Solution manual for FINITE 1st edition by berresford

Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions Section 1.1 {x | ≤ x < 6} {x | x ≤ 2} For (2, 3) and (4, –1), the slope is −1 − = −4 = −2 4−2 For (2, –1) and (2, 5), the slope is −( −1) 2−2 = +1 undefined Since y = 3x – is in slope-intercept form, m = and the y-intercept is (0, –4) Using the slope m = 3, we see that the point unit to the right and units up is also on the line The equation y = is the equation of the horizontal line through all points with y-coordinate Thus, m = and the y-intercept is (0, 4) The equation x = is the equation of the vertical line through all points with x-coordinate Thus, m is not defined and there is no y-intercept First, solve for y: 2x − 3y = 12 −3 y = −2 x + 12 = y x−4 Therefore, m = 23 and the y-intercept is (0, –4) y − (−2 ) = 5[x − (−1)] 10 y = –4 y + = 5x + y = 5x + Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions x = 1.5 11 First, find the slope −1 − −4 m= = = −2 7−5 Then use the point-slope formula with this slope and the point (5, 3) y − = −2(x − 5) 12 y − = −2x +10 y = −2x +13 13 Low demand: [0, 8); Average demand: [8, 20); High demand: [20, 40); Critical demand: [40, ∞) 14 a To find the linear equation, first find the slope of the line containing these points 14 − m= = =4 −1 Next, use the point-slope form with the point (1, 6): y − y1 = m (x − x1 ) 15 a Price = $50,000; useful lifetime = 20 years; scrap value = $6000 V = 50,000 −  50,000−6000  t ≤ t ≤ 20   20 = 50,000 − 2200t ≤ t ≤ 20 y − = (x − 1) b y − = 4x − y = 4x + To find the profit at the end of years, substitute into the equation y = 4x + y = 4(2) + = $10 million b Substitute t = into the equation V = 50,000 − 2200t = 50,000 − 2200(5) = 50,000 −11,000 = $39, 000 c The profit at the end of years is y = 4(5) + = $22 million c on [0, 20] by [0, 50,000] © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions Section 1.2 ( ⋅2 ) =( ⋅2 ) =( ) 251 = 25 = 5 27 =  )  (125 ( x ⋅x ) = ( x ) 2 −3 ( )−3 = 23 = 1   2 25 −3 =  z z⋅z  (ww2 )3 (w3 )3 w9 = = = w5 w w w3 w 10 xy ) (5= 11 Average body thickness = 0.4(hip-to-shoulder length)3/ = 0.4(16)3/ 12 2 2/3 = 0.4 2 ( 16 ≈ 25.6 ft  27= 125  ) =26 =64 ( 53=) 25 = x10 32 = −1 = ( 25 ) = 53 = 125 ( ) = ( 4) z  =  = = 1 = 23 ( ) ( ) (  z z3 z  = z ⋅z6 ⋅z   27 z= z ) 25 x y8 y = 25 x3 y x 25 x3 y C′ = x 32 0.6 C = C ≈ 2.3C To quadruple the capacity costs about 2.3 times as much © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions Section 1.3 Yes a No f ( x) = x − Domain = {x | x ≤ or x ≥ 1}, Range = {y | y ≥ –1} f (10) = 10 − = = b a Domain = {x | x ≥ 1} since f (x ) = x − is defined for all values of x ≥ h(x ) = x h (81) = 81 14 b a z+4 h(−5) = = −1 −5+4 h( z ) = is z+4 defined for all values of z except z = –4 b Domain ={z | z ≠ –4} since h( z ) = = 81 = Domain = {x | x ≥ 0} since h (x ) = x is defined only for nonnegative values of x 10 11 x2 − x − = (x − 7)( x + 1) = Equals Equals at x = at x = −1 x = 7, 12 x + 2x = 15 x + 2x − 15 = (x + 5)(x − 3) = 13 Equals Equals at x = −5 at x = x = −1 x + 40 = 18x x −18 x + 40 = x − x + 20 = ( x − )(x − 5) = Equals Equals at x = at x = x = −5, x = x = 4, x = 14 x − 50 x = x − 10 x = x ( x − 10 ) = 15 Let x = the number of board feet of wood C(x) = 4x + 20 Equals Equals = at x 0= at x 10 = x 0,= x 10 © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions 16 a To find the break-even points, set C(x) equal to R(x) and solve the resulting equation C(x) = R(x) 100x + 3200 = –2x2 + 300x 2x2 – 200x + 3200 = Use the quadratic formula with a = 2, b = –200, and c = 3200 x= 200 ± (−200) − 4(2)(3200) 2(2) 200 ± 14,400 200 ± 120 = 4 320 80 x= or x = 4 x = 80 or x = 20 The store will break even when it sells either 20 exercise machines or 80 exercise machines = b To find the number of exercise machines that maximizes profit, first find the profit function, P(x) = R(x) – C(x) P(x ) = −2 x + 300 x − (100 x + 3200 ) ( ) −2 x = + 200 x − 3200 Since this is a parabola that opens downward, the maximum profit is found at the vertex −200 −200 x= = = 50 2(−2) −4 Thus, profit is maximized when 50 exercise machines are sold per day The maximum profit is found by evaluating P(50) P(50) = –2(50)2 + 200(50) – 3200 = $1800 Therefore, the maximum profit is $1800 © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions Section 1.4 a f ( x) = x2 x −1 (−1) =− −1−1 b Domain = {x | x ≠ 1} a g(x ) = x + g(−5) = (−5) + = −3 = f (−1) = b Domain = all real numbers x + 2x − 3x = x x + 2x − = ( x (x ) x +18x = 12x 2x − 12x +18x = 2x x − 6x + = ( + 3)(x − 1) = ) 2x (x − 3) = Equals Equals Equals at x = at x = −3 at x = Equals Equals at x = at x = x = 0, x = −3,and x = x = and x = a f (g(x )) = [g(x )] = (7 x − 1) b g( f (x )) = 7[f (x )]− = x 5 ( )− = x5 −1 1 = g ( x) x + a f ( g ( x)) = b 1 g ( f ( x)) = [ f ( x)]2 + =   +  x © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions a For x = 3000, use ƒ(x) = 0.10x f (x ) = 0.10 x f (3000) = 0.10(3000) = $300 b For x = 5000, use ƒ(x) = 0.10x f ( x ) = 0.10 x = f ( 5000 ) 0.10(5000) = $500 c For x = 10,000, use ƒ(x) = 500 + 0.30(x – 5000) f (x ) = 500 + 0.30(x − 5000) f (10, 000) = 500 + 0.30(10,000 − 5000) = 500 + 0.30(5000) = $2000 d 10 First find the composition R(v(t)) 0.3 R(v (t )) = 2[v(t)] = 2(60 + 3t ) 0.3 Then find R(v(10)) R(v(10)) = 2(60 + 3(10))0.3 = 2(90)0.3 ≈ $7.714 million © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions Section 1.5 To find the value, evaluate P(1 + r ) when P = 1000, r = 0.10 and t = 8: n Value = P (1 + r ) n a = 1000 (1 + 0.10 ) = 1000 (1.10 ) = P 1 + r  Value  m mt 4(65) To find the value, evaluate P 1 + r   m when P = 1000, r = 0.10 and t = 8: = P 1 + r  Value  m mt To find the value, evaluate P 1 + r   m when P = 5810, r = 0.08, m = and t = 65: = 5810 1 + 0.08    260 = 5810(1.02) ≈ $1, 000,508.52 Yes, the trust fund will be worth $1,000,508.52 ≈ $2143.59 b mt mt 4(8) = 1000 1 + 0.10    = 1000(1.025)32 ≈ $2203.76 c To find the value, evaluate Pe rt when P = 1000, r = 0.10 and t = 8: Value = Pe r t = 1000e0.10(8) ≈ $2225.54 © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions To find the value, evaluate Pe rt when P = 560, r = 0.058 and t = 10: Value = Pe r t To find the value, evaluate Pe rt when P = 2000, r = 0.055 and t = 8: Value = Pe r t = 560e0.058(10) ≈ $1000.18 Yes, the bond will reach its “par” value of $1000 in ten years a n To find the amount, evaluate P(1 + r ) when P = 20,000, r = –0.035 and t = 4: = 2000e0.055(8) ≈ $3105.41 Since 2020 is 20 years afer the year 2000, the population of the world will be 5.89e0.0175x evaluated at x = 20 5.89e0.0175x = 5.89e0.0175(20) ≈ 8.36 billion 10 a Value = P (1 + r ) t = 20, 000 (1 − 0.35 ) = 20, 000 ( 0.65 ) 4 ≈ $3570 (rounded) b To find the amount, evaluate P(1 + r ) when P = 20,000, r = –0.035 and t = 0.5 (6 months): n Value = P (1 + r ) t = 20, 000 (1 − 0.35 ) = 20, 000 ( 0.65 ) 0.5 0.5 ≈ $16,125 (rounded) a b 100(25) ≈ 0.53 P(25) = 1− (0.9997) The probability meltdown within 25 years is approximately 53% P(40 ) = 1− (0.9997)100( 40) ≈ 0.70 The probability meltdown within 40 years is approximately 70% After 15 minutes, t = 15 60 = 0.25 and T (0.25) = 70 + 130e −1.8( 0.25 ) ≈ 153 degrees b 30 0.5 and 60 T (0.5) = 70 + 130e −1.8( 0.5 ) ≈ 123 degrees After 30 minutes, t = © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions 10 Section 1.6 log 100,000 = 5, since 10 = 100, 000 b log c log 10 = a log 16 = , since = 16 b log c log = a , since 101 = 10 = −1 , since −1 = a ln e5 = (Property 3) b ln c ln e = 1 = −1 , since e −1 = e e , since e1 = e f (x ) = ln(9x ) − ln9 = ln + ln x − ln9 = ln x , since 41 = = = ln x − ln + ln = ln x Let P represent the original value of the car Then 0.5P represents half its original value Use this fact and the formula, P(1 + r)t to write an equation P(1 − 0.30)t = 0.5 P Solve the equation for t 0.5 P P(1 − 0.30)t = (0.70)t = 0.5 log(0.70)t = log(0.5) t log(0.70) = log(0.5) log(0.5) t= log(0.70) t ≈ 1.9 The car will be worth half its original value in about 1.9 years b 1 = −2 , since 10 − = 100 100 f ( x ) = ln ( 4x ) + ln a f ( x ) = ln (e x ) − x − ln1 = 5x − x − = 3x Let P represent the original value of the car Then P represents two-thirds its original value Use this fact and the formula, P(1 + r)t, to write and solve an equation P(1 − 0.15)t = P (0.85)t = 3 () t log(0.85) = log ( ) log ( ) t= log(0.85)t = log 3 log(0.85) t ≈ 2.5 The printing press will be worth twothirds of its original value in about 2.5 years Let P represent the original value of the car Then 0.25P represents one-quarter its original value Solve the following formula for t P(1 − 0.30)t = 0.25 P (0.70)t = 0.25 log(0.70)t = log(0.25) t log(0.70) = log(0.25) log(0.25) t= log(0.70) t ≈ 3.9 The car will be worth one-quarter its original value in about 3.9 years © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Solution Manual for FINITE 1st Edition by Berresfor Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor Functions 11 The proportion of carbon-14 present after t years is e −0.00012t If the carbon used in the drawing contains 2.3% of its original carbon14 then e −0.00012 t = 0.023 ln(e −0.00012 t ) = ln(0.023) −0.00012t = ln(0.023) ln(0.023) t= −0.00012 t ≈ 31, 400 The estimated age of the cave paintings is 31,400 years 10 The proportion of potassium-40 remaining −0.00054t after t million years is e If the skeleton contained 99.91% of its original potassium-40, then e −0.00054 t = 0.9991 ln e −0.00054 t = ln(0.9991) −0.00054t = ln(0.9991) ln(0.9991) = ≈ 1.7 t −0.00054 The estimated age of the skeleton is 1.7 million years © 2012 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-FINITE-1st-Edition-by-Berresfor