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Solution Manual for Intermediate Algebra, 4th Edition

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    S OLUTIONS M ANUAL C T RIMBLE & A SSOCIATES I NTERMEDIATE A LGEBRA F OURTH E DITION Michael Sullivan, III Joliet Junior College Katherine R. Struve Columbus State Community College with contributions by Jessica Bernards Portland Community College Wendy Fresh Portland Community College

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    Contents Chapter R 1 Chapter 1 19 Chapter 2 120 Chapter 3 183 Chapter 4 282 Chapter 5 374 Chapter 6 470 Chapter 7 542 Chapter 8 675 Chapter 9 736 Chapter 10 822 Appendix A 862

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    1 Chapter R Section R.1 R.1 Exercises Answers will vary. Section R.2 Section R.2 Quick Checks 1. A set is a well-defined collection of objects. 2. The objects in a set are called elements. 3. Let D represent the set of all digits that are less than 5. set-builder: { } | is a digit less than 5 = D x x roster: { } 0,1, 2,3, 4 4. Let G represent the set of all digits that are greater than or equal to 6. set-builder: { } | is a digit greater than or equal to 6 = G x x roster: { } 6, 7,8,9 5. The statement is true because the order in which elements are listed in a set does not matter. 6. If every element of a set A is also an element of a set B , then say that A is a subset of B and write A ⊆ B . 7. False. The empty set can be denoted as { } or ∅ , but not as { } ∅ . 8. The statement ⊆ B A is true because all the elements that are in B are also in A . 9. The statement = B C is false because there are elements in B ( a and b ) that are not in C. There is also an element in C , namely d , that is not in B . 10. The statement ⊂ B D is false. In order for B to be a proper subset of D , it must be the case that all the elements that are in B are also in D . In addition, there must be at least one element in D that is not in B . Since = B D , this is not the case. 11. The statement ∅ ⊆ A is true. The empty set is a subset of every set. 12. The statement { } 5 0,1, 2,3, 4,5 ∈ is true because 5 is an element of the set. 13. The statement Michigan ∉ {Illinois, Indiana, Michigan, Wisconsin} is false because Michigan is an element of the set. 14. The statement 8 3 ∈ { | = p x x q where p and q are digits, 0 ≠ q } is true because 8 3 is of the form p q where 8 = p and 3 = q . 15. The whole numbers are numbers in the set {0, 1, 2, 3, 4, ...}. 16. The numbers in the set where and are integers, 0 , ⎧ ⎫ ⎪ ⎪ = ≠ ⎨ ⎬ ⎪ ⎪ ⎩ ⎭ p x x p q q q are called rational numbers. 17. The set of irrational numbers have decimal representations that neither terminate nor repeat. 18. False. If a number is rational, it cannot be irrational and vice-versa. 19. False. Decimals that terminate or repeat represent rational numbers, but decimals that do not repeat and do not terminate represent irrational numbers. 20. True. The set of rational numbers is a subset of the set of real numbers. 21. True. The set of rational numbers and the set of irrational numbers do not have any numbers in common. 22. 10 and 12 3 4 = are the only natural numbers. 23. 10, 0 0 3 = , and 12 3 4 = are the whole numbers. 24. 9 − , 10, 0 0 3 = , and 12 3 4 = are the integers.

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    Chapter R: Review ISM: Intermediate Algebra 2 25. 7 3 , 9 − , 10, 4.56 , 0 3 , 4 7 − , and 12 4 are the rational numbers. 26. 5.7377377737777… and π and 11 are the irrational numbers. Note that 5.7377377737777… is irrational because the decimal portion is non-terminating and non- repeating. 27. All numbers listed are real numbers. 28. a. To truncate, remove all digits after the third decimal place. So, 5.694392 truncated to three decimal places is 5.694. b. To round to three decimal places, examine the digit in the fourth decimal place and then truncate. Since the fourth decimal place contains a 3, do not change the digit in the third decimal place. Thus, 5.694392 rounded to three decimal places is 5.694. 29. a. To truncate, remove all digits after the second decimal place. So, 4.9369102 − truncated to two decimal places is − 4.93. b. To round to two decimal places, examine the digit in the third decimal place and then truncate. Since the third decimal place contains a 6, add 1 to the digit in the second decimal place. Thus, 4.9369102 − rounded to two decimal places is − 4.94. 30. { } 1 2 2, 0, ,3,3.5 − 0 3 −2 1 2 3.5 31. 3 6 < because 3 lies to the left of 6 on a number line. 32. 3 2 − < − because 3 − lies to the left of − 2 on a number line. 33. 2 0.6 3 = and 1 0.5 2 = 2 1 3 2 > because 2 3 lies to the right of 1 2 on a number line. 34. 5 0.714 7 ≈ 5 0.7 7 > because 5 7 lies to the right of 0.7 on a number line. 35. 2 0.6 3 = and 10 0.6 15 = 2 10 3 15 = because the numbers are the same. 36. 3.14159 π ≈ 3.14 π > because π lies to the right of 3.14 on a number line. R.2 Exercises 38. { } 1, 2,3 40. { } 3, 2, 1, 0,1, 2,3, 4,5 − − − 42. ∅ or { } 44. The statement ⊆ A C is true since all the elements of A are also elements of C . 46. The statement ⊂ A C is true because all the elements in set A are also in set C and there are elements in set C that are not in set A . 48. The statement ⊂ D B is false. In order for set D to be a proper subset of set B , all the elements of set D must be in the set B and there must be at least one element in set B that is not in set D . Since = B D , the statement is false. 50. The statement ∅ ⊂ B is true because the empty set is a subset of every set, and the set B is not empty. 52. { } 4.5 | is a rational number ∈ x x 54. { } 0 | is a counting number ∉ x x 56. a. 13 b. 0, 13 c. − 4.5656..., 0, 2.43, 13 d. 2

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    ISM: Intermediate Algebra Chapter R: Review 3 e. − 4.5656..., 0, 2, 2.43, 13 58. a. 15 b. 6 ,15 1 − c. 6 , 7.3,15 1 − d. 2 + π e. 6 , 2 , 7.3,15 1 − + π 60. a. To truncate, remove all digits after the second decimal place. So, 93.432101 − truncated to two decimal places is − 93.43. b. To round to two decimal places, examine the digit in the third decimal place and then truncate. Since the third decimal place contains a 2, do not change the digit in the second decimal place. Thus, 93.432101 − rounded to two decimal places is − 93.43. 62. a. To truncate, remove all digits after the second decimal place. So, 9.9999 truncated to two decimal places is 9.99. b. To round to two decimal places, examine the digit in the third decimal place and then truncate. Since the third decimal place contains a 9, add 1 to the digit in the second decimal place. So, 9.9999 rounded to two decimal places is 10.00. 64. 0 −5 4 5 3 1 2 − 2.5 66. 4 2 > since 4 lies to the right of 2 on a real number line. 68. 2 2 3 5 > since 2 3 lies to the right of 2 5 on a real number line. 70. 8 8 3 5 − < − since 8 3 − lies to the left of 8 5 − on a real number line. 72. Since sea level is an elevation of 0 ft, the Dead Sea has an elevation of 1349 − feet. 74. – $0.37 Cisco Systems reported a loss of $0.37 in a recent trading day. 76. − 7 The departure from the normal high temperature was –7ºF. 78. Answers will vary. 80. Answers will vary. 82. The sum must be irrational because irrational numbers have nonterminating/nonrepeating decimals, so the sum will have a nonterminating/nonrepeating decimal. 84. Since there is an infinite number of rational numbers, it would not be possible to list all the elements of the set of rational numbers. 86. The decimal 0.45 terminates, while the decimal 0 45 . has the block 45 repeating. Both numbers are rational. 88. There is no positive real number that is “closest” to 0. 90. Rounded: 19 2.7143 7 ≈ Truncated: 19 2.7142 7 ≈ Section R.3 Section R.3 Quick Checks 1. In the expression a ⋅ b , the expressions a and b are called factors. 2. False because |0| = 0, and 0 is neither positive nor negative. 3. 6 6 = because the distance from 0 to 6 on the real number line is 6 units.

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    Chapter R: Review ISM: Intermediate Algebra 4 4. 10 10 − = because the distance from 0 to 10 − on the real number line is 10 units. 5. ( ) 18 6 + − One number is positive, while the other is negative. Find the absolute value of each number: 18 18 = and 6 6 − = . Subtract the smaller absolute value from the larger and obtain 18 6 12 − = . Because the number with the larger absolute value is positive, the sum will be positive. So, ( ) 18 6 12 + − = . 6. 21 10 − + One number is positive, while the other is negative. Find the absolute value of each number: 21 21 − = and 10 10 = . Subtract the smaller absolute value from the larger and obtain 21 10 11 − = . Because the number with the larger absolute value is negative, the sum will be negative. So, 21 10 11 − + = − . 7. ( ) 5.4 1.2 − + − Both numbers are negative, so first find their absolute values: 5.4 5.4 − = and 1.2 1.2 − = . Now add the absolute values, 5.4 1.2 6.6 + = . Because both numbers to be added are negative, the sum is also negative. So, ( ) 5.4 1.2 6.6 − + − = − . 8. 6.5 4.3 − + One number is positive, while the other is negative. Find the absolute value of each number: 6.5 6.5 − = and 4.3 4.3 = . Subtract the smaller absolute value from the larger and obtain 6.5 4.3 2.2 − = . Because the number with the larger absolute value is negative, the sum will be negative. So, 6.5 4.3 2.2 − + = − . 9. 9 9 − + One number is positive, while the other is negative. Find the absolute value of each number: 9 9 − = and 9 9 = . Since the absolute values are the same, the difference will be 0. So, 9 9 0 − + = . 10. For any real number a , there is a real number − a , called the additive inverse, or opposite, of a such that a + ( − a ) = − a + a = 0. 11. For any real number a , − ( − a ) = a . 12. The additive inverse of 5 is 5 − because ( ) 5 5 0 + − = . 13. The additive inverse of 4 5 is 4 5 − because 4 4 0 5 5 ⎛ ⎞ + − = ⎜ ⎟ ⎝ ⎠ . 14. The additive inverse of 12 − is ( ) 12 12 − − = because 12 12 0 − + = . 15. The additive inverse of 5 3 − is 5 5 3 3 ⎛ ⎞ − − = ⎜ ⎟ ⎝ ⎠ because 5 5 0 3 3 − + = . 16. The additive inverse of 0 is 0 because 0 + 0 = 0. 17. ( ) 6 2 6 2 4 − = + − = 18. ( ) 4 13 4 13 9 − = + − = − 19. ( ) 3 8 3 8 11 − − = − + − = − 20. ( ) 12.5 3.4 12.5 3.4 9.1 − = + − = 21. ( ) 8.5 3.4 8.5 3.4 5.1 − − − = − + = − 22. ( ) 6.9 9.2 6.9 9.2 16.1 − − = − + − = − 23. True; the product of two negative real numbers is positive. 24. True. The commutative property of addition states + = + a b b a , and the commutative property of multiplication states ⋅ = ⋅ a b b a . 25. − 6(8) = − (6 ⋅ 8) = − 48 26. 12( − 5) = − (12 ⋅ 5) = − 60 27. 4 14 56 ⋅ = 28. − 7( − 15) = 7 ⋅ 15 = 105 29. − 1.9( − 2.7) = 1.9 ⋅ 2.7 = 5.13

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    ISM: Intermediate Algebra Chapter R: Review 5 30. The additive inverse of a , − a , is also called the opposite of a . The multiplicative inverse of a , 1 , a is also called the reciprocal of a . 31. The multiplicative inverse of 10 is 1 10 . 32. The multiplicative inverse of 8 − is 1 8 − . 33. The multiplicative inverse of 2 5 is 5 2 . 34. The multiplicative inverse of 1 5 − is 5 5 1 − = − . 35. A fraction is written in lowest terms if the numerator and denominator share no common factor other than 1. 36. 2 6 2 2 5 ⋅ = ⋅ 6 2 ⋅ 6 5 5 = ⋅ 37. 33 3 11 11 24 3 8 8 ⋅ = = ⋅ 38. 24 4 6 6 20 4 5 5 ⋅ − = − = − ⋅ 39. 4 9 is in lowest terms. 40. 5 12 5 12 3 25 3 25 5 3 4 3 5 5 5 ⋅ ⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ = 3 ⋅ 4 3 ⋅ 5 ⋅ 5 4 5 ⋅ = 41. 2 5 2 5 3 4 3 4 2 5 3 2 2 2 ⋅ ⎛ ⎞ − = − ⎜ ⎟ ⋅ ⎝ ⎠ ⋅ = − ⋅ ⋅ = − 5 3 2 ⋅ ⋅ 2 5 3 2 5 6 ⋅ = − ⋅ = − 42. 4 8 4 3 3 3 3 8 4 3 3 8 4 ÷ = ⋅ ⋅ = ⋅ = 3 ⋅ 3 4 ⋅ 2 1 2 ⋅ = 43. 10 3 5 12 10 12 3 5 10 12 3 5 2 5 = ⋅ ⋅ = ⋅ ⋅ = 3 ⋅ 4 3 ⋅ 5 ⋅ 8 = 44. 3 2 3 2 5 11 11 11 11 + + = = 45. 8 13 8 13 5 5 15 15 15 15 − − − = = = − 3 5 ⋅ 1 3 = − 46. 3 11 3 11 14 2 7 2 7 7 7 7 7 − − − ⋅ − − = = = − = − 47. 8 3 8 3 5 1 5 5 5 5 − − = = = 48. The least common denominator is the smallest number that each denominator has as a common multiple.

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    Chapter R: Review ISM: Intermediate Algebra 6 49. 20 2 2 5 15 3 5 = ⋅ ⋅ = ⋅ The common factor is 5. The remaining factors are 2, 2, and 3. 2 2 3 5 60 = ⋅ ⋅ ⋅ = LCD 3 3 3 3 3 9 20 20 3 20 3 60 ⋅ = ⋅ = = ⋅ 2 2 4 2 4 8 15 15 4 15 4 60 ⋅ = ⋅ = = ⋅ 50. 18 2 3 3 45 3 3 5 = ⋅ ⋅ = ⋅ ⋅ The common factors are 3 and 3. The remaining factors are 2 and 5. 3 3 2 5 90 = ⋅ ⋅ ⋅ = LCD 5 5 5 5 5 25 18 18 5 18 5 90 ⋅ = ⋅ = = ⋅ 1 1 2 1 2 2 45 45 2 45 2 90 ⋅ − = − ⋅ = − = − ⋅ 51. LCD = 2 2 5 3 60 ⋅ ⋅ ⋅ = 3 2 3 3 2 4 20 15 20 3 15 4 9 8 60 60 9 8 60 17 60 + = ⋅ + ⋅ = + + = = 52. LCD = 2 7 3 42 ⋅ ⋅ = ( ) 5 11 5 11 14 21 14 21 5 3 11 2 14 3 21 2 15 22 42 42 15 22 42 7 1 7 42 6 7 1 6 − ⎛ ⎞ − = + ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞ = ⋅ + ⋅ ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞ = + ⎜ ⎟ ⎝ ⎠ + − = − − ⋅ = = ⋅ = − 53. LCD = 5 5 6 150 ⋅ ⋅ = ( ) 4 7 4 7 25 30 25 30 4 6 7 5 25 6 30 5 24 35 150 150 24 35 150 59 150 − − ⎛ ⎞ − − = + ⎜ ⎟ ⎝ ⎠ − − ⎛ ⎞ = ⋅ + ⋅ ⎜ ⎟ ⎝ ⎠ − − ⎛ ⎞ = + ⎜ ⎟ ⎝ ⎠ − + − = = − 54. LCD = 2 3 3 5 90 ⋅ ⋅ ⋅ = ( ) 5 1 5 1 18 45 18 45 5 5 1 2 18 5 45 2 25 2 90 90 25 2 90 27 90 3 9 10 9 3 10 − − ⎛ ⎞ − − = + ⎜ ⎟ ⎝ ⎠ − − ⎛ ⎞ = ⋅ + ⋅ ⎜ ⎟ ⎝ ⎠ − − ⎛ ⎞ = + ⎜ ⎟ ⎝ ⎠ − + − = − = ⋅ = − ⋅ = − 55. The Distributive Property states that a ( b + c ) = ab + ac . 56. − ( a + b ) = − a − b 57. ( ) 5 3 5 5 3 5 15 + = ⋅ + ⋅ = + x x x 58. ( ) ( ) 6 1 6 6 1 6 6 − + = − ⋅ + − ⋅ = − − x x x 59. ( ) ( ) 4 8 4 4 8 4 32 − − = − ⋅ − − ⋅ = − + z z z 60. ( ) 1 1 1 6 9 6 9 3 3 3 6 9 3 3 2 3 + = ⋅ + ⋅ = + = + x x x x 61. − (11 p + 8) = − 11 p + ( − 8) = − 11 p − 8

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    ISM: Intermediate Algebra Chapter R: Review 7 62. − ( − 9 − 4 t ) = − ( − 9) − ( − 4 t ) = 9 + 4 t R.3 Exercises 64. 5 5 6 6 = 66. 7 7 2 2 − = 68. ( ) 6 10 6 10 4 − + = − + = 70. ( ) 9 3 6 + − = 72. ( ) ( ) 8.2 4.5 8.2 4.5 12.7 − − = − + − = − 74. − 5( − 15) = 5 ⋅ 15 = 75 76. 7( − 15) = − (7 ⋅ 15) = − 105 78. − 10.4( − 0.6) = 10.4 ⋅ 0.6 = 6.24 80. 25 5 5 40 5 8 8 ⋅5 = = ⋅ 82. 40 8 5 8 16 8 2 ⋅ = = ⋅ 5 8 ⋅ 5 2 2 = ⋅ 84. 2 15 2 15 2 3 5 2 5 8 5 8 5 2 4 ⋅ ⋅ ⋅ ⋅ = = = ⋅ ⋅ ⋅ 3 5 ⋅ ⋅ 5 2 ⋅ 3 4 4 = ⋅ 86. 7 12 7 12 7 3 4 7 3 35 3 35 3 5 7 ⋅ ⋅ ⎛ ⎞ − − = ⋅ = = ⎜ ⎟ ⋅ ⋅ ⎝ ⎠ 3 ⋅ 4 3 ⋅ 5 7 ⋅ ⋅ 4 5 = 88. 10 15 10 21 3 21 3 15 2 5 3 7 3 3 5 2 5 − ÷ = − ⋅ ⋅ ⋅ ⋅ = − ⋅ ⋅ ⋅ = − 3 ⋅ 7 3 3 ⋅ ⋅ 5 ⋅ 14 3 = − 90. 18 18 14 18 14 7 3 7 3 7 3 14 3 6 7 2 7 3 3 ⋅ = ⋅ = ⋅ ⋅ ⋅ ⋅ = ⋅ = 6 7 ⋅ ⋅ 2 7 ⋅ 3 ⋅ 12 = 92. 8 18 8 18 10 5 2 5 5 5 5 5 5 − ⋅ − = = − = − = − 2 5 ⋅ 2 = − 94. LCD 2 2 2 3 3 72 = ⋅ ⋅ ⋅ ⋅ = 3 7 3 9 7 4 8 18 8 9 18 4 27 28 72 72 27 28 72 1 72 − = ⋅ − ⋅ = − − = = − 96. ( ) 3 5 3 3 5 3 15 − = ⋅ − ⋅ = − y y y 98. ( ) ( ) 5 4 5 5 4 5 20 − − = − ⋅ − − ⋅ = − + x x x 100. ( ) 3 2 3 2 2 6 2 + ⋅ = ⋅ + ⋅ = + x y x y x y 102. ( ) 2 2 2 3 15 3 15 3 3 3 6 30 3 3 2 10 ⎛ ⎞ − + = − ⋅ + − ⋅ ⎜ ⎟ ⎝ ⎠ = − − = − − x x x x 104. − (11 − 8 k ) = − 11 − ( − 8 k ) = − 11 + 8 k 106. 4 12 4 12 16 − + = + =

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    Chapter R: Review ISM: Intermediate Algebra 8 108. 18 2 9 2 45 5 9 5 LCD 2 2 2 3 5 120 ⋅ = = ⋅ = ⋅ ⋅ ⋅ ⋅ = ( ) 18 23 2 23 45 24 5 24 2 24 23 5 5 24 24 5 48 115 120 120 48 115 120 163 120 − − = − − = − ⋅ − ⋅ = − − − + − = = − 110. 2 2 1 2 1 2 8 3 3 8 3 2 4 ⋅ ÷ = ⋅ = = ⋅ ⋅ 1 3 2 ⋅ ⋅ 1 12 4 = ⋅ 112. 5.4 10.5 5.1 5.1 − + = = 114. ( ) 5 9 45 45 45 − − ⋅ = − − = − = − 116. 4 8 4 7 8 3 3 7 3 7 7 3 28 24 21 21 4 21 4 21 − = ⋅ − ⋅ = − = = 118. 4 1 4 2 1 5 15 6 15 2 6 5 8 5 30 30 8 5 30 − = ⋅ − ⋅ = − − = 3 30 3 3 3 30 3 10 = = = = ⋅ 3 1 10 10 = ⋅ 120. 10 10 6 21 10 126 6 21 21 1 21 21 21 10 126 116 21 21 − + = − + ⋅ = − + − + = = 122. 20 20 4 5 4 4 5 5 1 5 5 4 ⋅ ⋅ = ⋅ = = 4 4 5 ⋅ ⋅ 16 16 1 = = 124. 7 0 − is undefined. 126. 0 0 5 = 128. ( ) 9 9 0 + − = Additive Inverse Property 130. 1, 0 = ≠ a a a Division Property 132. ( ) 3 4 3 12 − = − x x Distributive Property 134. 0 0 6 = Division Property 136. 83 − 49 = 83 + ( − 49) = 34 The difference in life expectancy from 1950 to 2016 is 34 years. 138. ( ) ( ) ( ) 400 20 45 60 105 150 400 150 20 45 60 105 550 230 550 230 320 − − − − + = + − + + + = − = + − = Paul’s balance at the end of the month was $320. 140. ( ) 535 8 535 8 543 − − = + = The difference between the highest and lowest elevations in Louisiana is 543 feet. 142. a. 1 2 3 5 1, 2, 1.5, 1.666667, 1 1 2 3 8 13 21 1.6, 1.625, 1.615385, 5 8 13 34 55 1.619048, 1.617647, 21 34 89 144 1.618182, 1.617978, ... 55 89 = = = ≈ = = ≈ ≈ ≈ ≈ ≈

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    ISM: Intermediate Algebra Chapter R: Review 9 b. The ratios approach a number that is roughly equivalent to 1.618. The golden ratio is 1 5 1.61803398875... 2 + ≈ c. Answers will vary. 144. ( ) ( ) , 6 2 6 2 8 8 = − − = + = = d P Q −2 6 0 146. ( ) ( ) , 1.6 9.3 1.6 9.3 10.9 10.9 = − − = + = = d P Q −12 6 0 −9.3 1.6 148. ( ) 7 7 6 5 7 , 6 6 5 5 1 5 5 30 7 37 37 5 5 5 5 ⎛ ⎞ = − − = + = ⋅ + ⎜ ⎟ ⎝ ⎠ = + = = d P Q 6 0 7 5 − −3 150. 9 9 3 3 1 1 , , 9, , , 3, , , 1 4 2 4 2 4 2 ⎧ ⎫ ± ± ± ± ± ± ± ± ± ⎨ ⎬ ⎩ ⎭ 152. We do not distribute across multiplication. 154. No, the order of subtraction is important. For example: 7 5 5 7 2 2 − ≠ − ≠ − 156. No, the order of division is important. For example: 2 4 4 2 1 2 2 ≠ ≠ 158. The fraction 1 4 is equivalent to 2 8 , so 2 parts in 8 added to 1 part in 8 results in 3 parts in 8. 160. 162. 164. 166. 168. Section R.4 Section R.4 Quick Checks 1. In the expression 6 5 , the number 5 is the base and 6 is the exponent or power. 2. False. ( ) 4 7 7 7 7 7 7 7 7 7 − = − ⋅ ⋅ ⋅ = − ⋅ ⋅ ⋅ 3. 3 4 4 4 4 64 = ⋅ ⋅ = 4. ( ) ( ) ( ) 2 7 7 7 49 − = − ⋅ − = 5. ( ) ( ) ( ) ( ) 3 10 10 10 10 1000 − = − ⋅ − ⋅ − = − 6. 3 2 2 2 2 8 3 3 3 3 27 ⎛ ⎞ = ⋅ ⋅ = ⎜ ⎟ ⎝ ⎠ 7. ( ) 2 8 8 8 64 − = − ⋅ = −

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    Chapter R: Review ISM: Intermediate Algebra 10 8. ( ) ( ) ( ) ( ) ( ) 3 5 5 5 5 125 125 − − = − − ⋅ − ⋅ − = − − = ⎡ ⎤ ⎣ ⎦ 9. 5 2 6 10 6 16 ⋅ + = + = 10. 3 2 5 6 6 30 36 ⋅ + ⋅ = + = 11. 4(5 + 3) = 4 ⋅ 8 = 32 12. 8(9 − 3) = 8 ⋅ 6 = 48 13. (12 − 4)(18 − 13) = 8 ⋅ 5 = 40 14. (4 + 9)(6 − 4) = 13 ⋅ 2 = 26 15. 3 7 10 4 9 13 + = + 16. 1 4 8 2 5 1 4 16 5 22 4 18 − + ⋅ + = − + + = − = 17. ( ) ( ) [ ] [ ] 25 2 8 3 8 25 2 5 8 25 10 8 25 2 50 − − = − ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ = − = = 18. ( ) ( ) 3 5 8 8 4 2 9 4 2 5 10 5 + = = = − 19. The order of operations is (1) parentheses, (2) exponents, (3) multiplication and division, (4) addition and subtraction. 20. 6 5 2 6 10 16 + ⋅ = + = 21. ( ) 3 9 4 12 4 48 + ⋅ = ⋅ = 22. 7 5 12 6 4 10 14 7 + = = + 23. ( ) [ ] 4 8 3 2 4 5 2 4 10 14 + − ⋅ = + ⋅ ⎡ ⎤ ⎣ ⎦ = + = 24. 4(6 2) 9 4 4 9 16 9 7 − − = ⋅ − = − = 25. 2 8 2 5 8 2 25 8 50 42 − + ⋅ = − + ⋅ = − + = 26. 3 5 3 3 2 5 3 3 8 15 24 9 ⋅ − ⋅ = ⋅ − ⋅ = − = − 27. 2 2 5(10 8) 5 2 5 4 20 − = ⋅ = ⋅ = 28. 2 2 3( 2) 6 3 3 4 3 4 6 3 3 16 12 18 48 30 48 18 − + ⋅ − ⋅ = ⋅ + ⋅ − ⋅ = + − = − = − 29. 2 4 6 4 36 2 3 2 2 3 2 4 36 6 2 40 8 5 + + = ⋅ + ⋅ + + = + = = 30. ( ) 3 2 2 3 3 7 2 8 5 3 7 2 3 3 49 2 27 3 49 54 3 5 3 5 15 − − − = − − ⋅ = − − ⋅ = − − = − − = − ⋅ = − R.4 Exercises 32. ( ) ( ) ( ) ( ) 3 4 4 4 4 64 − = − ⋅ − ⋅ − = − 34. ( ) 4 2 2 2 2 2 16 − = − ⋅ ⋅ ⋅ = − 36. ( ) 3 ( 7) [( 7)( 7)( 7)] 343 343 − − = − − − − = − − = 38. 4 3 3 3 3 3 81 4 4 4 4 4 256 ⎛ ⎞ ⎛ ⎞ − = − ⋅ ⋅ ⋅ = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 40. ( ) ( ) 2 2 3 2 9 4 13 − + − = + = 42. 5 3 12 15 12 3 − ⋅ + = − + = − 44. 2 + 5(8 − 5) = 2 + 5 ⋅ 3 = 2 + 15 = 17 46. ( ) [ ] ( ) 3 15 7 3 3 15 4 3 11 33 − − = − = = ⎡ ⎤ ⎣ ⎦

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    ISM: Intermediate Algebra Chapter R: Review 11 48. 12 4 8 4 2 4 2 2 2 1 2 − ⋅ ⋅ = = = − − − ⋅ 1 2 − ⋅ 4 = − 50. 6 2 5 3 12 15 3 3 ⋅ − ⋅ = − = − = 52. 3 45 15 4 3 12 9 12 21 5 5 ⋅ + ⋅ = + = + = 54. ( ) ( ) [ ] ( ) 2 25 2 10 4 2 25 2 6 2 25 12 2 13 26 − − = − ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ = − = = 56. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 5 2 5 2 3 5 2 3 25 6 25 31 − − − − = − − − = − − = − − = − 58. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 4 2 3 5 2 4 2 3 25 2 4 6 25 2 4 19 2 4 19 2 23 46 − + ⋅ − = − + ⋅ − = − + − = − + − = − + = − = − 60. 6(3 2 10) 6(6 10) 6( 4) 24 24 ⋅ − = − = − = − = 62. 2 2 2 3 2 9 18 18 6 16 4 16 4 12 4 4 ⋅ ⋅ = = = = − − − 3 6 ⋅ 3 2 2 = ⋅ 64. 2 3 3(5 2 ) 3(5 4) 3 9 3 2 27 2 3 9 2 3 + + ⋅ = = = ⋅ ⋅ ⋅ ⋅ 9 ⋅ 2 3 ⋅ 9 ⋅ 1 2 = 66. ( ) ( ) ( ) ( ) [ ] [ ] ( ) ( ) 4 3 2 8 6 4 3 2 2 5 14 2 2 3 5 14 2 5 4 3 4 5 14 10 4 7 4 5 4 + − + ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ = − + − ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ + = − = = 7 5 4 ⋅ ⋅ 7 5 = 68. 2 3 3 5 9 5 4 5 29 3 8 9 29 3 2 9 5 29 24 9 9 5 5 9 1 ⎛ ⎞ ⎛ ⎞ ⋅ = ⋅ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + − ⋅ ⎝ ⎠ − ⋅ ⎝ ⎠ ⎛ ⎞ = ⋅ ⎜ ⎟ − ⎝ ⎠ = ⋅ = 70. ( ) ( ) 5 2 2 4 5 4 32 4 5 16 32 20 16 12 16 28 + − + = + − + = − + = + = 72. 2 2 2 5(37 6 ) 7 2 4 5(37 36) 7 2 16 5 4 6 2 9 5 4 6 2 3 5(1) 14 16 12 9 5 4 5 2 3 9 − ⋅ − − ⋅ − + = + + ⋅ − + ⋅ − − = + − + − = + 5 3 2 3 3 9 15 2 9 9 13 9 = ⋅ − = − = 74. ( ) ( ) 2 2 4 3 4 3 4 3 16 1 5 2 5(7 5) 4 1 15 10 4 4 ( 2) 30 28 2 4 7 2 − − − − − − = = − − − − + ⋅ + 4 2 3 3 8 9 15 2 10 3 30 30 4 4 30 30 1 1 30 30 4 30 4 30 1 4 ⋅ − ⋅ − = = − − − ⎛ ⎞⎛ ⎞ = = − − ⎜ ⎟⎜ ⎟ − ⎝ ⎠⎝ ⎠ = 76. ( ) 2 3 5 4 − ⋅ − = 78. ( ) ( ) 3 5 6 3 24 + ⋅ − =

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    Chapter R: Review ISM: Intermediate Algebra 12 80. 2 4 3.1416 3 4 3.1416 9 12.5664 9 113.0976 113.10 ⋅ ⋅ = ⋅ ⋅ = ⋅ = ≈ The surface area is about 113.10 square centimeters. 82. 2 10 8 100 8 800 320 2.5 2.5 2.5 ⋅ ⋅ = = = The engine has a rating of 320 horsepower. 84. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 105 100 80 100 115 100 95 100 105 100 5 20 15 5 5 4 4 25 400 225 25 25 4 700 175 4 − + − + − + − + − + − + + − + = + + + + = = = 86. Answers will vary. One suggestion: 3 + 2 ⋅ 7 is equivalent to 3 + 7 + 7, which equals 17. Since 3 + 2 ⋅ 7 = 3 + 14 = 17, we can see that we multiply before adding. 88. 3 4 − is equivalent to 3 1 4 1 4 4 4 64, − ⋅ = − ⋅ ⋅ ⋅ = − whereas 3 4 4 4 4 64 ( ) ( )( )( ) . − = − − − = − 90. 3 6 216 375 216 159 3 3 5 125 125 125 125 ⎛ ⎞ − = − = − = ⎜ ⎟ ⎝ ⎠ 92. 2 3 3 2 9 8 1 1 3 2 1 6 7 − − = = + ⋅ + 94. ( ) 2 4 4 8 2.3 3.7 26.33 11 3 ⋅ − ⋅ ≈ −

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