Solution manual for college algebra 2nd edition by miller

Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller Chapter Equations and Inequalities Section 1.1 Linear Equations and Rational Equations e Linear; linear first solution conditional identity contradiction rational empty (or null); { } orφ a Linear; −2x = {−4} 12 − = 4+ x − 8= x {8} 11 −6x − = 20 −6x = 24 x = −4 {−4} 12 −8y + = 22 −8y = 16 −2x = −2 −2 x = −4 {−2} − x= c Linear;   −2 − x = −2(8)   x = −16 = − 12t = −12t 0= t {0} {− 16} 14 11 = − 2( 5p − 2) 11 = − 10p + d Nonlinear e Linear; x− 2= x − + = 8+ x = 10 11 = 11− 10p = −10p 0= p {10} {0} 4x 4x x {3} b Nonlinear c Linear; {48} y = −2 13 = − 3( 4t+ 1) = − 12t− b Nonlinear 10 a Linear; 12 = 12 = 3= 12 = 4+ x x 1  4(12) = 4 x  4  48 = x 12 = d Nonlinear 80 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 15 −6(v − 2) + = − (v + 4) −6v + 12 + = − v −  22     3 −6v + 15 = 5− v 22 4( y − 3) = 3 y + 2( y − 2) 4y − 12 = 3( y + 2y − 4) −5v = −10 v=2 {2} 4y − 12 = 3( 3y − 4) 4y − 12 = 9y − 12 16 −5(u − 4) + = 11− (u − 3) −5u + 20 + = 11− u + −5y = −5u + 22 = 14 − u {2} 17 {0} −4u = −8 u= 23 2.3 = 4.5x + 30.2 −27.9 = 4.5x −6.2 = x {− 6.2} 24 2.8 = p {2.8} 19 0.05y + 0.02( 6000 − y ) = 270 0.05y + 120 − 0.02y = 270 x− =1 5 1 6 x −  = 6(1) 3 6 x − 10 = x = 16 {16} 0.03y + 120 = 270 0.03y = 150 y = 5000 25 20 0.06x + 0.04(10,000 − x) = 520 0.06x + 400 − 0.04x = 520 w − = w +2 3 1 2  12 w −  = 12 w + 2 4 2 3  6w − = 8w + 24 −2w = 33 0.02x = 120 w =− x = 6000 {6000}  33  −   2 21 2( 5x − 6) =  x − 3( x − 10) 10x − 12 = 4( x − 3x + 30) 10x − 12 = 4( −2x + 30) 10x − 12 = −8x + 120 18x = 132 x= x− = 3 1 4 x −  = 4( 2) 2 4 x−6= x = 14 {14} 18 9.4 = 3.5p − 0.4 9.8 = 3.5p {5000} y=0 132 22 = 18 81 33 Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller 26 27  n+ n− 2  n+1  20 − = 20 − 1     10  5( n + 3) − ( n − 2) = 2( n + 1) − 20 p − = p −1 10 15 3 2 7  30 p −  = 30 p − 1 10  5  15  12p − = 14p − 30 −2p = −21 21 p=  21   2 5n + 15 − 4n + = 2n + − 20 n + 23 = 2n − 18 − n = −41 {41} 30 y −1 y y + + = +1  y −1 y   y+3  +  = 20  + 1 20  4    4( y − 1) + 5y = 10( y + 3) + 20 4y − + 5y = 10y + 30 + 20 9y − = 10y + 50 {−54} 28 t− t+ t− − = +2 10  t− t+   t−  30 − = 30 + 2     10  10(t− 2) − 6(t+ 7) = 3(t− 4) + 60 10t− 20 − 6t− 42 = 3t− 12 + 60 4t− 62 = 3t+ 48 t= 110 {110} − y = 54 y = −54 31 a T = −1.83a + 212 T = −1.83( 4) + 212 = 204.68°F x − x x +1 + = +2  x−6 x  x +1  21 +  = 21 + 2 7    7( x − 6) + 3x = 7( x + 1) + 42 ≈ 205°F T = −1.83a + 212 193 = −1.83a + 212 −19 = −1.83a 10.4 ≈ a 10.4 × 103 = 10,400 Approximately10, 400 ft 32 a C = 167.95x + 94 b 7x − 42 + 3x = 7x + + 42 10x − 42 = 7x + 49 3x = 91 91 x= 29 n = 41 C = 167.95( 9) + 94 = $1605.55 b C = 167.95x + 94  91    3 n+ n− n+1 − = −1 10 2445.30 = 167.95x + 94 2351.30 = 167.95x 14 = x 14 credit-hours 33 S = 14.2t+ 149 362 = 14.2t+ 149 213 = 14.2t 15 = t 2004 + 15 = 2019 In 2019 82 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 34 41 − ( − 2w ) = 4(w + 1) − 2w − 10 −6 + 2w = 4w + − 2w − 10 S = 18t+ 232 628 = 18t+ 232 396 = 18t 22 = t −6 + 2w = 2w − 0= Identity;  42 −5 + 3x = 3( x − 1) − −5 + 3x = 3x − 3− 2000+ 22 = 2022 In 2022 35 a C = 7x b C = 105 7x = 105 x = 15 The motorist will save money beginning on the 16th working day 36 a C = 2.25x C = 89 b 2.25x = 89 x ≈ 39.6 The commuter will save money on the 40th ride 37 a S1 = 45,000 + 2250x −5 + 3x = 3x − 0= Identity;  1 x + 3= x +1 43 1  1  4 x + 3 = 4 x + 1 2  4  2x + 12 = x + x = −8 Conditional equation; {−8} 44 b S2 = 48,000 + 2000x c S1 = S2 45,000 + 2250x = 48,000 + 2000x 250x = 3000 x = 12 yr 38 a S1 = 25,000 + 0.16x y − 5= y − 2  1  6 y − 5 = 6 y − 4 3  6  4y − 30 = y − 24 3y = y=2 Conditional equation; {2} b S2 = 30,000 + 0.15x c S1 = S2 25,000 + 0.16x = 30,000 + 0.15x 0.01x = 5000 x = $500,000 39 2x − = 4( x − 1) − 1− 2x 45 + = x−5 x+ x ≠ 5,x ≠ −4 46 − = x+1 x− x ≠ −1,x ≠ 47 − = 2x − 5x − x − =  5x − x  2 x −  2  2x − = 4x − − 1− 2x 2x − = 2x − −3 = −5 Contradiction 40 4( − 5n) + = −4n − − 16n 12 − 20n + = −4n − − 16n x≠ −20n + 13 = −20n − 13 = −8 Contradiction 83 ,x ≠ 0,x ≠ Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller 48 − = 2x − x 4x − − = 5 2x − x  4 x −  4  x ≠ 0,x ≠ 6,x ≠ 49 3 c = − c− c− 3  c   4(c − 3)  = 4(c − 3)  −    c − 3  c− 4 4c = 12 − 3( c − 3) 4c = 12 − 3c + y = 17 {} 7c = 21 c= ; The value does not check t= 25 7 d − = d−7 d−7 7   d  8( d − 7)  −  = 8( d − 7)    d − 8  d − 7 56 − 7( d − 7) = 8d 54 w +3 w −5 + 1= 4w w w +   w − 5 4w  + 1 = 4w    4w   w  w + + 4w = 4(w − 5) 5w + = 4w − 20 w = −23 {−23} = t− t − 1 = t− (t+ 1)(t− 1) 55 56 − 7d + 49 = 8d {}     (t+ 1)(t− 1)  (t+ 1)(t− 1)  t−11 = (t+ 1)(t− 1)   {2} x+ x− + 1= 6x x  x+   x − 7 6x  + 1 = 6x    6x   x  x + + 6x = 6( x − 7) 7x + = 6x − 42 x = −44 {−44} 53 − = 3t t 1   7 3t −  = 3t   3t  t t− = 21 {25} 51 − = 2y y 1   5 2y  − = 2y     2y   y y − = 10 {17} 50 52  t+ = t= 84 −15d = −105 d=7 ; The value does not check Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller = w +2 w −4 = w + (w + 2)(w − 2) 56     (w + 2)(w − 2)  (w + 2)(w − 2)  w 1+  = (w + 2)(w − 2)    w −2= w =7 {7} 11 − = x − x + x − 25 11 − = x − x + ( x + 5)( x − 5) 57   11   ( x + 5)( x − 5)  ( x + 5)( x − 5)  x − − x +  = ( x + 5)( x − 5)  {−4}   2( x + 5) − 1( x − 5) = 11 2x + 10 − x + = 11 x + 15 = 11 x = −4 10 − = c+ c− c − 10 − = c + c − ( c + 3)( c − 3) 58     10   ( c + 3)( c − 3)  (c + 3)(c − 3)  c + − c −  = (c + 3)(c − 3)    2( c − 3) − 1( c + 3) = 10 2c − − c − = 10 c − = 10 {19} c = 19 85 Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller {} −14 − = x − x − 12 x − x + −14 − = ( x − 4)( x + 3) ( x − 4) ( x + 3)    −14  ( x − 4)( x + 3)  x − x + − x −  = ( x − 4)( x + 3)  x +  )( ) ( )  )   (  ( −14 − ( x + 3) = 2( x − 4) −14 − x − = 2x − −17 − x = 2x − −3 = x ; The value −3 does not check {} 2 − = x + 5x + x + x + 2 − = ( x + 2)( x + 3) ( x + 2) ( x + 3)    2  ( x + 2)( x + 3)  x + x + − x +  = ( x + 2)( x + 3)  x +  )( ) ( )  )   (  ( − 2( x + 3) = ( x + 2) − 2x − = x + −4 − 2x = x + −2 = x ; The value −2 does not check 59 60 − = x − x − x − x + 3x + − = ( x − 2)( x + 1) ( x − 2)( x + 2) ( x + 2)( x + 1)     ( x + 2)( x − 2)( x + 1)  x − x + − x − x +  = ( x + 2)( x − 2)( x + 1)  x + x +  )( ) ( )( )  )( )   (  ( 5( x + 2) − 2( x + 1) = 4( x − 2) 5x + 10 − 2x − = 4x − 3x + = 4x − 16 = x {16} 61 86 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller − = x − 2x − x − 16 x + 6x + − = ( x − 4)( x + 2) ( x − 4)( x + 4) ( x + 4)( x + 2)     ( x + 4)( x − 4)( x + 2)  x − x + − x − x +  = ( x + 4)( x − 4)( x + 2)  x + x +  )( ) ( )( )  )( )   (  ( 4( x + 4) − 1( x + 2) = 2( x − 4) 4x + 16 − x − = 2x − 3x + 14 = 2x − x = −22 {−22} 62 63 64 3m = − m − m + 2m − m + 3m = − m − ( m + 4)( m − 2) m +  3m    ( m + 4)( m − 2)  m −  = ( m + 4)( m − 2)  m + m − − m +  )( )    (  5( m + 4) = 3m − 2( m − 2) 5m + 20 = 3m − 2m + 5m + 20 = m + 4m = −16 m = −4 { } ; The value −4 does not check 10 15n = − n − n − 2n − 24 n + 10 15n = − n − ( n − 6)( n + 4) n +  15n   10  ( n − 6)( n + 4)  n −  = ( n − 6)( n + 4)  n − n + − n +  )( )    (  10( n + 4) = 15n − 6( n − 6) 10n + 40 = 15n − 6n + 36 10n + 40 = 9n + 36 n = −4 { } ; The value –4 does not check 87 Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller 65 66 67 68 69 5x − = 3x − 5x − 3x + − x 5x −3 − = ( 3x + 1)( x − 2) 3x + x −  5x   −3  ( 3x + 1)( x − 2)  3x + x − − 3x + 1 = ( 3x + 1)( x − 2)  x −  )( )    (  5x − 1( x − 2) = −3( 3x + 1) 5x − x + = −9x − 4x + = −9x − 13x = −5 x=− 13  5 −   13 3x − = 2x + x − 2x + 1− x 3x −4 − = ( 2x + 3)( x − 1) 2x + x −  3x   −4  ( 2x + 3)( x − 1)  2x + x − − 2x + 3 = ( 2x + 3)( x − 1)  x − 1 )( )    (  3x − 2( x − 1) = −4( 2x + 3) 3x − 2x + = −8x − 12 x + = −8x − 12 9x = −14 14 x= −  14  −   9 W = K − T forK A = lw forl 70 A lw W + T = K orK = W + T = 71 Δs = s2 − s1 fors1 w w A A Δs− s2 = −s1 = l orl= w w s1 = s2 − Δs E = IR forR 72 Δt= tf − ti forti E IR = Δt− tf = −ti I I ti = tf − Δt E E = R orR = I I P = a + b + c forc P − a − b = c orc = P − a − b 88 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 73 7x + 2y = fory 2y = −7x + 2y −7x + = 2 −7x + y= ory = − x + 2 74 3x + 5y = 15 fory 5y = −3x + 15 5y −3x + 15 = 5 −3x + 15 y= ory = − x + 5 75 5x − 4y = fory −4y = −5x + −4y −5x + = −4 −4 5x − y= ory = x − 4 76 7x − 2y = fory −2y = −7x + −2y −7x + = −2 −2 7x − y= ory = x − 2 1 x + y = fory 77  1 6 x + y  = 6(1)  2 3x + 2y = 2y = −3x + 2y −3x + = 2 −3x + ory = − x + y= 2 78 x − y = fory  1 12 x − y  = 12( 2)  4 3x − 8y = 24 −8y = −3x + 24 −8y −3x + 24 = −8 −8 3x − 24 ory = x − y= 8 n 79 S = ( a + d ) ford n  2( S ) =  ( a + d )  2  2S = n( a + d ) 2S = na + nd 2S − na = nd 2S − na nd = n n 2S − na =d n 2S − na 2S d= −a ord = n n n 80 S =  2a + ( n − 1) d  fora n  2(S ) =  2a + ( n − 1) d   2  2S = n  2a + ( n − 1) d  2S = n ( 2a + nd − d ) 2S = 2an + n2d − nd 2S − n2d + nd = 2an 2S − n2d + nd 2an = 2n 2n 2S − n d + nd 2S − n2d + nd = a ora = 2n 2n 89 Section 1.2 Solution Manual for College Algebra 2nd Edition By Miller S Wave s x 11 = x − 10  x   11 = 9( x − 10)   9( x − 10)    x − 10   9 9x = 11( x − 10) 9x = 11x − 110 2x = 480 x = 240 km 31 Let x represent the price set by the merchant x − 0.25x = 180 + 0.40(180) 0.75x = 180 + 72 0.75x = 252 x = $336 32 Let x represent the price set by the bookstore x − 0.10x = 80 + 0.35( 80) 0.90x = 80 + 28 0.90x = 108 x = $120 33 a C = 110 + 60x C = 350 b 110 + 60x = 350 x 60x = 240 x = hr 34 a C = 2400 + 80x C = 5520 b 2400 + 80x = 5520 x 80x = 3120 x = 39 hr 35 Let x represent the height of the Washington Monument x = 444 4x = 2220 x = 555 ft 36 Let x represent the height of the light post x = 300 km 30 Let x represent the distance from the epicentre to the station Distanc Rat Tim P Wave s x x 4.8 x x − = 20 4.8  x x −  = 24( 20) 24  4.8  5x − 3x = 480 110 = 2x 55 = x x − 10 = 55 − 10 = 45 There were 55 Democrat and 45 Republican senators 27 Let x represent the number of deer in the population 30 = x 80 5x = 2400 x = 480 deer 28 Let x represent the number of bass in the lake 24 = x 40 4x = 960 x = 240 bass 29 Let x represent the distance from the epicenter to the station Distanc Rat Tim P Wave x s S Wave x s x x − = 40  x x 15 −  = 15( 40)  5 5x − 3x = 600 2x = 600 4.8 x x light post x ft man ft 100 40 ft 20 ft Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller x = 1.5 + x + x x = 20 40 + 20 x = 20 60 20x = 360 x = 18 ft 37 Let x represent the height of the pole Then, x is the length of the pole that is in the ground, and x is the length of the pole that is in the snow x − x − x = 1.5   24 x − x − x  = 24(1.5)   24x − 16x − 3x = 36 5x = 36 x = 7.2 The pole is 7.2 ft long, and the snow is 4.8 ft deep C = ( F − 32) 38 F = ( F − 32) 5  9( F ) =  ( F − 32)  9  9F = 5F − 160 4F = −160 F = −40 −40°C = −40°F 39 Let x represent the amount of 20% fertilizer solution (in litres) to be drained (and therefore the amount of water to be added) 40 L is the amount of the resulting 15% fertilizer solution Therefore, ( 40 − x ) is the amount of 20% fertilizer solution that is not drained Amount of Solution Pure fertilizer 0% Solution x 20% Solution 0( x) 0.20( 40 − x) 40 − x 15% Solution 40 0.15( 40) 0( x) + 0.20( 40 − x) = 0.15( 40) − 0.20x = −0.20x = −2 x = 10 10 L should be drained and replaced by water 40 Let x represent the amount of water (in litres) to be evaporated Therefore, ( 200 − x) is the amount of the final 25% salt solution 0% Solution 5% Solution 25% Solution 200 − x 200 Amount of Solution x 0( x) 0.05( 200) 0.25( 200 − x) Pure salt 101 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.2 Solution Manual for College Algebra 2nd Edition By Miller 0( x) + 0.05( 200) = 0.25( 200 − x) 10 = 50 − 0.25x −40 = −0.25x 160 = x 160 mL should be evaporated 41 The length of the lot is l= 128 + 2x The width of the lot is w = 60 + 2x P = 2l+ 2w 440 = 2(128 + 2x) + 2( 60 + 2x) 440 = 256 + 4x + 120 + 4x 440 = 8x + 376 64 = 8x 8= x The width of the easement is ft 42 The length of the play area is l= 78 + 2x The width of the play area is w = 36 + 2x P = 2l+ 2w 396 = 2( 78 + 2x) + ( 36 + 2x) 396 = 156 + 4x + 72 + 4x 396 = 8x + 228 168 = 8x 21 = x The width of the border is 21 ft 43 a The width of the kitchen is w The length of the kitchen is l= w + P = 2l+ 2w 48 = 2(w + 4) + 2w 48 = 2w + + 2w 48 = 4w + 40 = 4w 10 = w The kitchen is 14 ft by 10 ft b A = lw + 0.1lw = 1.1lw A = 1.1(14)(10) = 154 ft2 c C = 1.06(12)(154) = $1958.88 44 a The width of the porch is w The length of the porch is l= 2w + P = 2l+ 2w 64 = 2( 2w + 2) + 2w 64 = 4w + + 2w 64 = 6w + 60 = 6w 10 = w l= 2w + = 2(10) + = 22 The porch is 22 ft by 10 ft b A = lw + 0.1lw = 1.1lw A = 1.1( 22)(10) = 242 ft2 c C = 1.075( 5.85)( 242) ≈ $1521.88 45 Aliyah had 8000 − 0.28( 8000) = 8000 − 2240 = 5760 to invest Let x represent the amount she invested at 11% Then, ( 5760 − x ) is the amount she invested at 5% 11% Investment 5% Investment Total 5760 − x Principal x Interest (I = Prt) x ( 0.11)(1) ( 5760 − x)( 0.05)(1) 453.60 102 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller x ( 0.11) + ( 5760 − x)( 0.05) = 453.60 0.11x + 288 − 0.05x = 453.60 0.06x + 288 = 453.60 0.06x = 165.60 x = 2760 5760 − x = 5760 − 2760 = 3000 Aliyah invested $2760 in the stock returning 11% and $3000 in the stock returning 5% 46 Let x represent the amount Caitlin invested in the balanced fund Then, ( 2x) is the amount she invested in the stock fund Balanced Fund (3.5%) Principal x Interest (I = Prt) x ( 0.035)(1) Stock Fund (17%) 2x ( 2x)( 0.17)(1) Total 1125 x ( 0.035) + ( 2x)( 0.17) = 1125 0.035x + 0.34x = 1125 0.375x = 1125 x = 3000 2x = 2( 3000) = 6000 Caitlin invested $3000 in the balanced fund and $6000 in the stock fund 103 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.3 Solution Manual for College Algebra 2nd Edition By Miller larger number x + 25 = 4+ x x 1  x + 25   x = x +   x  x   x + 25 = 4x + 24 = 3x 8= x x+ 25 = 8+ 25 x = 12.8 8x = 89.6 x = 11.2 12.8 = y 12 12.8y = 96 y = 7.5 x = 11.2 ft and y = 7.5 cm x 1.2 = 48 0.96 1.04 0.96x = 1.248 x = 1.3 0.5 1.2 = y 0.96 1.2y = 0.48 y = 0.4 x = 1.3 m and y = 0.4 in 49 No If x represents the measure of the smallest angle, then the equation x + ( x + 2) + ( x + 4) = 180 does not result in an odd integer value for x Instead the measures of the angles would be even integers 50 No If x represents the number of each type of bill, then the solution to the equation 20x + 10x + 5x = 100 is not a whole number 51 Let x represent the smaller number Then, ( x + 16) is the 47 = 33 The numbers are and 33 53 Let x represent the tens digit of the number Then, (14 − x) is the ones digit 10(14 − x) + 1( x) = 10( x) + 1(14 − x) + 18 140 − 10x + x = 10x + 14 − x + 18 140 − 9x = 9x + 32 108 = 18x 6= x 14 − x = 14 − =8 The original number is 68 54 Let x represent the tens digit of the number Then, ( − x) is the ones digit 10( − x) + 1( x) = 10( x) + 1( − x) − 45 90 − 10x + x = 10x + − x − 45 90 − 9x = 9x − 36 126 = 18x 7= x 9− x = 9− =2 The original number is 72 m 1x1 + m 2x2 = 55 larger number x + 16 = 3+ x x 2  x + 16   x = x 3+   x  x   x + 16 = 3x + 14 = 2x 7= x x+ 16 = + 16 ( 30)( −1.2) + ( 20) x2 = 56 20x2 = 36 x2 = 1.8 m m 1x1 + m 2x2 = ( 64) x1 + (80)( 2) = = 23 The numbers are and 23 52 Let x represent the smaller number Then, ( x + 25) is the 64x1 = −160 x1 = −2.5 m 104 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 57 m 1x1 + m 2x2 = 17 (10)( −3.2) + m (8) = −6 −14 = i ⋅ i 14 = i2 84 8m = −32 m = kg 58 = ( −1) 22 ⋅ 21 = −2 21 m 1x1 + m 2x2 = m ( −10) + ( 6)( 7) = −10m = −42 m = 4.2 kg 18 = i2 150 = ( −1) 52 ⋅ = −5 Section 1.3 Complex Numbers 19 −1 i b real; imaginary conjugate −121 = i 121 = 11i −100 = i 100 = 10i −98 = i 98 = 7i −63 − i 63 = 3i −19 = i 19 10 −10 −15 = i 10 ⋅ i 15 20 −23 = i 23 21 11 − −16 = − i 16 = −4i −98 −2 −45 −5 −63 12 − −25 = − i 25 = −5i 13 − −9 = i ⋅ i = 2i⋅ 3i= 6i2 = 6( −1) = −6 14 22 −1 −36 = i ⋅ i 36 = 1i⋅ 6i= 6i2 = 6( −1) = −6 15 −10 −5 = i 10 ⋅ i i 98 = i 45 = i 63 = i 80 i 98 = = 49 = i 45 = = 9=3 63 =i = i = 3i 80 =i = i 16 = 4i 23 Real part: 3; Imaginary part: −7 24 Real part: 2; Imaginary part: −4 25 Real part: 0; Imaginary part: 19 26 Real part: 0; Imaginary part: 40 27 Real part: − ; Imaginary part: 4 28 Real part: − ; Imaginary part: = i2 50 = ( −1) 52 ⋅ = −5 16 −80 = −6 −15 = i ⋅ i 15 = i2 90 = ( −1) 32 ⋅10 29 −4 = 4⋅ 2i = 8i= + 8i = −3 10 105 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.3 Solution Manual for College Algebra 2nd Edition By Miller b i = i ⋅ i 30 −144 = 2⋅12i = 24i= + 24i 31 + −12 = + 3i or + 2i ( 29 28 = (1) ⋅ i1 = i c i = i ⋅ i 50 ) 32 − −24 = + −2 ior − 2i 48 = (1) ⋅ i2 = − + 3i = + i 14 14 14 = + i 14 + 5i 34 = + i 6 = + i −4 − 6i −4 −6 i 35 = + −2 −2 −2 = + 3i − 15i 15 36 = − i −3 −3 −3 = −3+ 5i 37 33 −41 = i−44 ⋅ i3 d i 42 a = (1) ⋅ i3 = − i i3 = b i = i ⋅ i 47 44 = (1) ⋅ i3 = − i 66 64 c i = i ⋅ i = (1) ⋅ i2 = − −27 = i−28 ⋅ i1 d i 43 a −18 + −48 −18 + 3i = 4 18 3i =− + 4 9 = − + 3ior − + i 2 −20 + −50 −20 + 2i 38 = −10 −10 −20 2i = + −10 −10 2 ior − i = 2− 2 14 − −98 14 − 2i 39 = −7 −7 14 2i =− + 7 = −2 + 2ior − + i 40 −10 + −125 −10 + 5i = 5 10 5i =− + 5 = −2 + 5ior − + i = (1) ⋅ i1 = i i37 = i36 ⋅ i1 −37 b i =i = i−40 ⋅ i3 = (1) ⋅ i3 = −i c i = i ⋅ i 82 80 = (1) ⋅ i2 = −1 −82 −84 d i = i ⋅ i = (1) ⋅ i2 = −1 44 a i = i ⋅ i 103 100 = (1) ⋅ i3 = − i −103 b i = i−104 ⋅ i1 = (1) ⋅ i1 = i c i = 52 − 52 d i = 45 ( − 7i) + ( 8− 3i) = ( + 8) + ( −7 − 3) i = 10 − 10i 46 ( − 10i) + ( 8+ 4i) 41 a i = 20 106 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller = ( + 8) + ( −10 + 4) i 55 2i( 5+ i) = 10i+ 2i = 10i+ 2( −1) = 14 − 6i 47 (15 + 21i) − (18 − 40i) = (15− 18) + 21− ( −40)  i = −2 + 10i 56 4i( + 5i) = 24i+ 20i2 = ( 250 − 80) + (100 − 25) i 57 = −3+ 61i 48 ( 250 + 100i) − ( 80 + 25i) −3 = 170 + 75i 1  5  49  + i −  + i    12  ( = 24i+ 20( −1) = −20 + 24i ) 11 − −7 = i ( 11 − i ) = i 33 − i 21 = i 33 − ( −1) 21  5   =  − + − i    12   5   =  − + − i  6   12 12  =− + i 12 =− + i 12 3    50  − i −  + i    10  −2 58 ( ) = 21 + i 33 13 + −5 = i ( ) 13 + i = i 26 + i2 10 = i 26 + ( −1) 10 = − 10 + i 26 59 ( 3− 6i)(10 + i) = 3(10) + 3( i) + ( −6i)(10) + ( −6i)( i) = 30 + 3i− 60i− 6i2 = 30 − 57i− 6( −1) = 36 − 57i 60 ( − 5i)( 8+ 2i)    1 =  − +− − i  10     7  4 =  − +− − i  10 10   24 24  =− − i 10 24 51 ( 2.3 + 4i) − ( 8.1− 2.7i) + ( 4.6 − 6.7i) = 2( 8) + 2( 2i) + ( −5i)( 8) + ( −5i)( 2i) = 16 + 4i− 40i− 10i2 = 16 − 36i− 10( −1) = 26 − 36i = ( 2.3 − 8.1+ 4.6) + ( + 2.7 − 6.7) i 61 ( − 7i) = ( 3) − 2( 3)( 7i) + ( 7i) = −1.2 + 0i  0.05   −0.12   0.07  52  + −  −0.03i  +0.08i  +0.05i       = ( 0.05 − 0.12 − 0.07) 2 = − 42i+ 49i2 = − 42i+ 49( −1) = − 42i− 49 = −40 − 42i + ( −0.03+ 0.08 − 0.05) i = −0.14 + 0i 53 − (16 + 24i) = −2 − 3i 54 − ( 60 − 30i) = −10 + 5i 62 (10 − 3i) = (10) − 2(10)( 3i) + ( 3i) 2 = 100 − 60i+ 9i2 ( = 100 − 60i+ 9( −1) = 100 − 60i− = 91− 60i )( 63 3− −5 + −5 107 ) Section 1.3 Solution Manual for College Algebra 2nd Edition By Miller ( )( ) = 3( 4) + 3( i 5) + ( −i 5) ( 4) + ( −i 5)( i 5) 69 a + 6i = 3− i + i b ( 3− 6i)( 3+ 6i) = ( 3) + ( 6) = + 36 b ( − 5i)( + 5i) = ( 4) + ( 5) = 16 + 25 = 12 − i − 5( −1) = 17 − i ( )( ) = ( + i )(10 + i 7) = 2(10) + 2( i 7) + i (10) 64 + −7 10 + −7 b ( − 8i)( + 8i) = ( 0) + ( 8) = + 64 ( ) b ( − 9i)( + i) = ( 0) + ( 9) = + 81 = 20 + 12i + 7( −1) 73 (10 − 4i)(10 + 4i) = (10) + ( 4) = 100 + 16 = 24 + 8i− 15i+ 35i2 74 ( 3− 9i)( 3+ 9i) = ( 3) + ( 9) = + 81 2 = −24 − 3i− 6( −1) = −18 − 3i 76 ( −5i)( 5i) = ( 5) = 25 2 77 = ( 2) − 2( 2)( i) + i2 + ( 2) ( )( + 3i = + 2( 2)( i) + i2 = − 4i+ i2 + + 4i+ i2 78 = + 2i ( ) + 3i ( ) + ( 3) 2 = 2+ 3= )( + 7i = = + 2( −1) = ) − 7i ( 5) + ( ) 2 = + = 12 = ( 3) − 2( 3)( 2i) + ( 2i) + ( 3) 2 = 90 75 ( 7i)( −7i) = = 49 = −24 + 9i− 12i− 6i2 2 = 116 = 24 − 7i+ 35( −1) = −11− 7i 66 −3( 8− 3i) − 6i( + i) 68 ( 3− 2i) + ( 3+ 2i) = 81 = 13+ 12i 65 4( + 2i) − 5i( 3− 7i) 2 = 64 72 a − 9i = 20 + 2i + 10i + 7i2 = 41 71 a − 8i +i i 67 ( − i) + ( + i) = 45 70 a + 5i = 12 + 3i − 4i − 5i2 2 + 2( 3)( 2i) + ( 2i) = − 12i+ 4i2 + + 12i+ 4i2 = 18 + 8i2 = 18 + 8( −1) = 10 108 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 79 + 2i ( + 2i)( + i) = − i ( − i)( + i) = 18 + 6i+ 6i+ 2i2 = ( 3) + (1) 18 + 12i+ 2( −1) = ( ) = 16 + 19 + 9i = 17 19 = + i 17 17 − 5i (8 − 5i)(13 − 2i) 81 = 13 + 2i (13 + 2i)(13 − 2i) ) )( ) − 5i ( 6) + ( ) − 5i 36 + − 5i = 41 i = − 41 41 −1 84 − 3i = − 3i + 3i = − 3i + 3i = ( 104 − 16i− 65i+ 10i2 (13) + ( 2) 104 − 81i+ 10( −1) = ( ( ( 4) + (1) 20 + 9i+ 1( −1) = = 121+ 16 98 − 73i = 137 98 73 = − i 137 137 −1 83 + 5i = + 5i − 5i = + 5i − 5i 20 + 5i+ 4i+ i2 110 − 40i− 33i+ 12i2 (11) + ( 4) 110 − 73i+ 12( −1) = 9+1 16 + 12i = 10 16 12 = + i 10 10 = + i 5 5+ i ( + i)( + i) = 80 − i ( − i)( + i) = 10 − 3i (10 − 3i)(11− 4i) = 11+ 4i (11+ 4i)(11− 4i) 82 ) ( ( 169 + 94 − 81i = 173 94 81 = − i 173 173 = ) )( ) + 3i ( 4) + ( 3) + 3i 16 + + 3i = 19 = + i 19 19 = 109 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.3 Solution Manual for College Algebra 2nd Edition By Miller 5⋅ i = 13i 13i⋅ i 5i 5i = = 13i 13( −1) 5i = =− i 13 −13 = 0− i 13 6( − i) = 86 7i 7i( − i) −6i −6i = = −7i −7( −1) 6 = − i= − i 7 −1 −1 = 87 3i −3 −1⋅ 3i − 3i = = 3i⋅ 3i 3i − 3i − 3i = = 3( −1) −3 85 = 88 −2 −11 = = 91 89 92 b2 − 4ac = 93 a = 4i x + 25 = −25 + 25 = 0=  b x2 + 25 = ( −5i) + 25 = 25( −1) + 25 = −25 + 25 = 0=  94 a x2 + 49 = ( 7i) + 49 = 49( −1) + 49 = −49 + 49 = 0=  b x2 + 49 = ( −7i) + 49 = 49( −1) + 49 = − ( 2)( 6) −49 + 49 = 0=  x2 − 4x + = 95 a = 4i ( −5) − 4( 2)( 4) = −32 = i 32 b2 − 4ac = ( 5i) + 25 = 25( −1) + 25 = = 16 − 48 90 ( 4) = −16 = i 16 11 11 i= + i 11 11 − 4( 2)( 5) = 16 − 32 11i −2 ⋅ 11i ( 4) = −4 = i = 2i 3i 3i = 0+ 3 −2 b2 − 4ac = ( −6) = 36 − 40 11i⋅ 11i −2 11i −2 11i = = 11( −1) 11i2 = b2 − 4ac = ( + i 3) − 4( + i 3) + = 2 − 4( 5)(10) + 4i + 3i2 − − 4i + = = 25 − 200 + 3( −1) − + = = −175 = i 175 − 3− = = 5i 0=  110 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller b logic for simplification would be −9 ⋅ −4 = i ⋅ i x2 − 4x + = ( ) ( ) 2− i − 2− i + = = i2 36 − 4i + 3i − + 4i + = = −1⋅ = −6 100 The product ( a + b)( a − b) + 3( −1) − + = − 3− = simplifies to a2 − b2 The product (a + bi)( a − bi) simplifies to 0=  96 a x2 − 6x + 11 = a2 − (bi) , which simplifies to (3+ i 2) − 6(3+ i 2) + 11= a2 + b2 101 Any real number For example: 102 Any complex number and its conjugate For example: + 5i and − 5i In general, for real numbers, a and b, ( a + bi)( a − bi) = a2 + b2 , which is a + 6i + 2i2 − 18 − 6i + 11 = + 2( −1) − 18 + 11 = 9− 2− = 0=  b real number 103 z ⋅ z = ( a + bi)( a − bi) = a2 + b2 x2 − 6x + 11 = ( ) ( ) 3− i − − i + 11 = 104 z2 − z2 − 6i + 2i − 18 + 6i + 11 = = ( a + bi) − ( a − bi) + 2( −1) − 18 + 11 = 2 = a2 + 2abi+ (bi) − a2 − 2abi+ (bi)    = a2 + 2abi+ b2i2 − a2 + 2abi− b2i2 9− − = 0=  = ( 4ab) i 97 ( a + bi)( c + di) 105 a x2 − = ( x + 3)( x − 3) = ac + adi+ bci+ bdi2 b x2 + = ( x + 3i)( x − 3i) = ac + ( ad + bc) i+ bd ( −1) 106 a x2 − 100 = ( x + 10)( x − 10) = ( ac − bd) + ( ad + bc) i 98 ( a + bi) = ( a) + 2( a)(bi) + (bi) = a2 + ( 2ab ) i+ b2i2 2 b x2 + 100 = ( x + 10i)( x − 10i) 107 a x2 − 64 = ( x + 8)( x − 8) b x2 + 64 = ( x + 8i)( x − 8i) = a + ( 2ab ) i+ b ( −1) ( 108 a x2 − 25 = ( x + 5)( x − 5) ) = a − b + ( 2ab ) i 2 b x2 + 25 = ( x + 5i)( x − 5i) 99 The second step does not follow because the multiplication property of radicals can be applied only if the individual radicals are real numbers Because −9 and −4 are imaginary numbers, the correct ( )( ) b x + = ( x + i 3)( x − i 3) 110 a x − 11 = ( x + 11)( x − 11) b x + 11 = ( x + i 11)( x − i 11) 109 a x2 − = x + x − 2 111 Section 1.4 Solution Manual for College Algebra 2nd Edition By Miller 111 113 112 114 10 Section 1.4 Quadratic Equation y2 + 7y − 18 = ( y + 9)( y − 2) = quadratic linear ± k 100 y + = or y − = y = −9 y=2 {−9,2} −b ± b2 − 4ac x = 2a b − 4ac ( x − 3)( x + 7) = x − = or x + = x=3 x = −7 { 11 8t(t+ 3) = 2t− 8t2 + 24t= 2t− 8t2 + 22t+ = ( 2t+ 5)( 4t+ 1) = 2t+ = or 4t+ = 2t= −5 4t= −1 t= − t= −  1 − ,−   4 } 3,− (t+ 4)(t− 1) = t+ = or t− = t = −4 t= {−4,1} 12 n + 5n = 24 y2 = 18 − 7y 6m ( m + 4) = m − 15 6m + 24m = m − 15 n + 5n − 24 = 6m + 23m + 15 = ( n + 8)( n − 3) = ( 6m + 5)( m + 3) = n + = or n − = n = −8 n= 6m + = or m + = m = −3 6m = −5 m =−   − ,− 3   {−8,3} 13 112 40p2 − 90 = Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller ( ) 10 4p2 − = 10( 2p − 3)( 2p + 3) = n − 4n + 2n − = 27 n2 − 2n − 35 = 2p − = or 2p + = 2p = 2p = −3 3 p= p=− 2  3  ,−   2 ( n + 5)( n − 7) = n + = or n − = 19 x2 = 81 32n − 162 = ( x = ± 81 = ±9 9,− ) 16n2 − 81 = { 2( 4n − 9)( 4n + 9) = 15 n= − w = ± 121 = ±11 {11,−11}  9  ,−   4 3x2 = 12x 21 5y2 − 35 = 5y2 = 35 y2 = 3x2 − 12x = 3x ( x − 4) = 3x = or x − = {0,4} x=0 { v2 = z = 25z z − 25z = { or z − 25 = } 5,− 4u2 = −64 u2 = −16 ( m + 4)( m − 5) = −8 u = ± −16 = ±4i {4i,−4i} m + 4m − 5m − 20 = −8 m − m − 12 = 24 8p2 + 72 = ( m + 3)( m − 4) = {−3,4} v=± 23 4u2 + 64 = z = 25 {0,25} 17 } 7,− 6v2 = 30 z( z − 25) = z= y=± 22 6v2 − 30 = x=4 16 } 20 w = 121 4n − = or 4n + = 4n = 4n = −9 n= n = −5 {−5,7} 14 ( n + 2)( n − 4) = 27 18 8p2 = −72 m + = or m − = m = −3 m =4 p2 = −9 p = ± −9 = ±3i {3i,−3i} 113 n= Section 1.4 Solution Manual for College Algebra 2nd Edition By Miller 25 ( k + 2) = 28 17 =± − 17 t= ± − i 17 = ± 2 17 i = ± 2  17  i  ±   t− k + = ± 28 k = −2 ± 28 = −2 ± {−2 ± 7} 26 3( z + 11) − 10 = 110 3( z + 11) = 120 ( z + 11) = 40 z + 11 = ± 40 47  1 30  a −  = −  3 47 a− = ± − z = −11± 40 {−11± 10} = −11± 10 47 ± − i 47 = ± 3 47 = ± i 3  47  i  ±   a= 27 2(w − 5) + = 23 2(w − 5) = 18 w − 5= ± w = 5± w = 5± w = + or w = − w =8 {8,2} 28 ( c − 3) = 49 w =2 1  31 x + 14x + n = x + 14x +  (14)  2  2 = x + 14x + ( 7) 2 c − = ± 49 = x2 + 14x + 49 c = 3± 49 = ( x + 7) c = 3± n = 49;( x + 7) c = 3+ or c = 3− c = 10 {10,−4} 2 c = −4 2 1  32 y + 22y + n = y + 22y +  ( 22)  2  2 = y + 22y + (11) 2 17  1 29  t−  = −  2 = y2 + 22y + 121 = ( y + 11) n = 121;( y + 11) 114 2 ... resulting 75% sand mixture 100% Sand 70% Sand 75% Sand 96 Full file at https://TestbankDirect.eu /Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution... 400 mph and the plane to New York City travels 460 mph 97 Full file at https://TestbankDirect.eu /Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.2 Solution Manual for College... 16(1.75)  16  2x + 5x = 28 7x = 28 x= The loop is mi 98 Full file at https://TestbankDirect.eu /Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution