| 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 |
.
2) f(x) =
a)
b)
.
3) 
.
| a) vert. sep: 25 | horiz. length: 500 | start at (0, 600) | open on right |
| b) $625.00 | c) [ $500, $1000 [ |
.
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 [
.
| 5) let x = # of bottles | y = # of cans | with 45 bottles, 15 cans | profit max = $5.25 |
.
6) a) x =  |
b) x = -4 | c) x < 3 or x > 11 | d) x = 0.8107 | |
.
| 7) a) y = (x + 7) 2 - 2 | b)  |
c) y = log 3 (½ x) - 5 | d) y = (3) ¼ x + 5 | |
.
| 8) a) (x + 2) 2 + (y - 3) 2 = 25 | b) 4x - 3y + 17 = 0 | c) ( -5, -1) and ( 1, 7) |
.
| 9) a) C (-1, -3); r = 6 | b) C (0, 0); r = 5 |
c) y =  (- x + 15) |
d) (5, 5) and (-5, -5) |
10) a)  |
b)  |
c)  |
d) 0 | e)  |
.
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 . | |
.
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 |
.
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.
.
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
.
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 =  |
.
.
.
.