Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ CHAPTER 1: REVIEW OF REAL NUMBERS CHAPTER PREP TEST 25 14 39 13 + = + = = 12 30 60 60 60 20 32 21 11 − = − = 15 20 60 60 60 4 ÷ = ⋅ = 15 15 44.405 73.63 7.446 54.06 i, iii, iv 10 a = 0.5 C b = 0.7 10 D c = 0.75 A d 89 = 0.89 100 : c, d –17: a, b, d 0.3412: b, d : c, d π –1.010010001: c, d 27 : b, d 91 6.12 : b, d A terminating decimal is a decimal number that has a finite number of decimal places – for example, 0.75 A repeating decimal is a decimal number that has a block of digits that repeats with no other digits between the repeating blocks; an example is 8.454545 … = 8.45 The additive inverse of a number is the number that is the same distance from zero on the number line, but which is on the opposite side of zero The absolute value of a number is a measure of its distance from zero on the number line The union of two sets will contain all the elements that are in either set The intersection of the two sets will contain only the elements that are in both sets 10 {x | x < 5} does not include the value 5, but {x | x < 5} does include the value B SECTION 1.1 CONCEPT CHECK : c, d Objective 1.1.1 –14: c, e 9: a, b, c, d 0: b, c 53: a, b, c, d 7.8: none –626: c, e 11 A number such as 0.63633633363333…, whose decimal notation neither ends nor repeats, is an example of an irrational number 12 The additive inverse of a negative number is a positive number 13 31: a, b, c, d –45: c, e –2: c, e 9.7: none 8600: a, b, c, d : none Exercises y ∈ {1, 3, 5, 7, 9} is read “y is an element of the set {1, 3, 5, 7, 9} ” 15 : b, d 0: a, b, d –3: a, b, d ʌ F G 14 In symbols the phrase “the opposite of the absolute value of n” is − n 15 –27 16 17 − 18 − 17 19 20 ʌ − 2.33 : b, d 4.232232223: c, d INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe 21 33 22 1.23 23 91 24 25 26 27 28 29 30 31 32 Replace a with each element in the set and evaluate the expression –a –(–3) = –(–2) = –(0) = 33 Replace c with each element in the set and evaluate the expression c Replace x with each element in the set and determine whether the inequality is true x –2 –4 > –2 False –1 > –2 True > –2 True The inequality is true for –1 and Replace y with each element in the set and determine whether the inequality is true y > –4 –6 > –4 False –4 > –4 False > –4 True The inequality is true for Replace x with each element in the set and determine whether the inequality is true x < –3 –6 < –3 True –3 < –3 False < –3 False The inequality is true for –6 −4 = =0 =4 34 −3 = =0 =7 35 Replace b with each element in the set and evaluate the expression –b –(–9) = –(0) = –(9) = –9 Replace m with each element in the set and evaluate the expression −m − −6 = −6 − −2 = −2 − =0 − = −1 − = −4 36 Replace w with each element in the set and determine whether the inequality is true w –1 –2 –1 True –1 –1 True –1 False –1 False The inequality is true for –2 and –1 Replace p with each element in the set and determine whether the inequality is true p –10 )DOVH –5 )DOVH 7UXH 7UXH The inequality is true for and Replace q with each element in the set and evaluate the expression q Replace x with each element in the set and evaluate the expression −x − −5 = −5 − −3 = −3 − =0 − = −2 − = −5 37 Yes, negative real numbers are numbers such that − x > 38 No Objective 1.1.2 Exercises 39 Two ways to write the set of natural numbers less than five are {1, 2, 3, 4} and {n| n < 5, n ∈ natural numbers} The first way uses the roster method, and the second way uses set-builder notation 40 The symbol for “union” is ∪ The symbol for “intersection” is ∩ The symbol y ∞ is called the infinity y symbol y INSTRUCTOR USE E ONLY O 41 © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter hapter 1: Review of Real Numbers 66 {x x ≤ −2} 67 {x 68 {x − < x < 4} {2, 4, 6, 8, 10, 12} 69 {x − ≤ x ≤ 7} 46 {1, 3, 5, 7, 9, 11, 13} 70 {x 47 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30} {–20, –16, –12, –8, –4} 71 48 {x − ≤ x < 6} 49 {–35, –30, –25, –20, –15, –10, –5} 72 {x 50 {6, 12, 18, 24, 30, 36} 73 51 {x x > 4, x is an integer} or {x x ≤ 4} 74 {x x < −2} {x x < −2, x is an integer} or 75 52 {x x > 5} 76 {x x ≥ −2} 53 {x x ≥ −2} 54 {x x ≤ 2} 79 [–1, 5] 55 {x < x < 1} 80 [0, 3] 56 {x − < x < 5} 57 {x ≤ x ≤ 4} 58 {x 59 {x − < x < 5} 42 Replace each question mark with “includes” or “does not include” to make the following statement true The set [−4, 7) includes the number −4 and does not include the number 43 {–2, –1, 0, 1, 2, 3, 4} 44 {–3, –2, –1} 45 60 {x < x < 8} ≤ x ≤ 4} < x ≤ 5} 77 (–2, 4) 78 (0, 3) 81 (– 82 (– @ 83 [–2, 6) ≤ x ≤ 2} 84 [3, 85 (– 86 (–1, 87 (–2, 5) < x < 3} 88 (0, 3) 61 {x ≤ x ≤ 3} 89 [–1, 2] 62 {x − ≤ x ≤ 1} 90 [–3, 2] 63 {x x < 2} 91 (– @ 64 {x 92 (– –1) 65 {x x ≥ 1} x < −1} 93 [3, INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe 94 [–2, 117 iii 118 ii 95 A ∪ B = {1, 2, 4, 6, 9}; A ∩ B = {4} 96 A ∪ B = {−1, 0, 1, 2}; A ∩ B = {0,1} 97 A ∪ B = {2, 3, 5, 8, 9, 10}; A ∩ B = ∅ 119 A ∪ B is {x − ≤ x ≤ 1} ∪ {x ≤ x ≤ 1} = {x − ≤ x ≤ 1} = A 98 A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}; A ∩ B = ∅ 120 A ∪ A is set A 99 A ∪ B = {– 4, –2, 0, 2, 4, 8}; A ∩ B = {0,4} 121 B ∩ B is set B Applying Concepts 1.1 100 A ∪ B = {–3, –2, –1, 0, 1}; A ∩ B = {-2,-1} 122 A ∪ C is {x – ≤ x ≤ 1} , which is set A 101 A ∪ B = {1, 2, 3, 4, 5}; A ∩ B = {3,4,5} 123 A ∩ R is {x − ≤ x ≤ 1} , which is set A 102 A ∪ B = {0, 1, 2, 3, 4, 5}; A ∩ B = {2,4} 103 {x x > 1} ∪ {x x < −1} 124 C ∩ R is {x − ≤ x ≤ 0} , which is set C 125 B ∪ R is the set of real numbers, R 126 A ∪ R is the set of real numbers, R 104 {x x ≤ 2} ∪ {x x > 4} 127 R ∪ R is the set R 128 R ∩∅ is the empty set ∅ 105 {x x ≤ 2} ∩ {x x ≥ 0} 129 B ∩ C is {x ≤ x ≤ 1} ∩ {x − ≤ x ≤ 0} , which contains only the number 106 {x x > −1} ∪ {x x ≤ 4} 130 –3 > x > means the numbers that are less than –3 and greater than There is no number that is both less than –3 and greater than Therefore, this is incorrect 107 {x x > 1} ∩ {x x ≥ −2} 131 108 {x x < 4} ∩ {x x ≤ 0} 132 109 {x x > 2} ∪ {x x > 1} 133 110 {x x < −2} ∪ {x x < −4} 134 111 (– @ ∪[4, 135 The answer is ii and iii For example: i 112 (–3, 4] ∪ [–1, 5) 113 [–1, 2] ∩ [0, 4] ii 114 [–5, 4) ∩(–2, 115 (2, ∪ (–2, 4] iii 116 (– @ ∪ (4, 5−4 ≤0 3−2 1≤ False 2−3 ≤0 5−4 −1 ≤ True 5−4 ≤0 2−3 −1 ≤ True INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ iv 4−5 ≤0 2−3 1≤ False Chapter hapter 1: Review of Real Numbers Objective 1.2.1 Exercises 11 Rewrite subtraction as addition of the opposite -8 – (-3) -8 + = -5 Projects or Group Activities 1.1 c 136 E = {1,3,5,7,9} 137 Ec = {1,4, 6, 8,9} 138 Ec = {even natural numbers} 139 Ec = {irrational numbers} 140 a E∪Ec = U 12 13 Negative because only the would be raised to the 8th power 14 -45 negative, only raised to the 5th power (-4)5 negative, -4 multiplied together an odd number (5) of times b E ∩ Ec = ∅ -46 negative, only raised to the 6th power (-4)6 positive, -4 multiplied together an even number (6) of times SECTION 1.2 Concept Check Students should paraphrase the rule: Add the absolute values of the numbers, then attach the sign of the addends 15 – (-7) = 2+7 = 16 192 ÷ (-32) = -6 Students should paraphrase the rule: Find the absolute value of each number, subtract the smaller of the two numbers from the larger, then attach the sign of the number with the larger absolute value 17 14 + (-26) = -12 18 (-10)(-27) = 270 19 29 + (-2) = 27 The word minus refers to the operation of subtraction, and the word negative indicates a number that is less than zero 20 24 – (-2) = 24 + = 26 21 (-8)(29) = -232 No, for example -5 + (-3) = -8 22 (-16)(32) = -512 If the product of two numbers is positive, both numbers would be negative or both numbers would be positive 23 (-20)(2) = -40 24 210 ÷ (-30) = -7 25 -16 + 33 = 17 26 -140 ÷ (-28) = a b If quotient of two numbers is negative, one number must be positive and one must be negative Yes For instance -3 – (-7) = -3 + = 27 13 + (-29) = -16 If the product of two numbers is zero, at least one of the numbers is zero 28 (-28)(32) = -896 37 29 -21-6 = -21 + (-6) = -27 30 -16 – 35 = -16 + (-35) = -51 (-5) 10 -2 · ? (8-2) · 31 (-30)(-3) = 90 -10 ? ·5 32 34 + (-6) = 28 -10 ? 30 33 -4 + (-8) = -12 -10 < 30 34 (-5)(12)= -60 35 48 ÷ -12 = -4 36 – = -3 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann Full file at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe 37 -23 – (41) = -23 + (-41) = -64 38 b The quotient − 416 ÷ 52 is negative c The quotient − 693 is positive d The quotient −87 ÷ is undefined −99 39 17 – 21 = -4 40 -27 + = -18 41 -42 + 16 or 16 + (-42) = -26 42 -4-14=-18 = 25 − 125 ÷ 43 -33-21 = -54 44 -21 + (-15) = -36 = 25 − 25 =0 Objective 1.2.2 Exercises 59 52 − (10 ÷ 2)3 ÷ = 52 − ( )3 ÷ 60 We need an Order of Operations Agreement to ensure that there is only one way in which an expression can be correctly simplified 45 = ⋅ ⋅ = 125 46 = ⋅ ⋅ ⋅ = 81 61 47 −2 = −(2 ⋅ ⋅ 2) = −8 27 – 12 ÷ – 52 27 – 12 ÷ – (5)(5) 48 −4 = −(4 ⋅ ⋅ 4) = −64 27 – – 25 49 ( −5) = ( −5)( −5)( −5) = −125 23 – 25 = -2 50 ( −8) = ( −8)( −8) = 64 62 -3 · 23 – 5(-6) 51 ⋅ = (2)(2) ⋅ (3)(3)(3)(3) = ⋅ 81 = 324 52 -3 · 23 – 5(1-7) -3 · (2)(2)(2) -5(-6) ⋅ 33 = (4)(4) ⋅ (3)(3)(3) = 16 ⋅ 27 = 432 -3 · -5(-6) 53 −2 ⋅ 32 = −(2)(2) ⋅ (3)(3) = −4 ⋅ = −36 -24 – (-30) 54 −3 ⋅ = −(3)(3) ⋅ (5)(5)(5) = −9 ⋅ 125 = −1125 -24 + 30 = 55 ( −2) ⋅ ( −3) = ( −2)( −2)( −2) ⋅ ( −3)( −3) = −8 ⋅ = −72 56 57 a 15 – (3)(3)(-2) ÷ 567 + (−812) is negative b −259 − (−327) is positive c The product of four positive numbers and three negative numbers is negative d 58 a 15 – 32(5-7) ÷ 15 – 32(-2) ÷ ( −4) ⋅ ( −2) = ( −4)( −4)( −4) ⋅ ( −2)( −2)( −2) = −64 ⋅ ( −8) = 512 63 The product of three positive numbers and four negative numbers is positive 15 – 9(-2) ÷ 15 – (-18) ÷ 15 – (-3) 15 + = 18 64 − 3(8 ÷ 4)2 = − 3(2)2 = − 3(4) = − 12 = −7 65 − (5 − 2)2 ⋅ = − (3)2 ⋅ = 16 − ⋅ = 16 − 27 = −11 The quotient is zero −91 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ 66 16 − 22 − 4−5 = 16 − 32 + 9+2 −1 = 16 − 11 = 16 + 11 177 = 11 Chapter hapter 1: Review of Real Numbers 73 2(8-11)-12÷3÷4+18·3÷6 =2(-3)-12÷3÷4+18·3÷6 = -6 – 12÷3÷4+18·3÷6 = -6 – ÷4 + 18 · ÷ = -6 – + 18 ·3 ÷6 = - – + 54 ÷ 67 5[(2 − 4) ⋅ − 2] = 5[( −2) ⋅ − 2] = 5[−6 − 2] = 5[−8] = −40 68 2[(16 ÷ 8) − ( −2)] + = 2[2 − ( −2)] + = 2[2 + 2] + = 2[4] + =8+4 = 12 ⎛ − ⎞ ÷ = 16 − ⎛ ⎞ ÷ 69 16 − ⎜ ⎟ ⎜ ⎟ ⎝3−6⎠ ⎝ −3 ⎠ = 16 − 4( −2) ÷ = 16 − ( −8) ÷ = 16 − ( −8) ⋅ = -6 -1 + = -7+9 =2 74 = -3(-4)3 ÷ · (-2) = -3(-64) ÷ · (-2) = 192 ÷ · (-2) = 48 · (-2) = -96 75 = – 12 – 18÷ = -12 - = 32 71 6[3 − ( −4 + 2) ÷ 2] = 6[3 − ( −2) ÷ 2] = 6[3 − ( −1)] = 6[3 + 1] = 6[4] = 24 – · – 18÷ (-3)2 = – · – 18 ÷ = 16 − ( −16) = 16 + 16 16 + ⎞ ⎛ 16 + ⎞ − 70 25 ÷ ⎛⎜ ⎟ − = 25 ÷ ⎜ ⎟ ⎝ −2 + ⎠ ⎝ −4 + ⎠ 24 = 25 ÷ ⎛⎜ ⎞⎟ − ⎝ ⎠ = 25 ÷ 5(6) − = 5(6) − = 30 − = 25 -3(5-9)3÷4 ·(-2) = -6 - = -8 76 -9 -2[4-(3-8)2]÷7-4 = -9 -2[4-(-5)2]÷7-4 = -9 -2[4- 25]÷7-4 = -9 -2[-21] ÷7 -4 = -9 + 42÷7 – = -9 + – = -3 – = -7 72 12 − 4[2 − ( −3 + 5) − 8] = 12 − 4[2 − (2) − 8] = 12 − 4[−8] = 12 − ( −32) = 12 + 32 = 44 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe 77 5+3[52+4(2-5)3]2 81 = 5+3[52+4(-3) ] = 5+3[52+4(-27)]2 = 5+3[52 – 108]2 = +3[56]2 = +3(3136) = + 9408 = 9413 78 -3(5-8)3 + (19-7) ÷ (1-3) = -3(-3)3+ 12 ÷ -2 = -3(-27) - = 81 - = 15 ÷ =3 = 75 79 28 ÷ (7-9)2·(1-3)4÷14 82 iii = 28 ÷ (-2)2 · (-2)4 ÷ 14 32 + 32 ÷ − = 32 + 32 ÷ − = 32 + − = 28 ÷ · 16 ÷ 14 83 ii = · 16 ÷ 14 − 2 (5 − 3)3 = − 2 (2)3 = 64 − 4(8) = 112 ÷ 14 =8 Applying Concepts 1.2 84 1122 = 81,402,749,386,839,761,113,321 The tens digit is 80 85 18 = 1,628,413,597,910,449 The ones digit is 86 533 = 116,415,321,826,934,814,453,125 The last two digits are 25 87 5234 has over 150 digits The last three are 625 88 (23)4 = 212 = 4096 2( ) = 81 = 2,417,851,639,229,258,349,412,352 They not equal each other, and the second expression is larger c (b ) c 89 Find b Then find a Projects or Group Activities 1.2 =7+1 =8 90 Abundant, 1+2+4+5+10 = 22 > 20 91 Perfect, 1+ 2+4+7+14 = 28 = 28 92 Deficient, example 32=9, 1+3 = < INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter hapter 1: Review of Real Numbers 11 SECTION 1.3 Concept Check The least common multiple (LCM) of or more numbers is the smallest number that is a multiple of all the numbers When adding fractions, the LCM of all the denominators is the least common denominator these numbers will have The greatest common factor (GCF) of or more numbers is the greatest number that divides evenly into all numbers When simplifying fractions, you divide both top and bottom by the GCF for the two numbers Yes, all integers are rational numbers because they can be written as fractions by writing the number with a denominator of Example: = 7/1 12 13 14 15 and -2.34 are not integers but are rational Yes, the smallest positive integer = 16 = No, positive rational numbers continue to grow forever No, zero does not have a reciprocal 17 18 Objective 1.3.1 Exercises The LCM of 8, 28, and is 56 , , The reciprocal of is 19 20 To find the quotient of two fractions, take the reciprocal of the second fraction and find the product 21 22 23 24 25 10 26 27 28 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file 10at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe 50 29 30 51 31 52 32 = 33 53 34 = 35 36 = 37 54 38 55 39 40 41 56 42 43 (reduce fractions for smaller numbers) 44 = 57 45 = 46 58 47 48 No For instance, there is no integer between and = Yes, add the two numbers and divide by = Objective 1.3.2 Exercises 49 59 Fractions = INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter apter 1: Review of Real N Numbers 11 66 60 = = 61 67 = 62 68 = 63 64 69 65 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file 12at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe = = 73 70 74 71 75 i, ii, and iv 72 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter apter 1: Review of Real N Numbers 13 0.5833… = 0.583 80 12 -60 100 -96 40 -36 40 -36 0.14545… 81 = 0.145 55 -55 250 -220 300 -275 250 -220 300 -275 25 76 ? needs to be a multiple of so that the fraction will reduce to an integer 0.1919… 82 Objective 1.3.3 Exercises -99 910 -891 190 -99 910 -891 19 0.625 77 = 0.19 99 -48 20 -16 40 -40 0.23076923… = 0.230769 83 13 -26 40 -39 10 -0 100 -91 90 -78 120 -117 30 -26 40 -39 0.3125 78 16 -48 20 -16 40 -32 80 -80 0.166… = 0.16 79 -6 40 -36 40 -36 0.07692307… 84 = 0.076923 26 -0 200 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file 14at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe -182 180 -156 240 -234 60 -52 80 -78 20 -0 200 -182 18 -4 -4 32 -32 96 -0.0009 0.21 -9 -0 18 -18 09 -09 85 86 87 0.0015 -0.0027 -0.0012 88 97 4.5 -0.013 45 0.31 x (-0.1) -0.031 -0 05 -00 58 -45 135 -135 89 -0.0008 +3.5 3.4992 90 91 0.0022 x (-0.8) -0.00176 98 0.02 -0.40 -0.38 0.0003 - 0.39 -0.3897 3.8 x(-3.9) 342 +1140 -14.82 100 -3.5 99 92 3.1 0.026 -.01 -35 31 -0 09 -00 91 -70 210 -210 -0 31 -31 93 -0.024 x -0.019 216 240 0.000456 94 0.0029 -0.003 -0.0001 95 -0.004 101 102 -0.0026 +0.028 0.0254 2.7 +0.007 2.707 INSTRUCTOR USE ONLY -0.018 01 © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter apter 1: Review of Real N Numbers 15 103 0.016 0.012 114 16 -00 19 -16 32 -32 104 -0.18 -0.007 0.187 105 0.4(1.21)+5.8 0.484+5.8 6.284 106 5.4-(0.09) 5.4-1 4.4 107 7-1.5625 5.4375 108 0.49-1.4 -0.91 109 6.44 110 2.8224-4.07 2.8224-14.8962 -12.0738 111 No 5/23 is a rational number, so its decimal representation either terminates or repeats 112 No By the Order of Opeartions Agreement, the correct expression to enter is (2/3)(3/4) The student must use the parentheses in order to get the correct answer 115 Applying Concepts 1.3 113 = INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file 16at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe 117 True 118 True 119 True 120 False; 121 False, 122 True 123 To find a number between two fractions, write equivalent fractions with common denominators and then select a number in between the two numerators For example: is between the two 116 SECTION 1.4 Concept Check Addition, multiplication Addition, multiplication A 1/c Distributive Property For instance, 3x and -4x No, the variable parts are not the same Yes, they have the same variable parts Objective 1.4.1 Exercises The fact that two terms can be multiplied in either order is called the Commutative Property of Multiplication 10 The fact that three or more addends can be added by grouping them in any order is called the Associative Property of Addition 11 The Multiplication Property of Zero tells us that the product of a number and zero is zero 12 The Addition Property of Zero tells us that the sum of a number and zero is the number 13 · = · 14 + 15 = 15 + 15 (3 + 4) + = + (4 + 5) Projects or Group Activities 1.3 16 (3 · 4) · = · (4 · 5) INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ 17 Chapter apter 1: Review of Real N Numbers =(-5)2|2| is undefined =(25)(2) 18 + = =50 19 3(x + 2) = 3x + 44 The volume of a cone with height h and base of radius r is V = π r h If r = in and h = in., then the exact volume of the cone is V = π ( )2 ( ) = ( )π in Use the π key on a calculator to approximate the volume to the nearest hundredth: V ≈ 12.57 in 20 5(y + 4) = 5y + 20 21 13 =0 −6 22 (x + y) + [– (x + y)] = 23 15 17 ( mn) = mn 24 x · = x 45 ab + dc (2)(3) + (–4)(–1) = + = 10 46 2ab – 3dc 25 2(3x) = (2 · 3) · x 26 ab + bc = bc + ab 2(2)(3) − 3( −4)( −1) = 4(3) − 3( −4)( −1) = 12 − ( −12)( −1) 27 A Division Property of Zero 28 The Inverse Property of Addition = 12 − 12 =0 29 The Inverse Property of Multiplication 30 The Commutative Property of Multiplication 31 The Addition Property of Zero 32 The Associative Property of Addition 33 A Division Property of Zero 47 b-a(2c+d) = 3-2[2(-1)+(-4)] =3-2[-2+(-4)] =3-2(-6) 34 The Distributive Property =3 + 12 35 The Distributive Property 36 The Addition Property of Zero = 15 37 The Associative Property of Multiplication 38 The Commutative Property of Addition 39 When the sum of a number n and its additive inverse is multiplied by the reciprocal of the number n, the result is zero 40 When the product of a number n and its reciprocal is multiplied by the number n, the result is n Objective 1.4.2 Exercises 41 “Evaluate a variable expression” means replace the variable expression with a numerical value and simplify the resulting expression 42 The value of a variable is a numerical value that replaces the variable wherever it appears in an expression The value of a variable expression is the number to which the expression simplifies when the value of the variable is substituted into the expression and the expression is simplified 48 b − (d − c )2 32 − [( −4) − ( −1)]2 = 32 − [( −4) + 1]2 = 32 − [−3]2 =9−9 =0 49 (b − a)2 + c [3 − 2(2)]2 + ( −1) = [3 − 4]2 + ( −1) = [−1]2 + ( −1) = + ( −1) =0 50 (b − d)2 ÷ (b − d) [3 − ( −4)]2 ÷ [3 − ( −4)] = (3 + 4)2 ÷ (3 + 4) 43 n2|m-n| if n=-5 and m=-3 = (-5)2|-3-(-5)| =(-5)2|-3+5| = (7)2 ÷ = 49 ữ =7 INSTRUCTOR USE ONLY â Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann Full file 18at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe 51 (bc + a)2 ÷ (d − b) [(3)( −1) + 2] ÷ ( −4 − 3) = [−3 + 2] ÷ ( −7) 2 57 = [−1] ÷ ( −7) =− 52 3(3) − 5( −1) − ( −5) + 14 = = = =2 3(2) − ( −1) − ( −1) + 2d2-a2c3 =2(-4)2 – (2)2(-1)3 58 =2(16) – (4)(-1) 2d − a b − 2c 2( −4) − −8 − = − 2( −1) − ( −2) −10 = 3+2 −10 = = −2 =32 + = 36 53 3b − 5c 3a − c b2-4ac =(3)2-4(2)(-1) = +8 = 17 54 59 a−d b+c − ( −4) 2+4 = = =3 + ( −1) + ( −1) 60 55 a2 + d 2 + ( −4) = + ( −4) = = 3ac − c2 −4 3(2)( −1) 6( −1) − ( −1)2 = − ( −1)2 −4 −4 −6 = − ( −1)2 −4 = − ( −1)2 = −1 = 61 − a a + 2d −2 + 2( −4) = −2 + ( −8) = −2 −6 = −2(6) = −12 62 d b − 2d −4 − 2( −4) = −4 − ( −8) 56 = −4 + = −4 11 = −4(11) = −44 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ 63 a − 4d 3b − c Chapter apter 1: Review of Real N Numbers 66 2(2) − 4( −4) − ( −16) = 3(3) − ( −1) − ( −1) + 16 = 9+1 20 = 10 =2 64 65 3d − b b − 2c 3( −4) − −12 − = − 2( −1) − ( −2) −15 = 3+2 −15 = = −3 −2bc + 19 bc + d ab − c −2(3)( −1) + 3( −1) + ( −4) −3 + ( −4) = −2(3)( −1) + 2(3) − ( −1) − ( −1) −7 6+1 −7 = −2(3)( −1) + = −2(3)( −1) + −1 = −2(3)( −1) + = −2(3)( −1) + = −6( −1) + = 6+1 =7 67 −3d ÷ ab − 4c 2b + c −3( −4) ÷ 2(3) − 4( −1) − ( −4) = −3( −4) ÷ 2(3) + ( −1) + ( −1) 6+4 = −3( −4) ÷ + ( −1) 10 = −3( −4) ÷ = −3( −4) ÷ = 12 ÷ =6 = −3( −4) ÷ 2(d – b) ÷ (3a – c) 2( − − 3) ÷ [3(2) − ( −1)] = 2( −7) ÷ [6 − ( −1)] = 2( −7) ÷ [6 + 1] = 2( −7) ÷ = −14 ÷ = −2 68 2d3 + 4a2b3 = 2(-4)3 + 4(2)2(3)3 = 2(-64) + 4(4)(27) = -128 + 432 = 304 69 −d − c a −( −4)2 − ( −1)3 (2) = −16 − ( −1)(2) = −16 + = −14 70 a c − d (2)2 ( −1) − ( −4)3 = 4( −1) − ( −64) = −4 + 64 = 60 71 −d + ac −( −4)3 + 4(2)( −1) = −( −64) + 8( −1) = 64 − = 56 72 b a 32 = INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file 20at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe 82 SA = s2 73 a = 4 = 256 SA = 6(14 ) SA = 1176 The surface area is 1176 ft 74 ab =8 75 V = LWH V = (14)(10)(6) V = 840 The volume is 840 πr h V = π(6)2 14 V = 168 π 76 V = 83 SA = s2 + ⎛⎜ ⎞⎟ bh ⎝2⎠ SA = + 2(4)(5) SA = 16 + 40 = 56 The surface area is 56 m in r= 1 d = (12) = 2 84 SA = πr 1 d = (2) = 2 SA = π(12 ) V ≈ 527.79 The volume is 1ʌ ft The volume is approximately 527.79 ft SA = π SA ≈ 12.57 The surface areD LV ʌ cm The surface area is approximately 12.57 cm 85 SA = πr + πrh sh V = (32 )5 V = 15 77 V = SA = π(6 ) + π(6)(2) The volume is 15 SA = 72 π + 24 π SA = 96 π SA ≈ 301.59 7KH VXUIDFH DUHD LV ʌ in The surface area is approximately 301.59 in ft 78 V = s3 V = 7.53 V = 421.875 86 SA = πr + πrl The volume is 421.875 m3 1 r = d = (3) = 1.5 πr 2 V = π(1.5)3 V = 4.5π V ≈ 14.14 7KH YROXPH LV ʌ cm The volume is approximately 14.14 cm 79 V = 80 V = πr h r= r= 1 d = (8) = 2 V = π(4 )8 V = 128 π V ≈ 402.12 The vROXPH LV ʌ cm The volume is approximately 402.12 cm 81 SA = 2LW + 2LH + 2WH SA = 2(5)(4) + 2(5)(3) l+ 2(4)(3) SA = 40 + 30 + 24 SA = 94 The surface area is 94 m r= 1 d = (3) = 1.5 2 SA = π(1.5)2 + π(1.5)(9) SA = 2.25π + 13.5π SA = 15.75π SA ≈ 49.48 7KH VXUIDFH DUHD LV ʌ ft The surface area is approximately 49.48 ft 87 Because b > a , the denominator will be negative The numerator will be positive because the product contains an even number of factors Therefore, the result will be a negative number 88 a No; with the irrational number π in the formula, V cannot be a whole number b No; volume is not measured in square units Objective 1.4.3 Exercises 89 If there are two terms with a common variable factor, the Distributive Property allows us to combine the two terms into one term Add the coefficients of the variable factor and write the sum as the coefficient of the common variable factor INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ 90 The four terms of the variable expression x − y + x − are 5x, −6y, 4x, and −8 The terms 5x and 4x are called like terms The coefficient of the y-term is −6 The constant term is −8 91 −5; 25; Commutative; Associative; 4y 92 5x + 7x = 12x 93 3x + 10x = 13x 21 117 4( − a − 2b) − 2(3a − 5b) = −4 a − 8b − a + 10b = −10 a + 2b 118 5(3a − 2b) − 3( −6 a + 5b) = 15a − 10b + 18 a − 15b = 33a − 25b 119 −7(2 a − b) + 2( −3b + a) = −14 a + b − 6b + a = −12 a + b 120 x − 2[ y − 2( x + 3[2 x + y ])] = x − 2[ y − 2( x + x + y )] 94 –8ab – 5ab = –13ab = x − 2[ y − 2(7 x + y )] = x − 2[ y − 14 x − 18 y] 95 12 ⎛⎜ ⎞⎟ x = x ⎝ 12 ⎠ 96 Chapter apter 1: Review of Real N Numbers = x − 2[−17 y − 14 x] = x + 34 y + 28 x = 31x + 34 y (3 y ) = y 97 –3(x – 2) = –3x + 121 x − 4[ x − 4( y − 2[5 y + 3])] 98 –5(x – 9) = –5x + 45 = x − 4[ x − 4( y − 10 y − 6)] = x − 4[ x − 4(–9 y − 6)] 99 (x + 2)5 = 5x + 10 100 – (x + y) = –x – y 101 – (–x – y) = x + y 102 5+2(3x-7) = + 6x – 14 = 6x-9 103 7-3(4a-5) = 7-12a + 15 = -12a+22 104 5v-3(2-4v) = 5v-6+12v = 17v – 105 -3m -2(4m+3) = -3m -8m-6 = -11m-6 106 -3+4(2z-9) = -3 + 8z – 36 = 8z – 39 107 -5-6(2y-3) = -5 -12y + 18 = -12y+13 108 4x – 3(2y – 5) = 4x – 6y + 15 109 –2a – 3(3a – 7) = –2a – 9a + 21 = –11a + 21 = x − 4[ x + 36 y + 24] = x − x − 144 y − 96 = −2 x − 144 y − 96 122 – 2(7x – 2y) – 3(–2x + 3y) = – 14x + 4y + 6x – 9y = – 8x – 5y 123 3x + 8(x – 4) – 3(2x – y) = 3x + 8x – 32 – 6x + 3y = 5x – 32 + 3y 124 110 3x – 2(5x – 7) = 3x – 10x + 14 = –7x + 14 111 2x – 3(x – 2y) = 2x – 3x + 6y = –x + 6y 112 3[a − 5(5 − 3a)] = 3[ a − 25 + 15a] = 3[16 a − 25] = 48 a − 75 113 5[−2 − 6( a − 5)] = 5[ −2 − a + 30] = 5[28 − a] = 140 − 30 a 114 3[ x − 2( x + y )] = 3[ x − x − y ] = 3[− x − y ] = −3 x − 12 y 115 5[ y − 3( y − x)] = 5[ y − y + x] = 5[−2 y + x] = −10 y + 30 x 116 −2( x − y ) + 2(3 y − x) = −2 x + y + y − 10 x = −12 x + 12 y 125 [8 x − 2( x − 12) + 3] = [8 x − x + 24 + 3] = [6 x + 27] = 2x + [14 x − 3( x − 8) − x] = [14 x − x + 24 − x] = [4 x + 24] =x+6 126 a The coefficient of a will be given by (31 − 88), which will be negative b The coefficient of b will be given by (−102 + 256), which will be positive c The constant term will be given by (73 − 73), which will be zero 127 3[5 − 2( y − 6)] = 3(5) − 3[2( y − 6)] = 15 − 6( y − 6) a No b Yes INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file 22at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe In order to get =1, you need a value for x so that the remainder is Applying Concepts 1.4 128 – 4(5x – y) = –20x + 4y The statement is correct; it uses the Distributive Property 129 4(3y + 1) = 12y + The statement is correct; it uses the Distributive Property 130 – 6x = 0x = The statement is not correct; it incorrectly uses the Distributive Property The correct answer is – 6x 131 + 3x = (2 + 3)x = 5x The statement is not correct; it mistakenly uses the Distributive Property It is in an irreducible statement That is, the answer is + 3x 132 3a – 4b = 4b – 3a The statement is not correct; it incorrectly uses the Commutative Property of Addition The correct answer is 3a – 4b = –4b + 3a 133 2(3y) = (2 · 3)(2y) = 12y The statement is not correct; it incorrectly uses the Associative Property of Multiplication The correct answer is (2 · 3)y = 6y 134 x ⋅ = x The statement is correct; it uses the Inverse Property of Multiplication 135 − x + y = y − x The statement is correct; it uses the Commutative Property of Addition SECTION 1.5 Concept Check 5=6 3=4 14-x Yes Objective 1.5.1 Exercises ten more than the product of eight and a number thirteen subtracted from the quotient of negative five and the cube of a number the difference between ten times a number and sixteen times the number 10 The sum of two numbers is 24 To express both numbers in terms of the same variable, let x be one number Then the other number is 24 − x 11 the unknown number: n the sum of the number and two: n + n – (n + 2) = n – n – = –2 12 the unknown number: n the difference between five and the number: 5–n n – (5 – n) = n – + n = 2n – 13 the unknown number: n 137 Commutative, Yes (3 5) = (2 3) 2 = n four-fifths of the number: n 5 12 17 n+ n= n+ n= n 15 15 15 one-third of the number: = True (7 8) = (5 7) 139 5 0 = = 14 the unknown number: n three-eighths of the number: n n n− n= n− n= n 24 24 24 one-sixth of the number: Yes, Associative 140 No, the difference between x and is x-2, where x less than is 2-x No, ten less than m is m-10 but ten less m is 10-m The “than” reverses the order (6x3) ÷ = 18 ÷ has a remainder of 138 5x The product of and x or times x (4x5)÷7 = 20÷7 has a remainder of b y+6 The sum of y and or more than y Projects or Group Activities 1.4 136 a = 15 ÷ = remainder of so x can be x=1 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ 15 the unknown number: n the product of eight and the number: 8n 5(8n) = 40n 16 the unknown number: n two-thirds of the number: n+ n n= n 3 number and three more than the larger number: [2(33 − x) + 6] − ( x + 3) = 66 − x + − x − = 69 − x 27 iii 28 i,ii, iii Objective 1.5.2 Exercises 17 the unknown number: n the product of seventeen and the number: 17n twice the number: 2n 17n – 2n = 15n 18 the unknown number: n six times the number: 6n the total of six times the number and twenty-two: 6n + 22 (6n + 22) = 3n + 11 19 the unknown number: n the square of the number: n2 the total of twelve and the square of the number: 12 + n 23 Chapter apter 1: Review of Real N Numbers n2 − (12 + n2 ) = n2 − 12 − n2 = −12 20 the unknown number: n the square of the number: n2 the difference between the number and seventeen: n – 17 n2 + 11 + (n − 17) = n2 + 11 + n − 17 = n2 + n − 21 the unknown number: n the sum of five times the number and 12: 5n + 12 the product of the number and fifteen: 15n 15n + (5n + 12) = 15n + 5n + 12 = 20n + 12 22 the unknown number: n twice the sum of the number and 11: 2(n + 11) 2(n + 11) – = 2n + 22 – = 2n + 18 23 Let the smaller number be x The larger number is 15 – x The sum of twice the smaller number and two more than the larger number: 2x + (15 – x + 2) = 2x + (17 – x) = x + 17 24 Let the smaller number be x The larger number is 20 – x The difference between five times the larger number and three less than the smaller number: 5(20 – x) – (x – 3) = 100 – 5x – x + = –6x +103 25 Let the larger number be x Then the smaller number is 34 – x The difference between two more than the smaller number and twice the larger number: [(34 – x) + 2] – 2x = 34 – x + – 2x = 36 – 3x 29 The length of a rectangle is eight more than its width To express the length and the width in terms of the same variable, let W be the width Then the length is W + 30 The width of a rectangle is one-third of its length To express the length and the width in terms of the same variable, let L be the length Then the width is L 31 Let A = money to be spent on high-speed rails Then the amount to be spent on the highway is 8A 32 If population of Houston is P, the population of New York is 4P 33 Distance from Earth to the moon: d Distance from Earth to the sun: 390d 34 Length of the longest road tunnel: L Length of the longest rail tunnel: L + 23.36 35 Amount of the first account: x Amount of the second account: 10,000 – x 36 Length of the longer piece: L Length of the shorter piece: – L 37 Flying time between San Diego and New York: t Flying time between New York and San Diego: 13 – t 38 Length of the shorter piece: L Length of the longer piece: 12 – L 39 The measure of angle B: x The measure of angle A is twice that of angle B: 2x The measure of angle C is twice that of angle A: 2(2x) = 4x 40 The width of the rectangle is W The length is three more than twice the width: 2W + Applying Concepts 1.5 41 One-half the acceleration due to gravity: g Time squared: t The product: gt 2 42 The product of m and a: ma 43 The product of A and v : Av 26 Let the larger number be x Then the smaller number is 33 – x The difference between six more than twice the smaller INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file 24at https://TestbankDirect.eu/ Chapter 1: Review iew of Real Numbers Numbe 44 The quotient of k and m: k m The square root of the quotient: k m 14 (-@ ∩ (0, 15 -10 – (-3) = -10 + = -7 Projects or Group Activities 1.5 45 The sum of twice a number and three 46 Four less than five times a number 16 –204 ÷ (–17) = 12 47 Twice the sum of a number and three 17 48 The product of five and four less than a number CHAPTER REVIEW EXERCISES 3 3 :− + = 4 18 3 − ÷ =− ⋅ 8 3/ ⋅ ⋅ 3/ Replace x with the elements in the set and determine whether the inequality is true x > –1 – > –1 False –2 > –1 False > –1 True > –1 True The inequality is true for and 20 -42 – (-3)2 = -16 – = -25 p ∈ {–4, 0, 7} 21 7-3(5-9)2 ÷ · (-2) = =− 19 –4.07 + 2.3 = –1.77 – p = 7-3(-4)2 ÷ · (-2) − − = −4 − =0 = 7-3(16)÷ · (-2) − = −7 = 7-48 ÷ · (-2) {–2, –1, 0, 1, 2, 3} = – 12 · (-2) {x x < −3} = + 24 = 31 {x − ≤ x ≤ 3} A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8} A ∩ B = {2, 3} [–3, 10 {x x < 1} 11 {x x ≤ −3} ∪ {x x > 0} 12 (–2, 4] 13 {x|x>4} ∪ {x|-2[` 22 INSTRUCTOR USE ONLY © Cengage Learning All Rights Reserved Full file at https://TestbankDirect.eu/ .. .Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter 1: Review... Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter hapter... Learning All Rights Reserved Full file at https://TestbankDirect.eu/ Solution Manual for Intermediate Algebra 8th Edition by Aufmann NOT FOR SALE Full file at https://TestbankDirect.eu/ Chapter 1: Review