Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file CHAPTER at https://TestbankDirect.eu/ REVIEW 12 2.3 3π 2.75 SECTION 1.1 The Language of Algebra a C = 2t + 15 25 b l = w c P = u – 13 a false b true 15 a –16 > –17 p 6.0 6.5 7.0 7.5 8.0 14 a false b true T = 30p 30(6.0) = 180 30(6.5) = 195 30(7.0) = 210 30(7.5) = 225 30(8.0) = 240 b – (–1.8) < 16 a false b true 17 − 18 = 18 w – 43 18 − − 6.26 = −6.26 ( x − 10 ) SECTION 1.3 Operations with Real Numbers SECTION 1.2 The Real Number System 19 –13 + (–14) = –27 a b 0, 20 –70.5 + 80.6 = 10.1 1 21 − − = − i − 2 =− − 4 =− a –5, 0, 15 b –5, 0, 2.4, 7, − , − 3.6, a − 3, π , 0.13242368 b all 22 –6 – (–8) = –6 + =2 −5, − 3, − , − 3.6 15 b 2.4, 7, π, , 0.13242368… a 23 (–4.2)( –3.0) = 12.6 24 − a b none 10 a b –5, 25 11 Chapter Review Full file at https://TestbankDirect.eu/ - 76 - 5 ⋅ =− 10 16 160 =− 32 − 2 = − 11 Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file at https://TestbankDirect.eu/ 9 41 − (5 − 9) = − (−4) = − 16 = −12 26 − ÷ 21 = − ⋅ 8 21 =− 56 42 + 6[−1 − 5(25 − 33 )] = + 6[−1 − 5(25 − 27)] = + 6[−1 − 5(−2)] 27 15 – 25 – 23 = –10 – 23 = –33 = + 6[−1 + 10] = + 6[9] = + 54 28 –3.5 + (–7.1) + 4.9 = –10.6 + 4.9 = –5.7 = 58 29 –3(–5)( –8) = 15(–8) = –120 43 −1.3 + (−2.7) = −4 = 2(4) =8 30 –1(–1)( –1)( –1) = 1(–1)( –1) = –1(–1) =1 31 (–3)5 = –243 44 ⎛ ⎞ ( −2 ) 32 ⎜ − ⎟ = ⎝ 9⎠ = 81 2 ) 16 + + 32 36 − ( ) 33 36 − 25 33 = 11 =3 34 –52 = – (5)(5) = –25 =2 ⎛ −6 ⎞ 45 ( −10 ) ⎜ ⎟ ( −1) = ( −1,000 )( 3)( −1) ⎝ −2 ⎠ = −3,000 ( −1) 36 − 100 = −10 38 ( (1) +32 36 − ( + 1) = 37 36 − = = 33 ( 0.4 ) = 0.064 35 ( − ) +32 = 3,000 9 = 25 25 = 46 −(−2 ⋅ 4) ÷ ⋅ = −(−8) ÷ ⋅ = −64 ÷ ⋅ = −8 ⋅ = −16 0.64 = 0.8 39 −6 + 2(−5) = −6 + 2(25) = −6 + 50 ( 47 ( x + y ) ( x2 − xy + y2 ) = ( −2 + 4) ( −2)2 − ( −2)( 4) + ( 4)2 = 44 = ( 2)( + +16) −20 40 − ( −3)( −2 ) − = −5 − ( −1) = −5 + =1 ( ) = ( 2)( 28) = 56 - 77 - Full file at https://TestbankDirect.eu/ Chapter Review ) Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file at https://TestbankDirect.eu/ 2 48 −b − b − 4ac −(−3) − (−3) − 4(2)(−2) = 2a 2(2) 63 (3x – 2y)1.2 = 3.6x – 2.4y 64 − + 16 = − 25 = 3−5 = −2 = =− ( ) 3 8c − 4c + = 6c − 3c + 4 65 8(6k) = 48k 66 (–7.5x)( –10y) = 75xy 67 –9(–3p)( –7) = –189p 68 15a + + 30a + = 45a + 16 69 3g2 + g2 – 3g2 – g2 = SECTION 1.4 Simplifying Algebraic Expressions Using Properties of Real Numbers 70 –m+4(m–12n) – (–8n) = –m + 4m – 48n + 8n = 3m – 40n 49 3(x + 7) = 3x + 21 71 50 t · = 5t 51 –x + x = 52 (27 + 1) + 99 = 27 + (1 + 99) 72 21.45l – 45.99l = –24.54l 53 ⋅ = ( = 40a3 − 16a 56 5(–9)(0)(2,345) = 74 57 (–3 · 5)2 = –3(5 · 2) 58 (t + z) · t = (z + t) · t 60 a b 102 =1 102 − 25 = −25 75 – x ≟ 2x + 24 – (–6) ≟ 2(–6) + 24 + ≟ –12 + 24 12 = 12 Yes, –6 is a solution 61 8(9x + 6) = 72x + 48 62 –(6y – 2) = –6y + Full file at https://TestbankDirect.eu/ 27 5 ( 2h + ) − ( h − 1) = h + − h + 4 4 4 32 = h+ 4 = h+8 SECTION 1.5 Solving Linear Equations Using Properties of Equality =0 5.88 = undefined Chapter Review ) = ⎡⎣10a3 − 4a ⎤⎦ 55 · 9.87 = 9.87 b ) ( 73 ⎡⎣ −2 a − a − 3a − 6a ⎤⎦ = ⎡⎣ −2a3 + 2a − 6a + 12a3 ⎤⎦ 54 + m = m 59 a 7 x− x= ⋅ x− ⋅ x 5 4 28 15 = x− x 20 20 13 = x 20 - 78 - Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file at https://TestbankDirect.eu/ (x − 3) = −12 76 82 − x − = x + 20 ? (− − 3) =− 12 ? (− 9) =− 12 − x − − x = x + 20 − x − 3x − = 20 − 3x − + = 20 + -15 ≠ -12 No, –6 is not a solution − x = 27 − 3x 27 = −3 −3 x = −9 77 x = −45 ⎛ x⎞ ⎜ ⎟ = ( −45 ) ⎝5⎠ x = −225 78 83 −4( y − 1) + ( −3) = 25 −4 y + + ( −3) = 25 −4 y + = 25 −4 y + − = 25 − −4 y = 24 t − 3.67 = 4.23 −4 y 24 = −4 −4 y = −6 t − 3.67 + 3.67 = 4.23 + 3.67 t = 7.9 79 − 3x − + x = x + 20 − x 84 0.0035 = 0.25 g 0.0035 0.25 g = 0.25 0.25 0.014 = g + ⎡⎣ − 13 ( x − 1) ⎤⎦ = 17 − 18 x + 3[ − 13x + 13] = 17 − 18 x + 3[ −13x + 15] = 17 − 18 x − 39 x + 45 = 17 − 18 x 80 −39 x + 50 = 17 − 18 x 0= x+4 −39 x + 50 + 18 x = 17 − 18 x + 18 x 0−4= x+4−4 −21x + 50 = 17 −4 = x −21x + 50 − 50 = 17 − 50 x = −4 81 −21x = −33 −21x −33 = −21 −21 11 x= 11 − x = −1 11 − x − 11 = −1 − 11 −5 x = −12 −5 x −12 = −5 −5 12 x= - 79 - Full file at https://TestbankDirect.eu/ Chapter Review Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file 85 at https://TestbankDirect.eu/ 89 4t + t + t − − = 6 ⎛ 4t + t + ⎞ ⎛ t − ⎞ − 6⎜ ⎟ = 6⎜ ⎟ ⎠ ⎝ ⎠ ⎝ 2( 4t + 1) − (t + 5) = t − 8t + − t − = t − ( x − 5) = ( x − 4) 15i ( x − ) = 15i ( x − ) 40 ( x − ) = ( x − ) 40 x − 200 = x − 24 40 x − 200 − x = x − 24 − x 7t − = t − 7t − − t = t − − t 34 x − 200 = −24 34 x − 200 + 200 = −24 + 200 6t − = −3 34 x = 176 6t − + = −3 + 6t = 34 x 176 = 34 34 88 x= 17 6t = 6 t =0 86 90 y 3y − 14 = − − ⎛ y ⎞ ⎞ ⎛ 3y − 14 ⎟ = 12⎜ − − 1⎟ 12⎜ ⎝ ⎠ ⎝ ⎠ y − 168 = −4 y − 12 y − 168 + y = −4 y − 12 + y 33.9 − 37.5 + 1.5 x = 0.9 − + x = − + x + = + 1.5 x = 4.5 1.5 x 4.5 = 1.5 1.5 x=3 13 y − 168 = −12 13 y − 168 + 168 = −12 + 168 13 y = 156 91 13 y 156 = 13 13 y = 12 87 33.9 − 0.5(75 − x) = 0.9 2( x − 6) = 10 + x x − 12 = 10 + x x − 12 − x = 10 + x − x –12 ≠ 10 − k = −0.06 no solution, ∅; contradiction − k − 0.06 = −1 −1 k = 0.06 92 − x + x − = −(3x + 1) − x + x − = −3 x − − x − = −3 x − 88 p = −10 4 i p = i−10 5 p = −8 − x − + 3x = −3 x − + 3x − = −1 all real numbers, ℝ; identity SECTION 1.6 Solving Formula; Geometry 93 10.5 + 12.5 + 3.5 + 4.5 = 31 ft Chapter Review Full file at https://TestbankDirect.eu/ - 80 - Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file 94 at https://TestbankDirect.eu/ C = dπ ≈ 17 ( 3.14159 ) 97 ≈ 53.41 cm A = π r2 ⎛d⎞ =π ⎜ ⎟ ⎝2⎠ ⎛ 17 ⎞ =π ⎜ ⎟ ⎝ 2⎠ = π ( 8.5 ) V = π r h for h 3 ⋅V = ⋅ π r 2h 3V = π r h 3V π r h = π r2 π r2 3V =h π r2 3V h= πr 2 = π ( 72.25 ) ≈ 3.14159 ( 72.25 ) ≈ 226.98 cm 98 95 V = π r 3 = π ( 7.5 ) = π ( 421.875 ) ≈ ( 3.14159 )( 421.875 ) ≈ 1,767.15 m3 96 a Mv0 + Iw2 for M Mv + Iw2 2⋅ K = 2⋅ 2 K = Mv0 + Iw2 K= K − Iw2 = Mv0 + Iw2 − Iw2 K − Iw2 = Mv0 2 K − Iw2 Mv0 = v0 v0 K − Iw2 =M v0 M= A=s = 102 = 100 in.2 b V = π r h Find the diameter: 10 – – = in So the radius is in V = π ( 42 ) (15 ) = π (16 )(15 ) = ( 240 ) π = 80π 99 K − Iw2 v0 = a + (n − 1)d for d = a + (n − 1)d − a = a + (n − 1)d − a − a = ( n − 1) d − a (n − 1)d = n −1 n −1 −a =d n −1 −a d= n −1 ≈ 251.3 in.3 - 81 - Full file at https://TestbankDirect.eu/ Chapter Review Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file 100 at https://TestbankDirect.eu/ x − y = 35 for y 103 x − y − x = 35 − x −5 y = 35 − x −5 y 35 − x = −5 −5 y = −7 + x y = x−7 Item stored number sold printers computers x x + 150 cost of storage for $1.50 $2.50 total cost of storage 1.50x 2.50(x+150) printer cost + computer cost=total storage cost 1.50 x + 2.50 ( x + 150 ) = 2,775 10 ⎡⎣1.5 x + 2.5 ( x + 150 ) ⎤⎦ = 10 [ 2,775] SECTION 1.7 Using Equations to Solve Problems 15 x + 25 ( x + 150 ) = 27,750 15 x + 25 x + 3,750 = 27,750 101 40 x + 3,750 = 27,750 40 x + 3,750 − 3,750 = 27,750 − 3,750 O’Hare x million Atlanta (x + 20.7) million Total 159.3 million 40 x = 24,000 40 x 24,000 = 40 40 x = 600 x + ( x + 20.7) = 159.3 x + 20.7 = 159.3 10 ( x + 20.7 ) = 10 (159.3) There are 600 printers in the warehouse 20 x + 207 = 1,593 20 x + 207 − 207 = 1,593 − 207 20 x = 1,386 20 x 1,386 = 20 20 x = 69.3 x + 20.7 = 90.0 104 O’Hare served 69.3 million passengers and Atlanta served 90.0 million 1st piece 2nd piece 3rd piece 4th piece Total length x x+3 (x + 3) + = x + (x + 6) + = x + 186 x + ( x + 3) + ( x + 6) + ( x + 9) = 186 x + 18 = 186 x + 18 − 18 = 186 − 18 102 x = 168 245 – 5c x 168 = 4 x = 42 x + = 45 x + = 48 x + = 51 The lengths of the cable are 42 ft, 45 ft, 48 ft, and 51 ft Chapter Review Full file at https://TestbankDirect.eu/ - 82 - Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file 105 at https://TestbankDirect.eu/ The angles are supplementary (sum is 180°) 1st angle 2nd angle Sum 108 x x − 15 180 375 − 550 = 550 − 550 x − 550 ⎛1 ⎞ x + ⎜ x − 15 ⎟ = 180 ⎝2 ⎠ x − 15 = 180 x − 15 + 15 = 180 + 15 x = 195 2 i x = i195 3 x = 130 x − 15 = 50 The angle measures are 50° and 130° 106 −175 = −550 x −175 −550 x = −550 −550 0.318 = x x ≈ 0.32 The markdown is 32% of the original registration fee 109 Let x = percent of decrease # sold in 2007 290,282 # sold in 2008 235,924 % decrease x · 290,282 # sold 06 = # sold 05 – % Decrease Let width = x Then the length = x + 13 P = L + 2W 235,924 = 290,282 − 290,282 x 235,924 − 290,282 = 290,282 − 290,282 x − 290,282 134 = ( x + 13) + ( x ) −54,358 = −290,282 x −54,358 −290,282 x = −290,282 −290,282 0.187 = x 18.7% = x 134 = x + 26 + x 134 = x + 26 134 − 26 = x + 26 − 26 108 = x The percent decrease is 18.7% 108 x = 4 27 = x 110 x = 27 Let x = percent of increase # sold in 2007 193,900 # sold in 2008 x + 13 = 40 The dimensions are 27 in by 40 in 198,309 % increase x · 193,900 # sold 08 = # sold 07 + % Increase 198,309 = 193,900 + 193,900 x 198,309 − 193,900 = 193,900 + 193,900 x − 193,900 SECTION 1.8 More About Problem Solving 107 Let x = percent of markdown new fee 375 original fee 550 markdown x · 550 new fee = original fee – markdown 375 = 550 − 550 x 4,409 = 193,900 x 4,409 193,900 x = 193,900 193,900 0.023 = x 99 is 80% of what? 99 = 0.80 ⋅ x 99 = 0.80 x 99 0.80 x = 0.80 0.80 123.75 = x 2.3% = x The percent increase is 2.3% x ≈ 124 It has been recorded about 124 days - 83 - Full file at https://TestbankDirect.eu/ Chapter Review Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file 111 at https://TestbankDirect.eu/ 113 Account Principal · Rate · Time = Interest 10% 9% Combined Interest x 25,000 – x 0.10 0.10x 0.09 0.09(25,000-x) 2,430 Amount · Strength = Pure concentrate 4% pesticide 0.04 0.04x x 12% pesticide 20 0.12 0.12(20) 10% solution x + 20 0.10 0.10(x + 20) 10% Int + 9% Int = Combined Int 0.10 x + 0.09( 25,000 − x) = 2,430 0.04 x + 0.12(20) = 0.10( x + 20) 0.04 x + 2.4 = 0.10 x + 0.10 x + 2,250 − 0.09 x = 2,430 100 [ 0.04 x + 2.4] = 100 [ 0.10 x + 2] 0.01x + 2,250 = 2,430 x + 240 = 10 x + 200 0.01x + 2,250 − 2,250 = 2,430 − 2,250 x + 240 − 10 x = 10 x + 200 − 10 x 0.01x = 180 −6 x + 240 = 200 0.01x 180 = 0.01 0.01 x = 18,000 −6 x + 240 − 240 = 200 − 240 −6 x = −40 −6 x −40 = −6 −6 20 x= x=6 25,000 − x = 7,000 She invested $18,000 at 10% and $7,000 at 9% 112 celebrity photographer Rate · Time = Distance t t 1.5 t – 1.5(t–1) Since the photographer will catch the celebrity, the distances for both will be the same (equal) Celebrity Distance = Photographer Distance gallons of the 4% solution are needed 114 Type of Coffee Mild Robust Mixture Pounds · Price = x 90 – x 90 7.50 8.40 7.90 Value 7.50x 8.40(90 – x) 7.90(90) mild value + robust value = mixture value 7.50 x + 8.40 ( 90 − x ) = 7.90 ( 90 ) t = 1.5(t − 1) t = 1.5t − 1.5 7.5 x + 756 − 8.4 x = 711 t − 1.5t = 1.5t − 1.5 − 1.5t −0.5t = −1.5 10 [ 7.5 x + 756 − 8.4 x ] = 10 [ 711] 75 x + 7,560 − 84 x = 7,110 −0.5t −1.5 = −0.5 −0.5 t =3 −9 x + 7,560 = 7,110 −9 x + 7,560 − 7,560 = 7,110 − 7,560 −9 x = −450 t −1 = −9 x −450 = −9 −9 x = 50 The photographer will catch the celebrity in minutes 90 − x = 90 − 50 = 40 He needs 50 pounds of mild coffee and 40 pounds of robust coffee Chapter Review Full file at https://TestbankDirect.eu/ - 84 - ... 1.6 Solving Formula; Geometry 93 10.5 + 12.5 + 3.5 + 4.5 = 31 ft Chapter Review Full file at https://TestbankDirect.eu/ - 80 - Solution Manual for Intermediate Algebra 5th Edition by Tussy Full... 56 - 77 - Full file at https://TestbankDirect.eu/ Chapter Review ) Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file at https://TestbankDirect.eu/ 2 48 −b − b − 4ac −(−3)... m = m 59 a 7 x− x= ⋅ x− ⋅ x 5 4 28 15 = x− x 20 20 13 = x 20 - 78 - Solution Manual for Intermediate Algebra 5th Edition by Tussy Full file at https://TestbankDirect.eu/ (x − 3) = −12 76 82 −