Solution manual for college algebra with intermediate algebra a blended course 1st edition by beecher

Chapter R Review of Basic Algebra 46 True, since −7 = −7 is true Exercise Set R.1 48 True, since −11 > −13 RC2 The correct answer is (b) See page R-2 in the text RC4 The correct answer is (f) See page R-3 in the text is true 50 x < −1 Ϫ1 RC6 The correct answer is (d) See page R-4 in the text √ 0, 1, 12, 25 −6, 0, 1, −4, 12, √ √ 25, − 52 x ≥ −1 Ϫ1 12 54 x < 3, 0.131331333133331 12 10 −3.43, −11, 12, 0, 56 x ≤ 11 ,− 34 13 12 All of them 58 14 {s, o, l, v, e} 60 16 16 {1, 3, 5, 7, 9, 11} 62 127 18 {−3, −2, −1} 64 20 {x|x is an integer greater than and less than 11} 22 {x|x is a rational number or x is an irrational number}, or {x|x is a real number} 24 {x|x is a real number and x ≤ 21}, or {x|x ≤ 21} = |0| = −15 74 | − 8| ≤ |8|, or | − 8| ≥ |8| since | − 8| = |8| 30 > −11 32 −6 < −3 Exercise Set R.2 34 −7 > −10 RC2 The opposite of a negative number is positive 11 < 250 RC4 The absolute value of a negative number is positive 38 −13.99 < −8.45 RC6 The sum of and a negative number is negative 27 14 40 − = −0.933 and − = −0.509433962 , so 15 53 27 14 x 44 t ≤ 10 68 465 72 | − 5| ≥ | − 2| 28 > −7 36 −13 66 16.4 70 26 18 > 13 2 −12 Copyright c 2017 Pearson Education, Inc 2 Chapter R: Review of Basic Algebra 55 12 −3 68 − 10 −10 70 12 −32 72 432 14 1.9 74 71.04 16 −11.19 18 − 20 − 76 78 − − 20 =− + 24 82 −9 − 84 21 41 =− 24 24 86 0.7 26 88 Not defined 28 30 64 125 80 −8 22 − + = − + = − 8 8 24 − + 65 14 90 92 Not defined 32 2x 94 34 −5 36 −20 38 46 10 96 − 98 − 65 40 100 42 44 44 0.8 This can also be expressed as follows: 1 10 10 = · = = , or 1.25 0.8 0.8 10 46 −17.7 102 8x 48 −34.8 104 13 50 − 52 − − 56 − − 106 − 13 20 − = =− + 8 10 54 − − = − + 15 − ÷ − 12 ÷ 7 = · − =− 10 − 12 =− · 10 − 10 =8 108 −3 12 22 − =− 15 15 110 110 112 −4000 36 35 =− + =− 63 63 63 114 58 −40 ÷ − = · − =− 60 −45 116 − ÷ − 5 =− · − = 62 35 118 − ÷ − =− · − 24 = 25 120 − ÷ − 5 =− · − = 64 −152 66 42.7 Copyright c 2017 Pearson Education, Inc Exercise Set R.3 122 24 124 −1.9 −1.9 10 19 = · =− , or − 0.095 20 20 10 200 126 −17.8 10 178 89 −17.8 = · =− = − , or − 5.5625 3.2 3.2 10 32 16 26 28 128 Not defined 130 10 ,− ; − , ; 1, −1; 0, does not exist; 6.4, , or 10 −6.4 1 ; −a, − 6.4 a 132 26 √ 134 3, π, 4.57557555755557 136 All of them 138 > 30 −3 32 −4 34 x6 = = 1 38 142 Use the definition of subtraction The number that 3 can be added to 11.7 to obtain −7 is −7 − 11.7, or 4 = −19 , or −19.45 −7 − 11 10 20 = 125 =1· = 125 1 125 16 16 = =1· 625 625 625 16 1 =− (−4)3 64 40 9−2 42 n−5 44 (−8)−6 46 [6 − 4(8 − 5)] = [6 − · 3] Exercise Set R.3 = [6 − 12] = −6 RC2 True; see page R-23 in the text 48 10[7 − 4(8 − 5)] = 10[7 − · 3] = 10[7 − 12] = RC4 False; see page R-22 in the text 10[−5] = −50 RC6 False; see page R-20 in the text 50 [9(7 − 4) + 19] − [25 − (7 + 3)] = RC8 True; see page R-22 in the text [9 · + 19] − [25 − 10] = [27 + 19] − 15 = 46 − 15 = 31 x4 52 [48 ÷ (−3)] ÷ t5 (3.8) − − = −16 ÷ − = 64 54 30 · 10 − 18 · 25 = 300 − 450 10 = 36 y 140 123 > −10 = −150 56 (9 − 12)2 = (−3)2 = 92 − 122 = 81 − 144 = −63 12 · · = 729 58 · − 32 − 23 = · − − = 56 − − = 39 14 (−7) · (−7) = 49 60 43 + 20 · 10 + 72 − 23 = 64 + 20 · 10 + 49 − 23 = 64 + 200 + 49 − 23 = 290 16 (0.1)(0.1)(0.1)(0.1)(0.1)(0.1) = 0.000001 18 (−3)(−3)(−3)(−3) = 81 62 (9 · + · 3)2 = (72 + 9)2 = (81)2 = 6561 2 2 16 · · · = 3 3 81 √ 22 64 5000 · (4 + 1.16)2 = 5000 · (5.16)2 = 20 5000 · (26.6256) = 133, 128 Copyright c 2017 Pearson Education, Inc 4 Chapter R: Review of Basic Algebra 66 (43 · − 14 · 7)3 + (33 · 34)2 = 3 110 2.3 (258 − 98) + (1122) = (160) + (1122) = 112 900 4, 096, 000 + 1, 258, 884 = 5, 354, 884 114 −79 68 18 − (2 · − 9) = 18 − (6 − 9) = 18 − (−3) = 21 116 −79 70 (18 − 2)(3 − 9) = 16(−6) = −96 118 23 72 [(−32) ÷ (−2)] ÷ (−2) = 16 ÷ (−2) = −8 120 74 30 · 20 − 15 · 24 = 600 − 360 = 240 − − 15 16 76 16 ÷ (19 − 15)2 − = 16 ÷ (4)2 − = 16 ÷ 16 − = − = −6 78 53 + 20 · 40 + 82 − 29 = 125 + 20 · 40 + 64 − 29 = 125 + 800 + 64 − 29 = 960 80 4000 · (3 + 1.14)2 = 4000 · (4.14)2 = 122 4000 · (17.1396) = 66, 558.4 = 8(7 − 3) 8·4 32 82 = = =8 4 84 = 64 43 = =8 8 = 86 53 − 72 = 125 − 49 = 76 = 88 10(−5) + 1(−1) = −50 − = −51 2(61 · 6−1 − 6−1 · 60 ) 1 6· − ·1 6 1− 10 2· = 6 124 12345679 · = 111, 111, 111 90 14 − 2(−6) + = 14 + 12 + = 33 12345679 · 18 = 222, 222, 222 12345679 · 27 = 333, 333, 333 92 −32 − ÷ − (−2) = −32 − − (−2) = According to this pattern, we have 12345679 · 36 = 444, 444, 444 −32 − + = −32 94 (3 − 8)2 = (−5)2 = 25 √ 126 (π) ≈ 5.047497267 √ ( 2)π ≈ 2.970686424 96 28 − 103 = 28 − 1000 = −972 √ Thus, (π) √ is larger than ( 2)π 98 × 103 − 5000 = × 1000 − 5000 = 2000 − 5000 = −3000 Exercise Set R.4 100 6[9 − (3 − 4)] = 6[9 − (−1)] = 6[9 + 1] = RC2 (c) 6[10] = 60 RC4 (e) 102 1000 ÷ (−100) ÷ 10 = −10 ÷ 10 = −1 104 106 108 · 15 · 16 2·3·5 = 3·2·8 /·3 /·5 = /·2 /·8 = = 20 − 62 20 − 36 −16 = = =− 92 + 32 81 + 90 45 RC6 (g) RC8 (f) 4|6 − 7| − · 4| − 1| − · 4·1−5·4 = = = · − 8|4 − 1| · − 8|3| 6·7−8·3 −16 − 20 = =− 42 − 24 18 t + 11, or 11 + t d − 0.203 18 + z, or z + 18 125 − + 12 · 53 − 32 + 12 · = = −32 ÷ (−16) ÷ (−4) ÷ (−4) 176 125 − + 60 = = −352 1 − − 2 d + c, or c + d 10 c ÷ h, or Copyright c c h 2017 Pearson Education, Inc Exercise Set R.5 12 t + s, or s + t 58 2.56y 2.56(3) 7.68 = = = 0.6 3.2x 3.2(4) 12.8 14 q − p 16 a + b, or b + a Exercise Set R.5 18 2z 20 −6t RC2 The statement · + · = 4(2 + 8) illustrates a distributive law 22 Let w represent the number of women attending Then we have 48%w, or 0.48w RC4 The multiplicative identity is the number RC6 In the expression 7(5 + x), and x are terms 24 d − $19.95 26 65t 30 x = = −5 y −6 5 32 = = = p+q 20 + 30 50 10 34 36 18 · 18m = =7 n 18 −18 −10 −10 x=5 45 25 25 x=0 0 4x − y = · − (−2) 5(x − 2) 5x − 5x − 10 x = −1 −15 −7 −15 = 12 + x = 4.6 13 21 13 = 14 x=0 −10 −2 −10 n3 − + p ÷ n2 = 23 − + 12 ÷ 22 5(x − 2) and 5x − 10 are equivalent; 5(x − 2) and 5x − are not equivalent; 5x − and 5x − 10 are not equivalent = − + 12 ÷ 4 a 4a = ·1= · = 3 a 3a 3 5y 15y = ·1= · = 10 10 10 5y 50y = 102 − 3(10 + 8) 10 · 18y 18y 36y = = · =2 18y · 18y 18y = 100 − 3(18) 12 −125 · 5t −125 5t 125 −625t = = · =− 15t · 5t 5t = 46 14 + y = 8−2+3 = 6+3 =9 40 7x − 2x 5x and 7x − 2x are equivalent; 7x + 2x and 5x are not equivalent; 7x + 2x and 7x − 2x are not equivalent = 12 − (−2) 38 5x x = −2 28 57y = 57(−8) = −456 30 7x + 2x a2 − 3(a − b) = 102 − 3(10 − (−8)) = 10 − 3(18) = 100 − 54 42 I = $18, 000(0.046)(2) = $1656 16 dc 44 A = (3.6 m)(1.9 m) = 6.84 m2 18 pq + 14 = 14 + pq; pq + 14 = 14 + pq = 14 + qp; 46 (−3)(−3)(−3)(−3)(−3) = −243 pq + 14 = qp + 14 48 (−5.3)2 = (−5.3)(−5.3) = 28.09 20 50 (4.5)0 = 1 52 (3x) = 3x (For any nonzero number a, a0 = 1.) s + qt = qt + s; s + qt = qt + s = tq + s; s + qt = s + tq (For any number a, a = a.) 22 (5 · p) · q 54 56 N = P + P (2.7%)(1), so we have N = P + 2.7%P , or N = P + 0.027P , or N = 1.027P Copyright c 24 + (p + q) 2017 Pearson Education, Inc 6 26 Chapter R: Review of Basic Algebra x8 28 256 74 Let x = Then = = = 16, but x2 = 22 = x 16 The expressions are not equivalent (4 + x) + y = + (x + y) = (x + y) + 4; (4 + x) + y = (x + 4) + y = x + (4 + y) = (4 + y) + x; (4 + x) + y = y + (4 + x) = y + (x + 4); Exercise Set R.6 others are possible 28 (8 · m) · n = · (m · n) = · (n · m) = (8 · n) · m; RC2 The terms 9y and 9c have different variables, so they are not similar terms The given statement is false (8 · m) · n = n · (8 · m) = n · (m · 8); RC4 −(5 − x) = −5 + x = x − 5; the given statement is true (8 · m) · n = · (m · n) = (m · n) · = (n · m) · 8; others are possible RC6 −(5c + 6d − w) = −5c − 6d + w = −5c − 6d − w; the given statement is false 30 3c + 32 7b − 7c 15a 34 −6c − 10d −3c 36 5xy − 5xz + 5xw 14x 38 P + P rt 14x 40 1 πr + πrs 4 10 −5x 12 −6x 42 5x, −9y, 12 14 14a − 9b 44 5a, −7b, −9c 16 7a + 9b 46 9(a + b) 18 9p + 12 48 22(x − 1) 20 −11.83a − 36.3b 50 6(y − 6) 52 a(b + 1) 3 22 − x + y − 34 4 54 3(x + y − z) 24 · 520 ft = 1040 ft; 56 4(a + 2b − 1) P = 2l + 2w = · 360 ft + · 160 ft = 58 4(2m + n − 6) 720 ft + 320 ft = 1040 ft 60 6(3a − 4b − 8) 26 5y 62 x(y − z + w) 64 28 −a − h(a + b) 2 66 Let x and y represent the numbers; x + y 68 n 70 P = 2(l + w) = 2(360 ft + 160 ft) = 30 −x + 8, or − x 32 −r + s, or s − r 34 −r − s − t 10 · 25 − · 23 = 10 · 25 − · 36 −9a + 7b − 24 = 250 − 24 38 4x − 8y + 5w − 9z = 226 72 Let a = −1 and b = Then (a−b)(a+b) = (−1−2)(−1+ 2) = −3 and a2 − b2 = (−1)2 − 22 = −3 Let a = and b = −2 Then (a−b)(a+b) = (3−(−2))(3− 2) = and a2 − b2 = 32 − (−2)2 = In fact, the expressions have the same value for all values of x, so they are equivalent Copyright c 40 x − 2y + z + 56.3w 42 6x + 44 5a − (4a − 3) = 5a − 4a + = a + 46 6x − − (9 − 3x) = 6x − − + 3x = 9x − 16 2017 Pearson Education, Inc Exercise Set R.6 48 −9(y + 7) − 6(y − 3) = −9y − 63 − 6y + 18 = = 7b − {5[12b − 32 − 9b − 10] + 14} −15y − 45 = 7b − {5[3b − 42] + 14} 50 8y − 4(5y − 6) + = 8y − 20y + 24 + = = 7b − {15b − 210 + 14} −12y + 33 52 = 7b − {15b − 196} −5t + (4t − 12) − 2(3t + 7) = 7b − 15b + 196 = −5t + 4t − 12 − 6t − 14 = −8b + 196 = −7t − 26 54 7b − {5[4(3b − 8) − (9b + 10)] + 14} 68 70 3d ÷ 2c = · ÷ · 10 1 − (10t − w) + (−28t + 4) + = −5t + w − 7t + + = −12t + w + 2 = 15 ÷ · 10 = 7.5 · 10 = 75 72 x2 ÷ 3(y − z) = 62 ÷ 3(8 − 10) = 62 ÷ 3(−2) 56 14b − [7 − 3(9b − 4)] = 14b − [7 − 27b + 12] = 36 ÷ 3(−2) = 14b − [19 − 27b] = 12(−2) = 14b − 19 + 27b = −24 = 41b − 19 74 16 58 7{−7 + 8[5 − 3(4 + 6)]} = 7{−7 + 8[5 − 3(10)]} = 76 7{−7 + 8[5 − 30]} = 7{−7 + 8[−25]} = 7{−7 − 200} = 7{−207} = −1449 60 [9(x + 5) − 7] + [4(x − 12) + 9] = [9x + 45 − 7] + [4x − 48 + 9] = 9x + 38 + 4x − 39 = 13x − 62 78 −16a + 24b − 32 [6(x + 4) − 12] − [5(x − 8) + 11] 80 16x − 8y + 10 = [6x + 24 − 12] − [5x − 40 + 11] 82 8(3a − 2b) = [6x + 12] − [5x − 29] 84 5(3p + 9q − 2) = 6x + 12 − 5x + 29 = x + 41 64 86 · (7 + 32 · 5) = 104 4{[8(x − 3) + 9] − [4(3x − 7) + 2]} 88 (2 − 7) · 22 + = −11 = 4{[8x − 24 + 9] − [12x − 28 + 2]} 90 = 4{[8x − 15] − [12x − 26]} −3[9(x − 4) + 5x] − 8{3[5(3y + 4)] − 12} = −3[9x − 36 + 5x] − 8{3[15y + 20] − 12} = 4{8x − 15 − 12x + 26} = −3[14x − 36] − 8{45y + 60 − 12} = 4{−4x + 11} = −42x + 108 − 8{45y + 48} = −16x + 44 66 − ÷ =− · 8 /·4 /·1 =− 2·4 /·3 /·3 =− = −42x + 108 − 360y − 384 3{[6(x − 4) + 52 ] − 2[5(x + 8) − 102 ]} = −42x − 360y − 276 = 3{[6(x − 4) + 25] − 2[5(x + 8) − 100]} 92 = 3{[6x − 24 + 25] − 2[5x + 40 − 100]} {x + [f − (f + x)] + [x − f ]} + 3x = {x + [f − f − x] + [x − f ]} + 3x = 3{[6x + 1] − 2[5x − 60]} = {x − x + x − f } + 3x = 3{6x + − 10x + 120} = x − f + 3x = 3{−4x + 121} = 4x − f = −12x + 363 Copyright c 2017 Pearson Education, Inc 8 Chapter R: Review of Basic Algebra 50 − Exercise Set R.7 x16 52 8−4 x16 y −20 z −8 = RC2 (b) RC4 (e) 54 RC6 (d) 56 8 −6 15 b15 a b =− 27 27a6 9−2 = 92 9−7 = 97 84 y 20 z 5−6 = 49 ·4 −4x4 y −2 5x−1 y −4 5x−1 y −4x4 y −2 = 54 y 24 44 x20 58 −200x3 y −5 8x5 y −7 −4 = x8 x8 , or 254 y 58 y 8 a 10 60 12 (9y)2 · (2y)3 = 81y · 8y = 648y 14 −18x7 y 16 −24x−12 y = − 24y x12 18 −36x−12n = − 36 x12n 20 15t−5a = 9−2 x−4 y 3−3 x−3 y 38 x8 y = 8x5 y −7 −200x3 y −5 3−4 x−4 y 3−3 x−3 y 5y −4x5 = = = 3xy 15 t5a 64 2x2 y −2 3x8 y = 3x6 y 9 = 32 m 1212 70 −4xb+5−(b−5) y 4+c−(c−4) = −4xb+5−b+5 y 4+c−c+4 = −4x10 y 1 = , or −2t 72 76pq y 11t 74 x3ab−3b 76 45c x15ac y 10bc = 2t m 78 −5xa+b−(a−b) y b−a−(b+a) = −5xa+b−a+b y b−a−b−a = 34 y −5x2b y −2a , or − 36 −3ab2 38 − 82 2.63 × 10−7 7a b 84 340, 000 = 3.4 × 105 ; 322, 000, 000 = 3.22 × 108 86 200 billionths = 200 × 0.000000001 = 0.0000002 0.0000002 ↑ places 42 520 44 9−12 = 5x2b y 2a 80 2.6 × 1012 7a−4 b2 7b2 =− 4 4a 16 40 = 29 39 x54 y 81 68 −3xa+1−(2−a) = −3xa+1−2+a = −3x2a−1 30 y −11t = = a48 b60 412 24 513 28 62 [(−4a−4 b−5 )−3 ]4 = [(−4)−3 a12 b15 ]4 = 4−12 a48 b60 = 66 11b+2−(3b−3) = 11b+2−3b+3 = 11−2b+5 −2 = x2 −25y 22 76 26 12−12 = 912 46 740 Small number, so the exponent is negative 48 32x15 y 20 0.0000002 = × 10−7 Copyright c 2017 Pearson Education, Inc Chapter R Summary and Review: Concept Reinforcement 88 92,400,000 112 = −6t − 5t + 13 + − 12t 90 0.000000000000000000000000000911 g = −23t + 21 92 3,240,000 tracks 114 94 33.8 × 10 −5 −6t − (5t − 13) + 2(4 − 6t) = 3.38 × 10 −4 54 − 38 · 24 − (16 − · 18) = 625 − 38 · 24 − (16 − · 18) 96 26.732 × 103 = 2.6732 × 104 = 625 − 38 · 24 − (16 − 72) = 625 − 38 · 24 − (−56) 98 1.5 × 103 = 625 − 38 · 24 + 56 100 × 10−5 102 First we find the number of seconds in 31 days 24 hr 60 60 sec 31 days × × × = 2, 678, 400 sec = day hr 2.6784 × 106 sec = 625 − 912 + 56 = −287 + 56 = −231 116 20 − (5 · − 8) = 20 − (20 − 8) = 20 − 12 Also, we have 818 = 8.18 × 102 =8 Now we find the number of hot dogs consumed 6 (8.18×10 )×(2.6784×10 ) = (8.18×2.6784)×(10 ×10 ) ≈ 21.9 × 10 = (2.19 × 10) × 108 118 (−3x−2 y )−3 (2x4 y −8 )−2 = 2.19 × 10 hot dogs 104 106 4.704 × 1013 = 0.8 × 10 = light years 5.88 × 1012 2π(6.71 × 107 ) ≈ 42.16 × 107 = (4.216 × 10) × 10 = 4.216 × 108 mi 120 (mx 108 hour = 60 minutes = 60(60 seconds) = 3600 seconds −bx x2 +bx n = (−3)−3 x6 y −15 2−2 x−8 y 16 = (−3)−3 x14 2−2 y 31 = 3−6 x28 2−4 y 62 = 16x28 729y 62 2 2 )(mbx n−bx ) = mx nx 122 (xa+b y a+b )c = xac+bc y ac+bc The amount of water discharged in one hour is Chapter R Vocabulary Reinforcement (4, 200, 000 ft3 /sec) × (3600 sec) 3 = (4.2 × 10 ft /sec) × (3.6 × 10 sec) The sentence x > −4 is an example of an inequality = (4.2 × 3.6)(106 × 103 ) ft3 = 15.12 × 109 ft3 In the notation 74 , the number is called the base = (1.512 × 101 ) × 109 ft3 A variable is a letter that can represent various numbers = 1.512 × 1010 ft3 In the expression 6x, the multipliers and x are factors year = 365 days = 365(24 hours) = 365 · 24 · 3600 seconds = 31,536,000 seconds The number 5.93 × 107 is written in scientific notation The amount of water discharged in one year is The product of reciprocals is (4, 200, 000 ft3 /sec) × (31, 536, 000 sec) The sum of opposites is = (4.2 × 106 ft3 /sec) × (3.1536 × 107 sec) The commutative law for addition states that a+b = b+a = (4.2 × 3.1536)(106 × 107 ) ft3 13 = 13.24512 × 10 ft = (1.324512 × 10 ) × 1013 ft3 14 = 1.324512 × 10 110 Chapter R Concept Reinforcement ft The statement is false For example, let a = and b = Then a − b = − = −1, but b − a = − = 1 · 70 × 1026 = 17.5 × 1026 = 1.75 × 10 × 1026 = 1.75 × 1027 oxygen atoms Copyright The statement is true See page R-2 in the text c 2017 Pearson Education, Inc 10 Chapter R: Review of Basic Algebra The statement is true We have −(−a) = a < 13 Zero is neither positive nor negative The given statement is false The statement is false because |0| = The statement is true See page R-15 in the text The statement is true See page R-12 in the text 13 + − We find the difference of the absolute values, 52 21 31 13 − = − = Since the negative number 28 28 28 has the larger absolute value, the answer is negative 31 13 =− + − 28 14 −8 − (−3) = −8 + = −5 The statement is true See page R-50 in the text 15 −17.3 − 9.4 = −17.3 + (−9.4) = −26.7 Chapter R Review Exercises 16 The rational numbers can be named as quotients of integers with nonzero divisors Of the given numbers, the rational numbers are 2, − , 0.4545, and −23.788 = 13 13 19 + = + = 4 4 The product of two negative numbers is positive We multiply the absolute values, 3.8(2.7) = 10.26, and make the answer positive (−3.8)(−2.7) = 10.26 14 The product of a negative number and a positive number is negative We multiply the absolute values and make the answer negative 2·9 2·3·3 2·3 · = = = · 14 · 14 3·2·7 2·3 =− Thus, − 14 19 > x The inequality x < 19 has the same meaning −13 ≥ is false since neither −13 > nor −13 = is true 7.01 ≤ 7.01 is true since 7.01 = 7.01 is true x > −4 We shade all numbers to the right of −4 and use a parenthesis at −4 to indicate it is not a solution x ≤ We shade all the numbers to the left of and use a bracket at to indicate it is also a solution The distance of −7.23 from is 7.23, so | − 7.23| = 7.23 10 The distance of − 9, or 0, from is 0, so |9 − 9| = 11 + (−8) We find the difference of the absolute values, − = Since the negative number has the larger absolute value, the answer is negative + (−8) = −2 19 −6(−7)(4) = 42(4) = 168 20 When a negative number is divided by a positive number, the answer is negative −12 −12 ÷ = = −4 21 When a negative number is divided by a negative number, the answer is positive −84 = 21 −4 22 When a positive number is divided by a negative number, the answer is negative 49 = −7 −7 23 24 12 −3.8 + (−4.1) The sum of two negative numbers is negative We add the absolute values, 3.8 + 4.1 = 7.9, and make the answer negative −3.8 + (−4.1) = −7.9 Copyright 13 18 − Since −3.9 is to the left of 2.9, we have −3.9 < 2.9 Ϫ4 − 17 (−3.8)(−2.7) We specify the conditions by which we know whether a number is in the set {x|x is a real number less than or equal to 46} − c 25 ÷ − 6·2 10 = · − 10 7 · =− =− 6·2 12 − 15 − ÷ − 16 2·5 8 · = 2·5 3 25 =− · − =− 16 15 Not defined: Division by 2017 Pearson Education, Inc 5·7 5·7 =− = · 10 6·2·5 = · 16 5·2·8 = = · 15 2·3·5 Chapter R Summary and Review: Review Exercises 26 −108 ÷ 4.5 = −108 = −24 4.5 11 44 2x − 14 2x − 2(x − 7) 2x + 14 27 When a = −7, then −a = −(−7) = x = −1 −16 −9 −16 12 28 When a = 2.3, then −a = −2.3 x = 10 13 34 29 When a = 0, then −a = −0 = x=0 −14 −7 −14 14 30 The values of 2x−14 and 2(x−7) are the same for the given values of x and, indeed, for any allowable replacement for x Thus, they are equivalent a · a · a · a · a = a5 factors 31 − − − = − There is no other pair of expressions that is the same for the given values of x Thus, there are no other equivalent expressions factors 32 a −4 33 34 3x 45 Since 9x = · 3x, we multiply by using as a name for 3x 3x 21x = · = 3 3x 9x = a = x−8 x8 23 − 34 + (13 · + 67) = − 81 + (13 · + 67) 46 −84x 7x(−12) 7x −12 −12 = = · = = −12 7x 7x · 7x 1 = − 81 + (65 + 67) 47 11 + a = a + 11 = −73 + 132 48 8y = y · = − 81 + 132 = 59 Commutative law of addition Commutative law of multiplication 49 (9 + a) + b = + (a + b) Associate law of addition 35 64 ÷ (−4) + (−5)(20) = −16 − 100 = −116 50 8(xy) = (8x)y 36 Let x represent the number Then we have 5x 51 −3(2x − y) = −3 · 2x − 3(−y) = −6x + 3y 37 Let y represent the number Then we have 28%y, or 0.28y 52 4ab(2c + 1) = 4ab · 2c + 4ab · = 8abc + 4ab 38 We have t − 53 5x + 10y − 5z = · x + · 2y − · z = 5(x + 2y − z) a 39 Let a and b represent the numbers Then we have − b Associate law of multiplication 54 ptr + pts = pt · r + pt · s = pt(r + s) 55 2x + 6y − 5x − y = (2 − 5)x + (6 − 1)y = −3x + 5y 40 Substitute −2 for x and carry out the calculations 5x − = 5(−2) − = −10 − = −17 56 7c − + 9c + − 4c = (7 + − 4)c + (−6 + 2) = 12c − 41 Substitute for x and 20 for y and carry out the calculations x−y − 20 −16 = = = −8 2 57 42 We usually consider length to be longer than width so we substitute 12 for l and for w and carry out the calculation (The result is the same if we substitute in the opposite order.) 58 4(x − 3) − 3(x − 5) = 4x − 12 − 3x + 15 = x + A = lw = 12 · = 84 ft2 = 9c − 4d + 60 16 36 x = 10 95 225 25 105 x=0 −5 25 25 7x − [4 − 5(3x − 2)] = 7x − [4 − 15x + 10] = 7x − [14 − 15x] = 7x − 14 + 15x x2 − (x + 5)2 (x − 5)2 x2 + −4 = 22x − 14 61 4m − 3[3(4m − 2) − (5m + 2) + 12] = 4m − 3[12m − − 5m − + 12] = 4m − 3[7m + 4] There is no pair of expressions that is the same for the given values of x, so there are no equivalent expressions Copyright Changing the sign of every term inside parentheses 59 12x − 3(2x − 5) = 12x − 6x + 15 = 6x + 15 43 Substitute to find the value of each expression x = −1 −(−9c + 4d − 3) c = 4m − 21m − 12 = −17m − 12 2017 Pearson Education, Inc 12 62 Chapter R: Review of Basic Algebra (2x4 y −3 )(−5x3 y −2 ) = 2(−5) · x4 · x3 · y −3 · y −2 69 First we convert 5.0 mils to dollars −5 = −10x y 5.0 × $0.001 = $0.005 10x7 =− y 63 64 65 Next we convert 0.005 and 13.4 million to scientific notation 0.005 = × 10−3 −15x2 y −5 −15 2−6 −5−(−8) = x y 10x6 y −8 10 = − x−4 y −5+8 = − x−4 y 3y =− 2x 13.4 million = 13, 400, 000 = 1.34 × 107 Finally we multiply to find the revenue (1.34 × 107 )(5 × 10−3 ) = (1.34 × 5)(107 × 10−3 ) = 6.7 × 104 The revenue will be $6.7 × 104 (−3a−4 bc3 )−2 = (−3)−2 (a−4 )−2 b−2 (c3 )−2 a−4(−2) b−2 c3(−2) = (−3)2 = a8 b−2 c−6 a8 = 9b c −2x4 y −4 3x−2 y −4 = = = 3x−2 y −2x4 y −4 2x + y = y + 2x by the commutative law of addition, so answer C is not correct 2x + y = y + 2x = y + x · by the commutative laws of addition and multiplication, so answer D is not correct We cannot express 2x + y as 2y + x using the commutative laws of addition and multiplication, so answer A is correct 72 (xy · x3y )3 = (xy+3y )3 = (x4y )3 = x12y 73 a−1 b = (2x )−1 (2x+5 ) = (2−x )(2x+5 ) = 2−x+x+5 = 25 = 32 34 (x−6 )4 (y 10 )4 = (−2)4 74 3x − 3y = 3(x − y), so (a) and (i) are equivalent 81x−6·4 y 10·4 = 16 (x−2 )5 = x−10 , so (d) and (f) are equivalent x(y + z) = xy + xz, so (h) and (j) are equivalent 81x−24 y 40 = 16 = 66 There are no other equivalent expressions 75 Answers may vary Five rational numbers that are not integers are , − , , −0.001, and 1.7 They are not integers because they are not whole numbers or opposites of whole numbers 81y 40 16x24 2.2 107 2.2 × 107 = × −3 −3 3.2 × 10 3.2 10 = 0.6875 × 1010 76 The quotient 7/0 is defined to be the number that gives a result of when multiplied by There is no such number, so we say the quotient is not defined = (6.875 × 10−1 ) × 1010 = 6.875 × (10−1 × 1010 ) = 6.875 × 109 67 − 4(−4) + 16 21 x − 4y = = = =7 3 3 Answer D is correct 71 2x+y = x·2+y by the commutative law of multiplication, so answer B is not correct 3x−2−4 y 6−(−4) −2 3x−6 y 10 −2 70 77 No; the area is quadrupled For a triangle with base b and height h, A = bh For a triangle with base 2b and height 1 2h, A = · 2b · 2h = 2bh = bh 2 (3.2 × 104 )(4.1 × 10−6 ) = (3.2 × 4.1)(104 × 10−6 ) = 13.12 × 10−2 = (1.312 × 10) × 10−2 = 1.312 × (10 × 10−2 ) 78 No; the area is quadrupled For a parallelogram with base b and height h, A = bh For a parallelogram with base 2b and height 2h, A = 2b · 2h = 4(bh) = 1.312 × 10−1 68 We divide 2.4 × 1013 2.4 1013 = × 12 12 5.88 × 10 5.88 10 ≈ 0.408 × 10 × 106 = 0.25 × 106 = 20 2.5 × 105 bills, and 2.5 × 105 bills would weigh 2.5 × 105 × 2.20 × 10−3 = 5.5 × 102 , or 550 lb Thus, it is not possible that a criminal is carrying $5 million in $20 bills in a briefcase 79 $5 million in $20 bills contains = (4.08 × 10−1 ) × 10 = 4.08 light-years Copyright c 2017 Pearson Education, Inc Chapter R Test 13 80 For 5n , where n is a natural number, the one’s digit will be Since this is not the case with the given calculator readout, we know that the readout is an approximation 13 −6 − (−5) = −6 + = −1 14 −18.2 + (−11.5) = −29.7 15 Chapter R Test 19 − − = 19 19 25 + = + = 4 4 16 (−4.1)(8.2) The product of a negative number and a positive number is negative We multiply the absolute values, (4.1)(8.2) = 33.62, and make the answer negative The irrational numbers are the numbers that cannot be named as quotients of integers with nonzero divisors That is, the irrational numbers are the numbers that are not √ rational They are and π We specify the conditions by which we know whether a number is in the set (−4.1)(8.2) = −33.62 15 − 16 The product of two negative numbers is positive We multiply the absolute values and make the answer positive 4·3·5 4·5 3 · 15 15 = = = · = 16 · 16 5·4·4 4·5 4 17 − {x|x is a real number greater than 20} Since −4.5 is to the right of −8.7, we have −4.5 > −8.7 a ≤ The inequality ≥ a has the same meaning Thus, − −6 ≥ −6 is true since −6 = −6 is true x > −2 We shade all numbers to the right of −2 and use a parenthesis at −2 to indicate that it is not a solution The distance of from is 0, so |0| = The distance of − − 15 16 = 18 −6(−4)(−11)2 = 24(−11)2 = (−264)2 = −528 −8 ≤ −6 is true since −8 < −6 is true Ϫ4 Ϫ3 Ϫ2 Ϫ1 7 7 = from is , so − 8 8 19 When a negative number is divided by a negative number, the answer is positive −75 −75 ÷ (−5) = = 15 −5 20 When a negative number is divided by a positive number, the answer is negative −10 = −5 21 − ÷ 10 + (−9) We find the difference of the absolute values, − = Since the negative number has the larger absolute value, the answer is negative + (−9) = −2 − 15 16 22 −459.2 ÷ 5.6 = 23 −3 =− · − 16 15 = 80 = 30 −459.2 = −82 5.6 Not defined: Division by 24 When a = −13, then −a = −(−13) = 13 11 −5.3 + (−7.8) The sum of two negative numbers is negative We add the absolute values, 5.3 + 7.8 = 13.1, and make the answer negative −5.3 + (−7.8) = −13.1 12 − + − 2 The sum of two negative numbers is negative We add 12 = and make the answer the absolute values, + = 2 negative = −6 − + − 2 25 When a = 0, then −a = −0 = 26 q · q · q · q = q4 factors 27 28 = a−9 a9 − (2 − 5)2 + ÷ 10 · 42 = − (−3)2 + ÷ 10 · 42 = − + ÷ 10 · 16 = − + 0.5 · 16 = 1−9+8 = −8 + =0 Copyright c 2017 Pearson Education, Inc 14 29 Chapter R: Review of Basic Algebra 7(5 − · 3) − 32 7(5 − 6) − 32 = 42 − 32 16 − 7(−1) − 32 = 7(−1) − = −7 − = 16 −16 , or − = 7 39 t + = + t 40 + (t + w) = (3 + t) + w 42 −2(3a − 4b) = −2 · 3a − · (−4b) = −6a + 8b 43 3πr(s + 1) = 3πr · s + 3πr · = 3πrs + 3πr 44 ab − ac + 2ad = a · b − a · c + a · 2d = a(b − c + 2d) x 31 Let x and y represent the numbers Then we have − 12 y 32 Substitute for x and −4 for y and carry out the calculations 45 2ah + h = h · 2a + h · = h(2a + 1) 46 6y − 8x + 4y + 3x = (6 + 4)y + (−8 + 3)x = 10y − 5x 47 4a − + 17a + 21 = (4 + 17)a + (−7 + 21) = 21a + 14 48 3x − 3y = · − 3(−4) = + 12 = 18 33 Substitute for b and 2.5 for h and carry out the calculations A = bh = (3)(2.5) = (1.5)(2.5) x(x − 3) x2 − 3x x = −1 4 x = 10 70 70 x=0 0 50 4x − [6 − 3(2x − 5)] = 4x − [6 − 6x + 15] = 4x − [21 − 6x] = 4x − 21 + 6x 51 52 −12x3 y −4 −12 3−7 −4−(−6) = x y 8x7 y −6 = − x−4 y −4+6 = − x−4 y 2 3y =− 2x (3a4 b−2 )(−2a5 b−3 ) = 3(−2) · a4 · a5 · b−2 · b−3 = −6a9 b−5 3x + 5x2 8x2 x = −1 x = 10 530 800 x=0 0 =− 53 6a9 b5 (5a4n )(−10a5n ) = 5(−10) · a4n · a5n = −50a4n+5n Although the expressions have the same value for x = 0, they are not the same for all of the given values of x, so they are not equivalent 9x as a name 36 Since 36x = · 9x, we multiply by using 9x for 3 9x 27x = · = 4 9x 36x −18x · −18x 3 −54x = = · = −36x −18x · −18x 2 38 pq = qp Changing the sign of every term inside parentheses = 10x − 21 The values of x(x−3) and x2 −3x are the same for the given values of x and, indeed, for any allowable replacement for x Thus, they are equivalent 37 −(−9x + 7y−22) = 9x−7y + 22 49 −3(x + 2) − 4(x − 5) = −3x − − 4x + 20 = −7x + 14 = 3.75 cm2 35 Associative law of addition 41 (4a)b = 4(ab) Associative law of multiplication 30 t + 9, or + t 34 Commutative law of addition = −50a9n 54 55 −60x3t −60 3t−7t = x = −5x−4t , or − 4t 12x7t 12 x (−3a−3 b2 c)−4 = (−3)−4 (a−3 )−4 (b2 )−4 c−4 = a−3(−4) b2(−4) c−4 (−3)4 12 −8 −4 a b c = 81 a12 = 81b8 c4 Commutative law of multiplication Copyright c 2017 Pearson Education, Inc Chapter R Test 56 −5a−2 b8 10a10 b−4 15 −4 = 10a10 b−4 −5a−2 b8 10 10−(−2) −4−8 a b −5 = [−2a12 b−12 ]4 = = (−2)4 (a12 )4 (b−12 )4 = 16a48 b−48 = 57 16a48 b48 0.00004.37 ↑ places Small number, so the exponent is negative 0.0000437 = 4.37 × 10−5 58 (8.7 × 10−9 )(4.3 × 1015 ) = (8.7 × 4.3)(10−9 × 1015 ) = 37.41 × 106 = (3.741 × 10) × 106 = 3.741 × 10 × 106 = 3.741 × 107 59 1.2 10−12 1.2 × 10−12 = × 6.4 × 10−7 6.4 10−7 = 0.1875 × 10−5 = (1.875 × 10−1 ) × 10−5 = 1.875 × (10−1 × 10−5 ) = 1.875 × 10−6 60 First we convert 0.002 to scientific notation 0.002 = × 10−3 Now we multiply to find the mass of Pluto (2 × 10−3 )(5.98 × 1024 ) = (2 × 5.98)(10−3 × 1024 ) = 11.96 × 1021 = (1.196 × 10) × 1021 = 1.196 × (10 × 1021 ) Answer C is correct = 1.196 × 1022 kg 61 (x−3 )−4 = x12 , so (b) and (e) are equivalent + 5x = 5x + = 5(x + 1), so (d), (f), and (h) are equivalent 5(xy) = (5x)y, so (i) and (j) are equivalent There are no other equivalent expressions Copyright c 2017 Pearson Education, Inc ... given values of x and, indeed, for any allowable replacement for x Thus, they are equivalent a · a · a · a · a = a5 factors 31 − − − = − There is no other pair of expressions that is the same for. .. 10−2 = 1.312 × (10 × 10−2 ) 78 No; the area is quadrupled For a parallelogram with base b and height h, A = bh For a parallelogram with base 2b and height 2h, A = 2b · 2h = 4(bh) = 1.312 × 10−1 68... shade all numbers to the right of −4 and use a parenthesis at −4 to indicate it is not a solution x ≤ We shade all the numbers to the left of and use a bracket at to indicate it is also a solution