Download ALGEBRA II WITH TRIGONOMETRY

ALGEBRA II WITH TRIGONOMETRY
MIDTERM EXAM REVIEW – PART I
ANSWER KEY
CHAPTER 1 PROBLEM SET
1)
a)
b)
x6z3
c)
x6
4y
d)
2)
3)
4)
5)
3xy3
Multiplicative Identity
Multiplicative Inverse
Commutative Property of Addition
Distributive Property
Associative Property of Multiplication
Additive Identity
a)
2
b)
b)
7)
a)
x=3
y=2
b)
x=
a)
c)
e)
g)
i)
6 – 2i
-4 – i
26 – 7i
15 – 8i
-i
k)

d)

a)
3 7
b)
5i 6
c)
16i
d)
e)
53 3  10 6
f)
30 14
g)
637
h)
6 6
i)
7 6 3
2
m)
9)
2
34
9
4
30  5 2
34
10)
3 + 5i
y=2
3
b)
d)
f)
h)
j)
5i
6
35  10i
53
a)
c)
no soution
x = 94
e)
t 
g)
L
i)
r = 2,
k)
A P
Pr
S
1 r
6
-6 + i
-16 + 8i
169
-i
-72i
13i
l)
6
14  23i
n)
29
b)
d)
x=1
x=2
f)
h 
h)
x
V
r2
b
2  4a
11
j)
x = -5,
x=4
l)
m=
a)
c)
x<3
5 < x  12
b)
d)
x  -3
-9  x  6
e)
-1 < x < 2
f)
-2  x 
g)
x>
5
5
12
k)
69
2 – 7i
3
c)
10
3
a)
1
16z 4 y12
a)
b)
c)
d)
e)
f)
30 
6)
8)
a) Rat. Real, Complex
b) Rat. Real, Complex
c) Rat. Real, Complex
d) Imag, Complex
e) Int, Rat, Real, Complex
f) Pure Imag, Imag, Complex
g)Whole, Int, Rat, Real, Complex
h) Irr, Real, Complex
j)
l)
h)
14
or x < 
4
3
3
x < -5 or x > 2
3
21
11
2
3
CHAPTER 2 PROBLEM SET
1)
a)
b)
c)
d)
D: {x  -5, -2, 0, 4, 7}
R: {y  -9, -3, 1, 9, 15}
Yes because there is exactly one x
mapped to each y
y = 2x + 1
y
x
e)
-5
-2
0
4
7
-9
-3
1
9
15
2)
y = -6
4)
a)
b)
c)
d)
f(-6) = -27
g(2) = -26
f(2a) = 8a – 3
f(g(-4)) = -443
a)
(
5)
6)
7)
3)
2
$1023.16 per year
, 0) and (0, 
1
b)
2
3
(6, 0) and (0, 4)
a)
2x + 3y = 12
b)
y=
c)
d)
e)
x+1
3
2x – y – 5 = 0
y = -7
x = -2
f)
y= 
a) f(5) = -3
2
x+
5
9)
a)
y = 12x + 55
a)
k=
b)
k = 63; y =
c)
w=
5
2
; y=
1
5
2
x
63
x
xy
3
w = 68
d)
y=
800
x2
y = 32
2)
a)
b)
coincide (many sol’n)
parallel (no soln)
3)
a)
(2, 3)
4)
a)
y-int: 4 slope: ½
shade below the solid line
y-int: 2 slope: -½
shade above the dotted line
y-int: -1 slope: 3
shade below dotted line
y-int: 3 slope: -1
shade above the solid line
b)
b)
5)
2.80x + 5.30y = 330
x + y = 100
80 peanuts, 20 cashews
6)
a)
 2

11
c)
not possible d)
2
b) f(-4) = 3
7c and 8)
See graphs on next page
a) y-int 2, slope -2/3
b) y-int 2, slope 5/2
c) y-int 2, slope -5/3,
shade below dotted line
d) y-int -4, slope 3,
shade below solid line
e) vertex (3, 5), opens up slopes 1 and -1
f) vertex (-3, 0), opens up slopes 2 and -2
88%
3.75 hours
CHAPTER 3 PROBLEM SET
1)
a)
(4, 2)
b)
no solution
)
1
3
10)
b)
c)
7)
8)
2

9 
b)
7) e) vertex (3, 5), opens up
x =slopes
1/3, y1=and
-4, -1
z=3
C(0, 0) = 0 (min)
C(0, 5) = 30 (max)
C(2, 4) = 30 (max)
C(6, 0) = 18
(6, -2)
 8 13


12 
2


4
6 
 4 
3


2
8
2

 

3
3
3
Chapter 2 Graphs
y
Problem 7c)
x
Problem 8:
a)
2x + 3y = 6
b)
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c)
5x + 3y < 6
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d)
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e)
f(x) = |x – 3| + 5
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5x - 2y = -4
6x – 2y  8
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f)
f(x) = |2x + 6|
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CHAPTER 4 PROBLEM SET
1)
2)
3)
b)
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
(x – 1) (x – 6)
(3x – 8) (3x + 8)
3x2 (x - 8) (x + 2)
3x (x – 1) (x – 7)
(2x + 3) (4x2 – 6x + 9)
(2x + 3)2
(x + 9) (x – 9)
(4x – 3) (16x2 + 12x + 9)
(5x2 + 2) (3x + 2)
(k + 3) (k – 3) (k + 4)
a)
x = 2 2
b)
x = 5
c)
x = 5  29
d)
x= 
e)
x=
f)
x=
a)
Vertex: (2, -1)
Axis of Symmetry: x = 2
Y-Intercept: (0, 3)
X-Intercepts: (1, 0) and (3, 0)
5  i 3
2
Vertex: (
1
4
, 
49
8
Axis of Symmetry: x =
Y-Intercept: (0, -6)
X-Intercepts: ( 
1
2
3
2
, 0) and (2, 0)
4)
a)
,2
1i 5
2
Discriminant: -31
2 imaginary roots
x=
c)
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1  i 31
16
Discriminant: 48
2 real roots
Discriminant: -92
2 imaginary roots
x=
d)
5  i 23
12
Discriminant: 0
1 real root
x=
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4
Vertex: (1, 4)
Axis of symmetry: x = 1
y-intercept: (0, 3)
x-intercepts: (3, 0) and (-1, 0)
d)
Vertex: (-3, -5)
Axis of symmetry: x = -3
y-intercept: (0, 13)

5 2 
x-intercepts:  3 
, 0


2
x = 3  2 3
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1
c)
b)
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)
5
3