y = 25 3. 5x + 2y = 9 8x -3y = 2 Lesson 5.3 Solving Linear systems by substitution The substitution method 1. Solve for either variable in one of the equations. If one of the variables in an equation has coefficient 1 or -1, choose that equation, since the substitution method is usually easier this way.

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2. Substitute for that variable in the other equation. The result is an equation with one variable. 3. Solve the equation in step 2. 4. Substitute the result in step 3 into equation in step 1. To find the value of the other variable. 5. Check the values in each of the original equations. Then write the solution set. Example Solve

x+y=9 3x + 2y = 22

using substitution method

Solution: X+y=9 (1) 3x + 2y = 22 (2) a. Solve for x in equation (1)

x+y=9 x = 9 –y

b. Substitute the expression 9 – y For x in equation (2) to get an equation in terms of y. c. Solve for y

3x + 2y = 22 3(9 – y) + 2y = 22 27 – 3y + 2y = 22 y=5

d. Substitute the value of y in equation (3). Then solve for x

x=9–y x = 9 -5 x=4

Practice Solve the following systems using the substitution method. 1. x + y = 10 x = 4y 2. 2x – y = 7 y=x–3 3. x = 5y -2 x – 4y = 1

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Lesson 5.4 Solving Linear Systems by Elimination Procedure in Elimination method 1. Write both equations in the standard form: ax + by = c 2. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either x or y is zero. 3. Add the new equations to eliminate a variable. The sum should be an equation with only one variable. 4. Solve the equivalent system by adding the two equations. 5. Substitute the result in step 4 into either of the given equations and solve for the other variable. 6. Check the solution in both equations. Then write the solution set. Example Solve the following linear systems by elimination 3x – 5y = 13 (1) 4x + 5y = -6 (2) Solution a. Both equations are already in standard form Equations (1) and (2) contain the terms -5y and 5y, respectively. Hence, we can eliminate y by adding (1) and (2) to form a new equation: 3x – 5y = 13 + 4x + 5y = -6 7x =7 x =1 Substitute x = 1 into equation (1) and solve for y. 3(1) – 5y = 13 3– 5y = 13 simplify - 5y = 10 subtract 3 on each side y = -2 divide both sides by -5 Current and Wind in Motion Problems 1. The current reduces the upstream rate and increases the downstream rate. 2. The headwind reduces the rate and the tailwind increases the rate. 3. Distance d, rate r, and time t are related by the formula d = rt Example A pilot flew his airplane 2 400 miles in 8 hours, flying with the wind. Flying against the wind over the same route, he returned in 10 hours. What was the rate of the plane and of the wind? Solution Let x = the rate (speed) of the plane in still air and YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Y = the rate (speed) of the wind Then, x + y = the rate of the plane with the wind and x – y = rate of the plane against the wind Tabulate the relationships Rate x time = Distance With the wind x+y 8 8(x+y) Against the wind x - y 10 10(x – y) Equations 8(x + y) = 2 400 10(x – y )= 2 400

(1) (2)

Solve the system: Eq. (1) x 5 : 40x + 40y = 12 000 Eq.(2) x 4: 40x – 40y = 9 600 80x = 21 600 x = 270 Find y by substituting 270 in place of x in Eq. (1) 8(x + y) = 2 400 8(270 + y) = 2 400 2160 + 8y = 2 400 8y = 240 y = 30 The airplane travels at 270 mph in still air and the speed of the wind is 30 mph.

Lesson 5.5 Solving Systems of Linear Inequalities A linear inequality in x and y can be written in one of the following forms; Ax + By > C, Ax + By < C, Ax + By Where A, B and C are real numbers and A and B are both not equal to 0. Some examples of linear inequalities are: 3x – y > -2 x<2 x + 3y The solutions of linear equation in x and y can be expressed in ordered pairs ( x, y). as with linear equations, solutions of linear inequalities can be also expressed in ordered pairs. An ordered pair (x,y) is a solution of an inequality in x and y if a true statement results when the variable in the inequality is replaced by the coordinates of the ordered pair. Example Determine whether each ordered pair is a solution of 2x + 3y YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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a. (-2,1) b. (1,2) c. (3,-1) Solution: In each case, substitute the x-coordinate for x and the y-coordinate for y in the inequality 2x + 3y . A true statement will be obtained if the ordered pair is a solution. a. For (-2,1): 2x +3y 2(-2) + 3(1) -4 + 3 -1 Since -1 6 is true, (-2, 1) is a solution of 2x + 3y b. For (1,2) 2x + 3y 2(1) + 3(2) 2+6 8 Since 8 is false, (1,2) is not a solution of 2x + 3y c. For (3,-1): 2x + 3y 6 2(3) + 3(-1) 6–3 6 3 Since 3 is true, ( 3, -1) is a solution of 2x + 3y Practice: Verify if each ordered pair is a solution of inequality. 1. X – 3y 2. 4x + y > 2 3. 5x + 2y 11 4. 2x – 2y

Chapter 6 Triangles and Triangle Congruence Lesson 6.1 Writing a Proof A conditional statement has two parts: a hypothesis, denoted by p, and a conclusion, denoted by q. In symbols, the statement, “ if p then q is written as p q. If it is 9:30 a.m. then it must be daytime P q

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A conditional statement may be true or false. To show that a conditional statement is false you only need to find one example (called a counterexample) in which, the hypothesis is satisfied and the conclusion is not satisfied. The converse of the conditional statement is formed by interchanging the hypothesis and conclusion. Example decide if the statement and its converse are both true. If m⦟ A = 100, then ⦟ A is an obtuse angle Solution The given statement is true because 100> 90. Its converse, “ if ⦟A is an obtuse angle, then m⦟A = 100” is false. Some obtuse angles do not measure 100. If a conditional statement and its converse are both true, you can combine them to form one biconditional statement or a biconditional. The parts of a biconditional statement are connected by the phrase if and only if. Conditional statement: if p, then q Converse: if q, then p Biconditional: p if and only if q

p p p

q q

Example Write the statement as a biconditional: “ Complementary angles measures 90”. Solution: Conditional Statement: if two angles are complementary, then the sum of their measures is 90. Converse: if the sum of the measures of two angles is 90,then they are complementary. Biconditional: Two angles are complementary if and only if the sum oftheir measures is 90. Other statements related to a conditional are inverse and its positive Conditional statement: if p, then q p p Inverse: if not p, then not q Contrapositive: if not q, then not p Example Write the inverse, converse, and contrapositive of each of the following statements. Tell whether each is true or false. Hypothesis : a parallelogram has a right angle Conclusion: it is a right triangle Inverse: if a parallelogram has no right angle, then it is not a rectangle.(true) Converse: if it is a rectangle, then it is a parallelogram with a right angle.(true) Contrapositive: if it is not a rectangle, then it is a parallelogram with no right angle. (false) YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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A definition is a statement of the meaning of a word, or term, or phrase which made use of previous defined terms A postulate is a statement which is accepted o be true without proof A theorem is any statement that can be proved true. A corollary to a theorem is a theorem that follows easily from a previously proved theorem. A proof is a logical argument in which each statement you make is backed up by a statement that is accepted as true. Definitions, postulates and properties: A. Definitions 1. Betweenness 2. Midpoint 3. Segment bisector 4. Right angle 5. 6. 7. 8.

Acute angle Obtuse angle Perpendicular line segment Complementary angles

9. Supplementary angles 10. Linear pair 11. Angle bisector 12. Congruent segments 13. Congruent angles B. Properties of equality 1. Addition Property of equality 2. Multiplication property of equality 3. Subtraction property of equality 4. Reflexive property of equality 5. Symmetric property 6. Transitive property of equality C. Law of substitution

If –then form If A-B-C, If A is the midpoint of ̅̅̅̅, ̅̅̅̅ If ̅̅̅̅ bisect ̅̅̅̅ at B. If angle A is a right angle. ⦟ if ̅̅̅̅ ̅̅̅̅ ⦟BAC is a right ⦟. ⊥ If ⦟A is an acute ⦟, ⦟A < 90 If ⦟ A is an obtuse ⦟, m ⦟ A > 90 If ̅̅̅̅ ⊥ ̅̅̅̅ ⦟ BAC is a right angle If ⦟ A and ⦟ are complementary ⦟s, m⦟ A + m⦟ B = 90 If ⦟ A and ⦟ are supplementary ⦟s, m⦟ A + m⦟ B = 180 If ⃗⃗⃗⃗⃗ and ̅̅̅̅ are opposite rays and ⃗⃗⃗⃗⃗ is any other ray, ⦟ ⦟ form a linear pair If ⃗⃗⃗⃗⃗ bisect ⦟ BAC, ⦟ ⦟ ̅̅̅̅, If ̅̅̅̅ If ⦟ A ⦟ , ⦟ ⦟ If a = b and c = d, If a = b and c = d, If a = b and c = d, For every real number a, a = a If a = b If a = b and b = c If a + b = c and b = x

D. Postulates 1. The angle Postulate(AAP)

addition

2. PCAC Postulate

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If T is in the interior of ⦟ PQR ⦟ = m⦟PQT + m⦟TQR If l1 || l2 , ⦟ ⦟

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3. Supplement Postulate

If ⦟1 and ⦟2 form a linear pair, ⦟ ⦟ angles

E. Theorems 1. Vertical Angle Theorem (VAT) 2.

4. 5.

If ⦟1 and ⦟2 are vertical angles, ⦟ ⦟ The supplement theorem If ⦟ 1 ⦟ 3 , ⦟ 1, and ⦟ 2 are supplementary angles, and ⦟ 3 and ⦟ 4 are supplementary angles, The Complementary Theorem ⦟ ⦟ (CT) If ⦟ 1 ⦟ 3, ⦟ 1 and ⦟ 2 are complementary angles, and ⦟ 3 and ⦟ 4 are complementary angles, ⦟ ⦟ PAIC theorem If l1 || l2 ⦟ ⦟ ⦟ ⦟ PSSIAS theorem If l1 || l2 ⦟ ⦟ are supplementary angles and ⦟b and ⦟ c are supplementary angles.

6. The sum of the measures of the m⦟A + m⦟B + m ⦟C = 180 angles of a triangle theorem 7. The Exterior Angle Theorem

m⦟ACD = m⦟A + m⦟B

These definitions, postulates, and theorems when used in writing proofs can be derived from the figures. Example Which statement justifies the conclusion? 1. ⦟A ⦟ m⦟A m⦟B 2. m⦟1 = m ⦟a + m ⦟b and m⦟a = m⦟c Law of substitution

answer: definition of congruent angles m⦟1 = m⦟c + m⦟b answer:

Practice: Write the following in conditional statements in if-then form. 1. Complements of congruent angles are congruent. 2. Any two right angles are congruent. 3. The sum of the degree measures of the angles of a triangle is 180. 4. The acute angles of a right triangle are complementary.

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Lesson 6.2 Congruent Triangles Congruent triangles are congruent if and only if their corresponding parts are congruent. Example ̅̅̅̅, ̅̅̅̅ If ̅̅̅̅ ⦟A ⦟ ,⦟ Then,

̅̅̅̅, ̅̅̅̅ ̅̅̅̅, ⦟ ⦟ ⦟

Example Given: , m⦟ O = 55 and SX = 4 Find: m⦟X and WO Solution: Because ⦟X ⦟ , m⦟X = m⦟O = 55 ̅̅̅̅̅ ̅̅̅ = 4 Because Practice Complete each statement to make it true If a. ⦟T _____ b. ⦟A _____ c. ̅̅̅̅̅ d. ̅̅̅̅

Lesson 6.3 Proving Triangle Congruence (The SSS Postulate) Postulate 18 The SSS Postulate If three sides of one triangle are congruent to the corresponding sides of the other triangle, then the two triangles are congruent. Explanation: If ̅̅̅̅

̅̅̅̅ , ̅̅̅̅

̅̅̅̅ and ̅̅̅̅ ̅̅̅̅ by the SSS Postulate Postulate 18 is often called the side-side-side pattern. Example Which triangles are congruent based on the markings in the given figure? G R E

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Solution the SSS postulate, ̅̅̅̅

̅̅̅̅ ̅̅̅̅

̅̅̅̅

Practice For each pair of congruent triangles, name all the corresponding sides and angles. Use the symbol to indicate each correspondence. a. b. c. Lesson 6.4 The SAS and ASA Congruent Postulates and SAA Theorem Postulate 19 The SAS postulate If two sides and the included angle of one triangle are congruent to the corresponding two sides and the included angle of another triangle, then the two triangles are congruent. If ̅̅̅̅ Then

̅̅̅̅ ⦟

̅̅̅̅

⦟

̅̅̅̅

Postulate 20 The ASA postulate If two angles and the included side of one triangle are congruent to the corresponding two angles and the included side of another triangle, then the two triangles are congruent. A D

B If ⦟A

C ̅̅̅̅

⦟

E ̅̅̅̅

⦟

F , then

Theorem 35 The SAA or AAS theorem If two angles and non-included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent A B

C D

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If ⦟ A

⦟

̅̅̅̅

̅̅̅̅, then

Theorem 36 Perpendicular Bisector Theorem If a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment. P A Q B If l is perpendicular bisector of ̅̅̅̅ and P is on l, then ̅̅̅̅

̅̅̅̅

Practice Use the given information to complete each statement. If the triangles cannot be shown to be congruent from the given information, write not congruent. W A 1. Why? K

2. Why?

E O

Lesson 6.5 Congruent Right Triangles Theorem 37 The L-L theorem (The Leg-Leg Theorem) If two legs of one right triangle are congruent to the corresponding legs of another right triangle, then the two triangles are congruent. A P

C R

If ⦟C and ⦟R are right angles, ̅̅̅̅

Q ̅̅̅̅

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̅̅̅̅

̅̅̅̅ Page 49

Then Given:

⦟C and ⦟R are right angles ̅̅̅̅ ̅̅̅̅ ̅̅̅̅ ̅̅̅̅

Prove: Proof: Statements Reasons 1. ⦟C and ⦟R are right angles 1. Given 2. ⦟C ⦟R 2. Any two right ⦟s are congruent. ̅̅̅̅ ̅̅̅̅ 3. 3. Given ̅̅̅̅ ̅̅̅̅ 4. 4. SAS Postulate Theorem 38 The L-AA Theorem ( The Leg-Acute Angle Theorem) If a leg and an acute angle of one right triangle are congruent to the corresponding leg and an acute angle of another, then the two triangles are congruent. A P B

Q C

If ⦟C and ⦟R are right angles, ̅̅̅̅ Then,

R ̅̅̅̅ and ⦟B

⦟Q,

Given: ⦟C and ⦟R are right angles ̅̅̅̅ ̅̅̅̅ ⦟B ⦟ Q Prove: Proof: Statements 1. ⦟C and ⦟R are right angles 2. ⦟C ⦟ ̅̅̅̅ ̅̅̅̅ ⦟ 3. ⦟ 4.

Reasons 1. Given 2. Any two right ⦟s are congruent 3. Given 4. ASA Postulate

Theorem 39 The H-AA Theorem (The Hypotenuse-Acute Theorem)

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If the hypotenuse and an acute angle of one right triangle are congruent to the corresponding hypotenuse and an acute angle of another, then the two triangles are congruent.

If ⦟C and ⦟R are right angles, ̅̅̅̅ Then,

̅̅̅̅

⦟

Given: ⦟C and ⦟R are right angles ̅̅̅̅ ̅̅̅̅ ⦟B ⦟ Q Prove: Proof: Statements 1. ⦟C and ⦟R are right angles ̅̅̅̅ ̅̅̅̅ ⦟B ⦟ Q 2. ⦟C ⦟ 3.

Reasons 1. Given 2. Any two right ⦟s are congruent 3. SAA theorem

Theorem 40 The H-L Theorem (The Hypotenuse and leg of one right triangle are congruent to the corresponding hypotenuse and leg of the other and then the two triangles are congruent. A P

B If ⦟C and ⦟R are right angles, ̅̅̅̅ Then,

̅̅̅̅

Example Write the statement and reasons to prove that ̅̅̅̅̅ Given: ̅̅̅̅̅ ⊥ ̅̅̅̅, ̅̅̅̅ ⊥ ̅̅̅̅̅ ̅̅̅̅̅ ̅̅̅̅

E M

Prove: ̅̅̅̅̅

̅̅̅̅̅ P

X A

̅̅̅̅̅

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

Sketch the triangles separately and mark any given information and any other information you can get from the diagram.



Show that

to prove that ̅̅̅̅

̅̅̅̅̅

Solution a. b. c. d. e.

Statements ̅̅̅̅̅ ⊥ ̅̅̅̅ ⦟E is a right angle

Reasons a. Given b. Definition of perpendicularity c. Definition of right triangle d. Given e. Definition of

̅̅̅̅ ⊥ ̅̅̅̅̅ ⦟ perpendicularity

f. Definition of right

triangle ̅̅̅̅̅ ̅̅̅̅ g. ̅̅̅̅̅̅̅̅̅̅̅̅̅̅ ̅̅̅̅̅ h. ̅̅̅̅̅ i. ̅̅̅̅̅ j. ̅̅̅̅

g. given h. Reflexive Property i. H-L Theorem J. CPCTC

CPCTC – Corresponding Parts of Congruent Triangle Theorem 41 If a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle. A Q P B C If ̅̅̅̅ bisects ⦟ ABC, Q lies on ̅̅̅̅, ̅̅̅̅ ⊥ ̅̅̅̅̅̅

̅̅̅̅ ⊥ ̅̅̅̅

̅̅̅̅

̅̅̅̅.

Theorem 42 If a point is equidistant from the sides of an angle, then the point lies on the bisector of the angle. If ̅̅̅̅ ⊥ ̅̅̅̅ ̅̅̅̅ ⊥ ̅̅̅̅

̅̅̅̅

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̅̅̅̅

⦟

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Practice Use the given information to decide whether there is a SSS, SAS, SAA, H-L, L-L, H-AA or L-AA congruence. Are the triangles necessarily congruent? ̅̅̅̅ ̅̅̅̅ ̅̅̅̅ 1. Given: ̅̅̅̅ ⦟ ̅̅̅̅ 2. Given: ̅̅̅̅ ̅̅̅̅ ⊥ ̅̅̅̅ 3. Given ⦟VHO ⦟ O is the midpoint of ̅̅̅̅ S E A

Lesson 6.6 Theorems on isosceles Triangles Theorem 43 The Isosceles Triangle theorem If two sides of a triangle are congruent, then the angles opposite them are also congruent. A

C If ̅̅̅̅

B ̅̅̅̅

⦟

Draw an auxiliary line to form two triangles which can be proven congruent Given: Prove: ⦟B

̅̅̅̅

⦟C

Proof: 1. 2. 3. 4. 5. 6.

Statements ̅̅̅̅ ̅̅̅̅ Draw ̅̅̅̅ as angle bisector of ⦟A intersecting ̅̅̅̅ at D. ⦟1 ⦟ ̅̅̅̅ ̅̅̅̅ ⦟

⦟

Reasons 1. given 2. Every angle has exactly one bisector 3. Definition of an angle bisector 4. Reflexive Property 5. SAS postulate 6. CPCTC

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Corollary An equilateral triangle is also equiangular An equilateral triangle has three 600 angles. Theorem 44 The Converse of the Isosceles Triangle Theorem If two angles of the triangle are congruent, then the sides opposite them are also congruent. ⦟ , then ̅̅̅̅

If ⦟ C

̅̅̅̅

Given: ⦟ B Prove: ̅̅̅̅

B ⦟ ̅̅̅̅

Proof: Statements Let ⃗⃗⃗⃗⃗ be a bisector of ⦟ BAC ⦟BAD ⦟ ⦟B ⦟ C ̅̅̅̅ ̅̅̅̅

1. 2. 3. 4. 5. 6. ̅̅̅̅

̅̅̅̅

Reasons 1. Every angle has a bisector. 2. Definition of angle bisector 3. Given 4. Reflexive Property 5. SAA theorem 6. CPCTC

Corollary An equiangular triangle is also equilateral. Practice ̅̅̅̅ ̅̅̅̅ , m ⦟ R is 8 less than twice a number and m⦟ X is 15 more 1. In than the same number. Find m⦟ E. 2. In isosceles with base ̅̅̅̅, m⦟R = 4x – 15 and m⦟O = 2x + 25. Find the measures of the base angles.

Lesson 6.7 Triangle Inequality Theorem 27 The Triangle Inequality Theorem YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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In a triangle the sum of the lengths of any two sides is greater than the length of the third side. A

AC + CB > AB AC + AB > BC

C AB + BC > AC

Example Can a triangle be constructed with sides of lengths 3 cm, 7 cm, 8 cm? Solution: 3+7>8 7+8>3 3+8>7 The sum of any two lengths is greater than the third length. Therefore, a triangle can be constructed Theorem 28 Unequal Sides Theorem If one side of a triangle is longer than the second, then the angle opposite the longer side is larger than the opposite the second side. If BC > AC then m⦟A > m⦟ B C 4

Theorem 29 Unequal Angles theorem If one angle of a triangle is larger than the second angle, then the side opposite the larger angle is longer than side opposite the second angle. C 450 700 A

If m⦟A > m⦟C, then BC > AB Corollary 29a

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The perpendicular segment from a point to a line is the shortest segment from the point to the line. P

If ̅̅̅̅ ⊥ m and ̅̅̅̅ , ̅̅̅̅, and ̅̅̅̅ are segments from P to m different from ̅̅̅̅ , then PR > PQ, PS > PQ, and PT > PQ.

Theorem 30 The Hinge theorem If two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle. A

D 30 E

800 B

Theorem 31 Converse of the Hinge Theorem If two sides of one triangle are congruent to the other two sides of another triangle respectively, and the third side of the first triangle is longer than the third side of the second, then the angle opposite the longer side is larger than the angle opposite the third side of the second triangle. If

with ̅̅̅̅

̅̅̅̅ ̅̅̅̅

̅̅̅̅

Then m⦟A > m⦟D

Practice For each of the following, provide a valid conclusion in (a) and the corresponding reason in (b) 1. In (a) _________ (b) _________

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2. In

EX > XP > EP (a) _________ (b) _________

3. In

⦟ ⦟ (a) _________ (b) _________

⦟

Lesson 6.8 Conditions that Guarantee Parallelism Postulate 15 The Parallel Postulate Given appoint and a line containing it, there is exactly one line through the given point parallel to the given line. Postulate 16 Given a point and a line not containing it, there is exactly one line through the given point perpendicular to the given line. Postulate 17 CACP Postulate Given two lines cut by a transversal, if corresponding angles are congruent, then the two lines are parallel. Theorem 17 AICP Theorem Given two lines cut by a transversal, if alternate interior angles are congruent, then the lines are parallel.

Theorem 18 SSIAS Theorem Given two lines cut by a transversal, if same side interior angles are supplementary, then the lines are parallel. Theorem 19 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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AI-CA theorem Given two lines cut by a transversal, if alternate interior angles are congruent, the corresponding angles are congruent. Theorem 20 The three Parallel lines Theorem In a plane, if two lines are both parallel to the third line then they are parallel. Theorem 21 The Perpendicular to a third Line Theorem If two coplanar lines are perpendicular to the third line, then they are parallel to each other. Practice Find the value of x that will make l1‖l2. 1. m⦟b = 5x – 24 and m⦟f = 6x - 46 2. m⦟d = 3x + 14 and m⦟e = 2x + 6 3. m⦟c = 2x + 24 and m⦟e = 3x + 12 4. m⦟a = 5x + 17 and m⦟h = 3x – 5

Lesson 6.9 Ways of Proving That Quadrilaterals Are Parallelograms Theorem 50 Each diagonal of parallelogram divides the parallelogram into two congruent triangles A B

If If

ABCD is a parallelogram with diagonal ̅̅̅̅ , then ABC ABCD is a parallelogram with diagonal ̅̅̅̅ then ABD

CDA.

Theorem 51 Opposite sides of a parallelogram are congruent A B

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If ABCD is a parallelogram, then ̅̅̅̅

̅̅̅̅ and ̅̅̅̅

̅̅̅̅ .

Theorem 52 Opposite angles of a parallelogram are congruent

If ABCD is a parallelogram, then angle A Theorem 53 Any two consecutive angles of a parallelogram are supplementary. If ABCD is a parallelogram, then angle A and angle B are supplementary angles. Theorem 54 The diagonals of a parallelogram bisect each other. ̅̅̅̅ ̅̅̅̅ If ABCD is a parallelogram, then ̅̅̅̅

̅̅̅̅

Theorem 55 If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. If ̅̅̅̅

̅̅̅̅

̅̅̅̅ , then ABCD is a parallelogram.

Theorem 50 Each diagonal of parallelogram divides the parallelogram into two congruent triangles A B

If If

ABCD is a parallelogram with diagonal ̅̅̅̅ , then ABC ABCD is a parallelogram with diagonal ̅̅̅̅ then ABD

CDA.

Theorem 51 Opposite sides of a parallelogram are congruent A B

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If ABCD is a parallelogram, then ̅̅̅̅

̅̅̅̅ and ̅̅̅̅

̅̅̅̅ .

Theorem 52 Opposite angles of a parallelogram are congruent

̅̅̅̅

Theorem 55 If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. ̅̅̅̅ ̅̅̅̅ ̅̅̅̅ , then ABCD is a parallelogram. If ̅̅̅̅ Theorem 56 A quadrilateral is a parallelogram if both pairs of opposite sides are congruent If ̅̅̅̅

̅̅̅̅ and ̅̅̅̅

̅̅̅̅ , then ABCD is a parallelogram.

Theorem 57 A quadrilateral is a parallelogram if both pairs of opposite angles are congruent. If angle A angle C and angle B angle D then ABCD is a parallelogram. Theorem 58 A quadrilateral is a parallelogram if two sides are both parallel and congruent. If ̅̅̅̅ ̅̅̅̅ and ̅̅̅̅ ̅̅̅̅ , then ABCD is a parallelogram. Theorem 59 If an angle of a quadrilateral is supplementary to both consecutive angles, then the quadrilateral is a parallelogram. If angle A and angle B are supplementary angles and angle A and angle D are supplementary angles, then ABCD is a parallelogram.

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̅̅̅̅ ̅̅̅̅ If ̅̅̅̅ Then AC > DF

̅̅̅̅

⦟

Test Decide whether the information given is enough to determine that a quadrilateral is a parallelogram. 1. 2. 3. 4. 5. 6. 7.

All four sides are congruent. Diagonals are congruent. Consecutive angles are supplementary. Diagonals bisect each other. Two pairs of opposite sides are congruent. Both pairs of opposite sides are congruent. Two pairs of consecutive sides are congruent.

Chapter 7 Statistics and Probability The three common measures of Central Tendency are the mean, median and mode. The mean (commonly called average) of a set of n number is the sum of all numbers divided by n. The median is the middle number when the number in a set of data is arranged in descending order. When there are even numbers of element, the median is the mean of the two middle numbers. The mode is the number that occurs most often in a set of data. A set of data can have more than one mode. If all the numbers appear the same number of times, there is no mode for that set of data. Formula Mean (grouped data) ̅=∑ Where ̅ = mean F = frequency Xm = class mark (average of lower interval and upper interval) ∑ = sum of the product of frequencies and class marks N = total frequency Example Calculate the mean of 40 students given scores f 98 – 100 2 95 – 97 1 92 – 94 1 89 – 91 6 86 – 88 6 83 – 85 5 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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80 – 82 77 – 79 74 – 76 71 - 73

9 2 3 5

Solution In order to use the formula, we still have to add the columns for class mark (Xm) and fXm scores 98 – 100 95 – 97 92 – 94 89 – 91 86 – 88 83 – 85 80 – 82 77 – 79 74 – 76 71 - 73

̅=∑

f 2 1 1 6 6 5 9 2 3 5 N = 40

Xm 99 96 93 90 87 84 81 78 75 72

fXm 198 96 93 540 522 420 729 156 225 360 ΣfXm = 3339

= 83.475

The mean score of 40 students in a Math Quiz is approximately equal to 83.475

The median of a Grouped Data The first step in the computation of the median of a grouped data is to determine the class interval which contains the ( )

score. This can be located under the column <cf

of the cumulative frequency distribution. The class interval that contains the ( ) is called the median class of the distribution. To calculate median, we use the

score

Formula The Median of a grouped data ̃ = Where

) ̃ = Mean

= the lower boundary or true lower limit of the median class N = total frequency YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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= cumulative frequency before the median class = frequency of the median class I = size of the class interval

Cumulative Frequency Distribution of a 30-point Math Quiz

Median class

Scores f 28 – 29 1 26 – 27 3 24 – 25 3 22 – 23 3 20 – 21 6 18 – 19 6 16 – 17 8 14 – 15 6 12 – 13 10 10 – 11 14 N = 60

<cf 60 59 56 53 50 44 38 30 24 14

Solution score = ( ) = 30 th score The class interval that contains the 30th score is 14 – 15 . XLB = 13.5 Cfb = 24 Fm = 6 I=2 ̃=

) = 13.5 + (

) = 13.5 + 2 = 15.5

This means that 50% of the students got scores below 15.5 or if the passing score is 50 percent of the total number of points, almost one-half of the class failed in that particular quiz. The Mode of a Grouped Data In the computation of the mode given a frequency distribution, the first step is to get the modal class. The modal class is that class interval with the highest frequency. To compute for the mode, we use the formula ̂ = XLB + * + Where XLB = lower boundary of the modal class = difference between the frequency of the modal class and the frequency of the class interval preceding it = difference between the frequency of the modal class and the frequency of the class interval succeeding it I = size of the class interval YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Example Find the mode of the data scores 98 – 100 95 – 97 92 – 94 89 – 91 86 – 88 83 – 85 80 – 82 77 – 79 74 – 76 71 - 73 Modal class: 80 – 82 XLB = 79.5 =9–2=7 =9–5=4 I=3 ̂ = XLB + * + = 79.5 + *

f 2 1 1 6 6 5 9 2 3 5 N = 50

Practice Find the mean, median, and mode for each set of data. x f 65 – 69 7 60 – 64 8 55 – 59 9 50 – 54 15 45 – 49 12 40 – 44 7 35 – 39 4 30 – 34 8

Lesson 7.2 Measures of Variation and Dispersion Four measures of variation 1. Range 2. Mean Deviation 3. Variance 4. Standard Deviation Range is the difference between the highest score (h.s.) and the lowest score (l.s.). in the grouped data, the range is the difference between the upper boundary (u.b.) of the highest class interval and the lower boundary (l.b.) of the lowest class interval. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Formula: Range = h.s. – l.s. Range = u.b. – l.b. Example The range of the set of scores:{11,12,13,15} is 15 – 11 = 4 The range of the frequency distribution scores 98 – 100 95 – 97 92 – 94 89 – 91 86 – 88 83 – 85 80 – 82 77 – 79 74 – 76 71 - 73

f 2 1 1 6 6 5 9 2 3 5 N = 50

Range = u.b. – l.b. = 100.5 = 70.5 = 30 The Mean Deviation (M.D.) The Mean Deviation is a measure of variation that makes use of all the scores in a distribution. This is more reliable than the range. Formulas Mean Deviation (ungrouped data) M.D. = Where

∑

̅̅̅

X represents the scores in the distribution ̅ is the mean N is the number of observation

Grouped Frequency Distribution ∑

̅̅̅

∑

̅̅̅

M.D. = or M.D. = Where f represents the frequency observation X. Steps in calculating Mean Deviation (M. D.) 1. Compute the mean of the distribution. 2. Add the entries in column |X - ̅|. this means, subtract the mean from each score and get the absolute value (Deviation from the mean). 3. Get the total of the score under the heading |X - ̅|. 4. Divide the sum obtained in step 3 by N. Example Find the mean deviation of ungrouped frequency distribution x 10 12 12 14 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Calculate the mean ̅=∑ = Add the column |x - ̅| X |x - ̅| 10 2 12 0 12 0 14 2 ̅=4 ̅

M.D. =

= =1

Example Calculate the mean deviation of grouped frequency distribution ̅ . Add the entries under column fXm and Add the entries in column Xm, , ̅. ̅ ̅ Class interval f Xm fXm 134 – 139 10 136.5 1365 18.6 186 128 – 133 9 130.5 1174.5 12.6 113.4 122 – 127 8 124.5 996 6.6 52.8 116 – 121 1 118.5 118.5 0.6 0.6 110 – 115 5 112.5 562.5 5.4 27 104 – 109 2 106.5 213 11.4 22.8 98 – 103 9 100.5 904.5 17.4 156.6 92 – 97 5 94.5 472.5 23.4 117 86 - 91 1 88.5 88.5 29.4 29.4 ̅ = 705.6 N = 50 Σ = 5895 Σ Calculate the mean using the formula ̅ = ̅. ∑ ̅= = = 117.9 Next, find the mean deviation Using the formula M.D. =

̅̅̅

∑

. Then fill the column

̅ and

= 14.11

The Variance and the Standard Deviation The variance is defined as the quotient of the sum of the squared deviations from the mean divided by N – 1 while the standard deviation is the square root of the variance. Formula Variance (S2) and Standard Deviation S2 =

∑

and

S=√

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∑

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Steps 1. 2. 3. 4. 5.

Calculate the mean. Get the difference between each score and the mean. Get the square of the difference. Get the sum of the squared deviations in step 3. Divide the sum of the squared differences by n – 1. The number obtained is called the variance. 6. Take the square root of the variance, the standard deviation. Example Find the standard deviation of the prices of 250-gram powdered soap 90 73 78 79 83 95 77 79 74 82 Solution ̅ X (X-̅ 90 9 81 73 -8 64 78 -3 9 79 -2 4 83 2 4 95 14 196 77 -4 16 79 -2 4 74 -7 49 82 1 1 ̅ = ΣX = 810 Σ( X - ̅ Σ 428 ̅= = 81 Variance = S2 = S2 =

∑

= 47.55

Standard Deviation = S = √ S=√

∑

= 6.89

The variance and Standard Deviation of Grouped Data Formula of raw score method Variance ∑

(∑

)

Standard Deviation √

∑

(∑

)

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Example Find the variance and standard deviation of the following distribution. X F Xm Fxm f 30 – 34 4 32 128 1024 4096 25 – 29 5 27 135 729 3645 20 -24 6 22 132 484 2904 15 – 19 2 17 34 289 578 10 – 14 3 12 36 144 432 N = 20 Σf xm = 465 Σ = 11655 Substitute: Σf xm = 465 Σ = 11655 In the formula: ∑

(∑

)

√ Practice A. Find the range of each set of ungroup data 1. 7, 23, 33, 38 2. 28, 22, 27, 30 B. Find the range of each of the following grouped data x f 30 -34 9 25 -29 8 20-24 7 15 -19 6 10-14 5 x 35 – 38 31 – 14 27 – 30 23 – 26 19 – 22 15 – 18 11 - 14

f 6 3 2 4 8 2 3

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Find the variance and standard deviation x f 45 -51 5 38 – 44 97 31 – 37 116 24 – 30 82 17 – 23 19 10 - 16 6

Lesson 7.3 Fundamental Principle of Counting The fundamental Principle of Counting states that we can find the number of ways that two or more separate tasks can happen by multiplying the number of ways each task can happen separately. If one thing occur in m ways and the second thing can occur in n ways, and a third thing can occur in p ways, and so on, then the sequence of things can occur in m x n x p x … ways Example Three coins are tossed. How many outcomes are possible Solution Each coin can land in one of two ways. Head or tail x x 2 x 2 x 2 = 8 There are 8 possible outcomes Practice Answer each question 1. A multiple choice test has four questions. Each question can be answered with a, b, c, or d. how many outcomes are possible? 2. Four coins are tossed. How many possible outcomes are possible? 3. Five dice are rolled. How many outcomes are possible?

Lesson 7.4 Probability Experiment – an activity with observable results. Outcomes – the result of an experiment. Sample space – the set of all possible different outcomes of an experiment Event – a collection of outcomes

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When a coin is tossed, getting a head is an event. The said event consist of the outcome “head”. In throwing a die, getting a “4” is an event. This means that the outcome is 4. Any activity that involves a chance such as rolling a die is an experiment. The result of an experiment is an outcome. The set of all different outcomes of an experiment is called a sample space of the experiment. In describing the sample space of an experiment, you will write all the possible outcomes of the experiment. Example Find the possible outcomes for each experiment a. Tossing a coin once Solution The sample space for this experiment can be written as S = {H,T}. Thus, n(S) = 2 There are two possible outcomes in the sample space b. Throwing a coin and a die together If the coin comes up head, there are six possible outcomes for the coin and the die. H1,H2,H3,H4,H5,and H6. If the coin comes up tail, there are six possible outcomes for the coin and the die: T1, T2, T3, T4, T5, and T6 Thus, the sample space for the experiment can be written as S = { H1,H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} Then, n(S) = 12 The following rules are to be followed whenever we are to analyze an experiment 1. Identify an outcome. 2. Find all outcomes. 3. Count the total number of outcomes. Practice Answer each problem 1. There are three red pen, 4 blue pens, 2 black pens and 5 green pens in a drawer. Suppose you choose a pen at random. a. What is the probability that the pen chosen is red? b. What is the probability that the pen chosen is blue? c. What is the probability that the pen chosen is red or black? 2. Suppose you draw one card at random from a standard deck of 52 playing cards. Find the probability of each outcome. a. The king of hearts YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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b. The green of spades c. A red card d. Not a red card 3. Two dice are rolled. What is the probability of getting a. A 4 and a 6? b. not 3 and a 5? c. The same number on both die? d. A sum of 8?

End

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