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Solutions Math 11A Exam Review 2011
Chapter 1 – Investigating Equations in 3-Space
1. a) (c: total charges, t: time in hrs) c = 6 + 2t, c = 7t, c = 20 b) 1.2 hrs, $8.40
c) $20
2. a) (b: base fee, v: per visit fee) 60.5 = b + 17v, 40.5 = b + 9v
b) b = $18, v = $2.50
3. a) (4, –3)
b) (-25/3 , -17/6)
c) (1.5, 4, -3)
4. (2, 8)
5. (4, -3)
6. Answers vary
7. a) 4a+2b+c=85, 9a+3b+c=95, 25a+5b+c=96
b) y=-3.17x2+25.83x+46
c) 66.67%
e) Mary
Chapter 3 – Sinusoidal Functions
1. a) neither b) sinusoidal c) neither d) periodic
2 a) Rx, VS ½, VT 5, HS 4, HT –30o b) VT 1, VS 3, HS 1/2, HT 45o
3. a) –1/4 (y + 5) = cos [1/3 (x – 10o)] b) amplitude 4, period 1080o, sinusoidal axis y = –5
4.
x
y
5. Examples: 1/10(y+30)=cos[18(x–3o)], 1/10(y+30)=sin[18(x+2o)],
20
15
–1/10(y+30)=sin[18(x–8o)], –1/10(y+30)=cos[18(x–13o)]
65
9
6. a) example: 1/120 y = cos [120o (x–2)]
b) y = –60
110
3
155
9
200
15
Chapter 4 – Trigonometric Equations
2. a)
 3
b)
2
3
3
c)
 3
1
d)
2
2
3. a)
4. a) 30o, 110o, 150o b) 30o+120ok, 110o+120ok, kI
5. a) x = 180ok, 135o+180ok, kI
b) x = 30o, 150o, 90o
3
3

2 2
b)
1
20 3  9 6
c)
4
18
c) x = 75o+180ok, 105o+180ok, kI
7. x  30360 k,150360 k, k I
8. a) x  150360 k, 210360 k, k I
b) x  60 ,120 ,180
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
 5
,
4 4
x  180 k, k I
x
x  330 ,  20 , 30 ,150
x  210 , 330
x    2k, k I
x  45

x   k, k I
4
x  50.95 ,  90 ,  270 ,  309.05
x  58
.  2k, 36
.  2k, k I
x  22.7445 k, k I
12xy
, x  y
 x  y x  y
1
, x  7,5,1,0,7
 x  7 x  5 x  1
2x  5
, x  31
, ,4
x

1
  x  3 x  4
9. a)
b)
c)
d)
2 x 2  7 x  10
, x  2,11
,
 x  1 x  2 x  1
10a) tan x + cot x = (sec x)(csc x)
sin x cos x  1  1 




cos x sin x  cos x  sin x 
multiply by (sin x)(cos x)
sin2 x + cos2 x = 1
b) csc2 x – (cos2 x)(csc2 x) = 1
csc2 x (1 - cos2 x) = 1
csc2 x ( sin2 x) = 1
1 =1
c) (sin x)(cos x) – (sec x)(sin x) = (- sin2 x)(tanx)
Divide by sin x
cos x – sec x = (- sin x)(tan x)
Multiply by cos x
cos2 x - 1 = - sin2 x
cos2 x + sin2 x = 1
d) 5 sin2 x + 3 cos2 x = 3 + 2 sin2 x
5 sin2 x – 2 sin2 x = 3 – 3 cos2 x
3 sin2 x = 3 ( 1 - cos2 x)
3 sin2 x = 3 sin2 x, 1 = 1
cos 4 x  sin 4 x
 cos x  sin x
cos x  sin x
(cos2 x - sin2 x)( cos2 x + sin2 x) = cos2 x - sin2 x
cos2 x + sin2 x = 1
e)
sec x sin x
= cot x
sin x cos x
cos x
 1  1  sin x
=


sin x
 cos x  sin x  cos x
Multiply both sides by (sin x)(cos x)
f)
1 – sin2 x = cos2 x
sin2 x + cos2 x = 1
g) (sec2 x)( csc2 x) = sec2 x + csc2 x
1
1
 1  1 
+

 2  =
2
2
2
 cos x  sin x  cos x sin x
Multiply by (cos2 x)( sin2 x)
1 = cos2 x + sin2 x
11
a)
5
36
b) 2571
. 
12.
 32
2
3  9 2
b)
17
a)
13. 6.28 cm
(cos x)(sec x)
 cot x
tan x
cos x  1 
 cos x  = 1
tan x
tan x
h)
1
1
=
tan x
tan x
1=1
14. a)
b)

1
 d  6  cos 36  x  2
7
1886
.
m

c)
d) 561
.  10k, 8.39  10k, k I
1
9
 y  100  cos   x  512.6 
150
35


b)
10252
. m
15.a)
c)
374.8 m
`
Chapter 5 - Statistics
1.
2.
3.
4.
mean 16.9
median 15 standard deviation 5.2436
68% 95% 99.7%
a. 81.5% b. 68% c. 50% d.97.35%
Suggested Solutions
Simple random – randomly select burgers from they supplier and sample them. (use random # generator to select)
Stratified random – randomly sample 10% of all burgers produced in each of its stores and test.
Cluster – test all the burgers in 150 (a portion of the population) of the McDonalds stores across the country.
Systematic- Check every 100th burger produced by the manufacturer.
5.
When collecting a repeated sample of the same size and examining the distributions of these sample means
a. The means when graphed will form a normal or bell shaped curve
b. The mean of the sample means will equal the population mean =
c. The standard deviation of the sample means is equal to the standard deviation of the population divided by the square
root of the sample size.
6.
a) The mean of the sample means equals the population mean = 64.
b) the standard deviation of the sample means = 0.814
7.
23.3178 to 27.5482
8.
9.
8.1578 to 8.9363
0.3892
Chapter 6 – Trigonometry and its Applications
1.
2.
33.2 cm2
a
b
c


, a 2  b 2  c 2  2bc cos A
sin A sin B sin C
354
. 
a  236
. , B  511
.  or128.9 , C  939
.  or16.1
3.
4.
5. B 
6. 34.74 m
7. 1050m