MATH 536 MOCK EXAM SOLUTIONS

1) V(x) = 2 | x - 5 | + 30

 a) V(0) = \$40 b) min = \$30 c) 5 months d) V(12) = \$44 e) 5 months

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2) f(x) =

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b)

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3)

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 a) vert. sep: 25 horiz. length: 500 start at (0, 600) open on right b) \$625.00 c) [ \$500, \$1000 [

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4)a)

Domain: R , x ! -1

Range: R , y ! 3

f(x) > 0 ]- º , - 7/3 ] 4 ] -1, º [

f(x) < 0 [-7/3, -1 [

b) f(x) < 0 [-7/3, -1 [

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 5) let x = # of bottles y = # of cans with 45 bottles, 15 cans profit max = \$5.25

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 6) a) x = b) x = -4 c) x < 3 or x > 11 d) x = 0.8107

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 7) a) y = (x + 7) 2 - 2 b) c) y = log 3 (½ x) - 5 d) y = (3) ¼ x + 5

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 8) a) (x + 2) 2 + (y - 3) 2 = 25 b) 4x - 3y + 17 = 0 c) ( -5, -1) and ( 1, 7)

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 9) a) C (-1, -3); r = 6 b) C (0, 0); r = 5 c) y = (- x + 15) d) (5, 5) and (-5, -5)
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 10) a) b) c) d) 0 e)

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 11) a) 45' = 0.75 0 so 57.75 0 = 57.75 = 1.01 R b) 308.57 0 . c) cm = 13.35 cm d) 12.5 R .

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12) y = 4 sin (¼ x + o /8) + 3 becomes y = 4 sin ¼ ( x + o/2) + 3

 ampl: 4, up first period: 8o phase shift: - o/2 k = 3 max = 7 min = -1

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13) a) after 1st year, the car is worth 0.7(\$25,000) = \$17,500

so V(t) = 17,500(0.85) t

we want V(t) < 10,000 t 17,500(0.85) t < 10,000

The car will be worth less than \$10,000 after 4.44 years.

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14)

 Olivier: \$3000(1.075) t Lauren: \$4000(1.07) t .

\$3000(1.075) t = \$4000(1.07) t

log 3 + t (log 1.075) = log 4 + t (log 1.07)

t (log 1.075 - log 1.07) = log 4 - log 3

t = = 61.71 years

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15)

 ellipse: a = 6, b = 3 c = circles: center are ( r = Foci at set x = parabola: vtx. at (0, , P(! hyperbola: a = 6, c = 8 so b 2 = 28 so Foci (! 8, 0), asymptotes: y =

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