Advanced Algebra 2 – Final Review Packet KG3 2011
2. Evaluate (7  y)  10x when x 

 1 
1

x 1
5. Solve the equation. 5 2  x  4 
 2 
4


6. An awards dinner costs $225 plus $5 for each
person making reservations. The total bill is $735.

How many people
made reservations?
7. For 1980 through 1990, the population, P, (in
thousands), of Hawaii can be modeled by P = 17(t +
56.6) where t = 0 represents 1980. What was the
population in 1987?

22. In 1980 the Wincom river was 45 feet below the
bridge. Because of silt build-up in the river bottom,
1
the river was only 27 feet below the bridge by
2
1987. Write an equation for the distance of the river
from the bridge, d, with t = 0 representing 1980. If
nothing is done about the silt, what year will the
 the bridge?
river reach
24. For the scatter plot shown, state whether x and y
have a positive correlation, a negative correlation,
or no correlation.
y
10
5
8. Solve the inequality. Then graph your solution.
6x  5  25
–10
5
a solution of the inequality
2
5x  4  3(x  7) ?
–5
9. Is x =
12. Solve the inequality. Then graph your solution.
3x  2 5
13. Solve the inequality. Then graph your solution.
2x  3  5

14. Solve the inequality. Then graph your solution.
2x  5  1
18. Determine whether the relation is a function.
(0, 4), (1, 4), (2, 5), (3, 6), (4, 6)
19. Determine whether the relation is a function.
(4, 0), (4, 1), (5, 2), (6, 3), (6, 4)
 1
20. Find f  . f (x)  18x 2  12x  3
 3

10
x
–10
Solve:
11. x  3  5 or x  4  14

5
–5
10. Is x = –7 a solution of the inequality
5x  7  3(x  7)?


21. Find the slope of the line passing through (3, -1)
and (6, 4).
3. Solve the equation.  x  3  7x  8

4. Solve the equation. 
5(3  4x) 7  (4  x)


2
1
and y  .
7
7
Page 1 of 18

25. For the following data:
A. Make a scatter plot of the data.
B. Approximate the best fitting line for the data.
C. Find an equation of your line of best fit.
x
1
2
3
4
5
6
7
8
y 1.75 4.1 4.95 7 8.15 11.1 11.95 14
26. Graph the inequality in a coordinate plane.
7
x7
3
1

 x  1, x  0
27. Graph the function. f x    2

 2x  1, x  0
28. Graph the equation. y = 2x  5
29. Graph the equation. y  x  2  2

 P (in thousands), of a town can
30. The population,
be modeled by P  2 t  8  4 , where t = 0
 During which two years does the
represents 1990.



Advanced Algebra 2 – Final Review Packet KG3 2011
44. Solve the system: 3x  4 y  2z  1
town have a population of 8000?
2x  2y  4z  12
31. Solve the system graphically:
4 x  3y  2
x  3y  2z  6
 x  2y  2
45. Solve the linear system.
x  2y  4z  12
–x
+ z = 1
x+ y +z = 4
32. Solve the linear system: 3x + 3y = 5
y x
33. Solve the system.
y  4 x  4

y  x  5

34. Solve the linear system.
7x – 4y = 12
3x + 2y = 4
35. The Modern Grocery has cashews that sell for
$3.50 a pound and peanuts that sell for $2.00 a

pound. How much of each must Albert, the grocer,

mix to get 60 pounds of mixture that he can sell for

$3.00 per pound. Express the problem as a system
of linear equations and solve using the method of
your choice to find the solution of the problem.
36. Sketch the graph of the system of linear
inequalities.
x  2
y4
43. A florist sells bouquets of daffodils and jasmine.
The florist pays $0.42 each for daffodils and $0.87
each for jasmine. He must also pass along his
overhead costs of $0.82 per bouquet to pay his rent,
refrigeration, and workers’ salaries. Write an
equation that models the total cost (TC) of a

bouquet of flowers as a function of the number of
daffodils and jasmine used. Complete the table
which details the total cost for several different

numbers of daffodils and jasmine.
daffodils
jasmine 02 2 3 4 5

3
4
5

Page 2 of 18

46. Solve the linear system.
2x + y – z = –1
x – 2y
= –7
x+y+ z = 4
47. Solve the linear system.
2x  3y  z  1
x yz3
3x  y  z  15
48. Tasty Bakery sells three kinds of muffins:
chocolate chip muffins at 35 cents each, oatmeal
muffins at 40 cents each, and cranberry muffins at
45 cents each. Charles buys some of each kind and
chooses three times as many cranberry muffins as
chocolate chip muffins. If he spends $9.60 on 23
muffins, how many oatmeal muffins did he buy?
49. Write the augmented matrix for the linear
system.
 x  4 y  2z  3

5x  7y  z  4
3x +2y  2z  8

50. Does the parabola open up or down?
y  4  6x  2x 2
51. Does the parabola open up or down?
y  7  5x  3x 2
52. Sketch the graph of the equation.
y  x2  4x 1
53. The surface of a cube is 380 square inches. How
long is each edge? (Round to two decimal places.)
Advanced Algebra 2 – Final Review Packet KG3 2011
70. Solve the equation by completing the square.
x 2  2x  35  0
x
x
71. Solve the equation by completing the square.
x2  4x  2  0

x
72. Find the value of c that makes x 2  10x  c a
perfect square trinomial. Write the new expression
as the square of a binomial.
54. Solve by factoring: x  18x  81  0
2

55. The base of a triangle is five feet longer than the
height. The area of the triangle is 75 square feet.

Find the height and base of the triangle.

74. Write the quadratic equation in vertex form.
What is the vertex? y  3x 2  48x  201
56. Find the zeros of the equation.
x 2  2x  15 = y
75. Write the equation in the form
y  a(x  h) 2  k. Then graph the equation.

y = x 2  2x + 2
57. Solve: 2x 2  5  3

58. Solvethe equation.  x 2  4  2x 2  5
76. Use the quadratic formula to solve the equation.
x 2  2x  1  0



60. The height, h (in feet), of a falling object on
Mars is given 
by h  6t 2  s , where t is the time
in seconds and s is the initial height in feet. If an 
object were dropped from a height of 237 feet, how
long would it take to travel half the distance to the

ground? (Round to two decimal places.)
77. Solve: 9x 2  60x  100
80. A rock is thrown from the top of a tall building.
The distance, in feet, between the rock and the

ground
t seconds after it is thrown is given by
d  16t 2  2t  445 . How long after the rock is
thrown is it 440 feet from the ground?
61. The height, h (in feet), of a falling object on
Mars is given by h  6t 2  s , where t is the time
in seconds and s is the initial height in feet. If an

object were dropped from a height of 125 feet, how
long would it take to reach the ground? (Round to

two decimal places.)
81. Graph: y  x 2  x
82. Sketch the graph of the inequality.
y  2x 2  8x  6

62. Solve: (6x  4)2 = 77
 Solve the equation. 4 x 2  5  7
64.
83. An arrow shot into the air is 208t  4.9t 2
meters above the ground t seconds after it is
released. During what period of time is the arrow
above 176.4 meters? Round your answer to the

nearest .01 second.

Write the expression as a complex number in
standard form.

65. i2  i
84. Write a quadratic function in vertex form that
has the given vertex and passes through the given
point. Vertex: (–9, –8); Point: (–11, 0)
63. Solve the equation. 2x 2  x  3  0

Simplify:
85. (2s3 t 4 u)2
66. (3  7i)(1 2i)


Page 3 of 18
67.
4i 3
4i 3
86. (3x 2 ) 3

Advanced Algebra 2 – Final Review Packet KG3 2011


x
102. Which equation’s graph passes through the
points 1, 0, 3, 0, 6, 0, and 0, 18?
[A] f (x )  x 3  10x 2  27x  18
[B] f (x )  x 3  4 x 2  15x  18
f (x ) 
x 3  4 x 2  15x
 [C] 
  18
3
2
 [D] f (x )  x  10x  27x  18

103. Simplify: 8 4 3


104. Evaluate. 16 5/4
3
12
88. Solve for x. 4 2  4 x 1  4 3  4 5

89. Evaluate the polynomial when w = 3:
3w 3 5w 2  2w  5
Graph:
90. y  x 4  x 2  3


x 1
87. Solve for x. 3  3  3
3
Page 4 of 18
91. Multiply: (x  3)(x 2  2x  4)


105. Use a calculator to evaluate (16) 1/3 . Round the
result
to three decimal places.
92. Factor: 2x 7  10x 5  28x 3

93. Factor
completely with respect to the integers.
4
10x
  160
106. Rewrite 71/5 using radical notation.

107. Evaluate
a calculator.

94. Factor completely with respect to the integers.
2x 3  3x 2  4 x  6
96. Find all real zeros of the function.
g(x)  2x 3  x 2  6x  3
110. x1 3  x1 4

98. Write a polynomial function that has the zeros
2, –2, and –1 and has a leading coefficient of 1.
Then graph the function to show that 2, –2, and –1
are solutions.

100. Graph the function y  x 3  4 x 2  x  6 .

101. You are given a piece of cardboard 14 inches
long and 12 inches wide. You want to create an
 by cutting x-inch squares out of
open topped box
the corners and folding up the sides so the edges
you just cut form right angles. What is the
maximum volume of the box (rounded to the

nearest tenth of a cubic inch)? What are the
approximate dimensions of the box (rounded to the
nearest quarter-inch)?
(Remember: 0  x  6 )


9804 to three decimal places using
Simplify:
251/6
109. 2/3
25
95. Find all real-number solutions.
x 3  6x 2  12x  8  0
99. Solve for x: x 4  12x 2  11 = 0
7

111. Write the expression in simplest form.
16xy 2
27z 5
112. The surface area of a tennis ball is 21.43 in 2 .
The surface area of a billiard ball is 13.32 in 2 . Find
the ratio of the volumes of a tennis ball to a billiard
4
 =  r 3 .
ball. Surface Area = 4 r 2 and Volume
3

113. Let f (x)  1 x 2 and g(x)  1 x . Find
f (x)  g(x).

114. A large city is growing by a rate of 0.5%

 3,260,000 residents of the
annually. If there were
city in 1997, predict how many (to the nearest
thousand) will be living in the city in 2002. Use
y  3,260,000(2.7) 0.005t , where t  0 represents
1997.

Advanced Algebra 2 – Final Review Packet KG3 2011
Page 5 of 18
125. The sales of a certain product after an initial
release can be found by the equation
s  14 7t  45 , where s represents the total
sales and t represents the time in weeks after
release. How many weeks will pass before the
product sells about 200 units? Round your answer
to the nearest week.
115. What is the equation for the inverse of the
function f (x) = 3x + 1?
116. Find the inverse of the relation.
(1, 7), (2, 5), (3, 3), (4, 1)

117. Find the inverse of the relation.
(1, 5), (2, 3), (3, 1), (4, –1)

Solve the equation. Check for extraneous solutions.
126. 3 x  5  5
118. Find the inverse of the relation.
(1, 1), (2, 2), (1, 3), (4, 4)

119. Write an equation for the inverse of the
relation. y  11x  9
128. Find the mode of the set of data.
10, 18, 19, 13, 18, 19, 10, 13, 19, 12
129. Graph: f (x)  3x
120. Sketch the graph of the function and its inverse
on the same coordinate plane. f (x)  2  2x

130. Find the value of $1000 deposited for 10 years
in an account paying 7% annual interest

compounded
yearly.
131. The projected worth (in millions of dollars) of
a large company is modeled by the equation
x
y 2261.03 . The variable x represents the
number of years since 1997. What is the projected
annual percent of growth, and what should the
company be worth be in 2003?


132. A company had total sales of $3,200,000 in
1985. Each year between 1985 and 1991 the sales
increased by 12%. Approximate the sales for 1991
to the nearest $100,000.
121. Sketch the graph of the function and its inverse
on the same coordinate plane. f (x)  2  4x
133. A piece of equipment costs $85,000 new but
depreciates 15% per year in each succeeding year.
Find its value after 10 years.

134. Evaluate: log 3 9
Find the inverse of the function.
135. y  log 8 x

136. y  log1/5 x
Refer to the function g(x)  1 x  3 .
124. What is the domain of g(x)?


138. Evaluate log 6 30 to three decimal places.
 139. Evaluate log 782 to three decimal places.
12


Advanced Algebra 2 – Final Review Packet KG3 2011
Page 6 of 18
140. Graph the function. State the domain and
range. y  log 2 (x 1)
158. Perform the operations and simplify.
3x  4
2

2
x  16
x4
142. Expand the expression. log 3 (x 2 y 3 )

145. Solve:
1
= 27 3x 5
9

2
159. Simplify: x  4
6
+ 6
x

146. Expand using the properties of logarithms:
3
5xy
loga 5
z
154. Sketch the graph of the function. f (x) 
4
3

= 0
160. Solve:
f 3
f 4

161. Find the distance between point A8, 2 and
point C7, 9, then find the midpoint of AC.

x2
x 2


162. Write the standard form of the equation of the

circle that passes through the point (3, 4) with its

center at the origin.
163. The pool at a park is circular. You want to find
the equation of the circle that is the boundary of the
pool. Find the equation if the area of the pool is 900
square feet and (0, 0) represents the center of the
pool.
155. Identify all horizontal and vertical asymptotes
x3
of the graph of the function. f (x)  3
x 8
164. Graph:
x 2  10x  21
x7
156. Divide:

2
x7
x 9
165. Graph the equation and identify the
x2
y2

= 1
asymptotes:

4
49
9
157. The length of 
a rectangle is
m, while its
y4


2
width is
m. Which of the following is true?
y
166. Determine the foci and vertices of the graph of
x2
y2
= 1.
+
16
36

18
m
[A] perimeter:
y(y  4)

[B] area:



11y  8
m
[C] perimeter:
y(y  4)
[D] area:

18
m2
y(y  4)

36
m2
y(y  4)

x2 y2

1
16 64
167. Write the equation in standard form and
classify the conic section.
2x 2  2y 2  12x  20y  66 = 0
168. Solve the system by substitution:
3x  4 y10
y 2x  3
170. Find the common difference of the arithmetic
sequence.
1 2 5
, , , 1, . . .
2 3 6
Advanced Algebra 2 – Final Review Packet KG3 2011
171. Find the common difference of the arithmetic
sequence.
3
7
 , 1, , 6, . . .
2
2
Page 7 of 18
185. A photographer points a camera at a window in
a nearby building forming an angle of 42 with the
camera platform. If the camera is 54 m from the
building, how high above the platform is the
window, to the nearest hundredth?

172. Find the common difference of the arithmetic
sequence.
–3.6, –3.9, –4.2, –4.5, . . .
x
42°
176. Find the common ratio of the geometric
sequence.
2, –8, 32, –128, . . .
177. In a financial deal, you are promised $700 the
first day and each day after that you will receive
65% of the previous day’s amount. When one day’s
amount drops below $1, you stop getting paid from
that day on. What day is the first day you would
receive no payment and what is your total income?
178. Expand 2s  3t  .
3
179. Half of a circle is inside a square and half is
outside, as shown. If a point is selected at random
 the square, find the probability that the point
inside
is also inside the circle.
r
2r
180. Eight balls numbered from 1 to 8 are placed in
an urn. One ball is selected at random. Find the
probability that it is NOT number 3.
181. A and B are independent events.
P(A) = 0.6
P(B) = 0.8
Find PA and B.
183. A fair coin is tossed 14 times. What is the
probability of obtaining exactly 1 head? Express the
 answer both in terms of n Ck and as a four-place
decimal.

54 m
187. Given triangle ABC with a = 17, C = 21°, and
B = 26°, find c. Round the answer to two decimal
places.
188. Solve ABC with A = 68  , b = 32, and c =
29.


Advanced Algebra 2 – Final Review Packet KG3 2011
[1]
[2]
5
2
[15] 1  x  4
5
x
8
x
–2 –1 0 1 2 3 4 5 6
4
7

[3]

[4]
[5] 102
[6] 1,081,200
[7] x < 5


20
9
x
-1
0
Page 8 of 18
1
2
3 4
5
x
3
x
[16] x < 2 or x > 3

–1 0

 
1
2
[9] No
[10] Yes
[11] [A] x  8 or x  10
[12] x  16 or x  2
[13] x  1 or x  7
7

1

x

[14]

3
7
–
3
–3 –2 –1 0
1
2
2
3
4
5
x
y

[24]
y = 1.75x
x
y
3
2
1

3
1
[17] It is.
[18] It is not.
[19] 3
[20] –3
[21] 13
5
[22] d = 45  2 t; 1998
[23] No correlation
[8] x<1
–3 –2 –1 0
x
x
–1
[25]
–2
–3
1 2
4 5
x
Advanced Algebra 2 – Final Review Packet KG3 2011
[35] (3, –8)
[36] 1320 ,  134 
[37] 9

[38] [C] 350
x  y  60
[39] 3.50x + 2.00y
 180
x = 40 pounds of cashews
y = 20 pounds of peanuts
f(x )
4
3
2
1
–4 –3 –2
[26]
1
2
x
–2
f( x )
5
4
1
[27]
–6 –5 –4 –3 –2
x
–1
Page 9 of 18
f(x )
y
4
3
3
2
1
2
1
–1
1
2
x
3 4
–2
[28]
–3
[40]
[41]
y
10
–1
–2
TC  0.42x  0.87y  0.82
daffodils
10 x
–10
[29]
[30] (3, –3)
–10
0

jasmine
y
3
2
1
–5
1 x
–2
–2
–3
[31]
[32] 1996, 2000
y
5
5x
–5
[33]
(–2, 2)
[34] no solution
–5
1 2 3 x
2
3
4
2
3.40 3.82 4.24
3
4.27 4.69
4
5.14
5
6.01 6.43 6.85
4.66
5.11 5.53
5.56 5.98 6.40
[42] [B] the value of z is  2
 (2, –1, 3)
[43]
[44](–1, 3, 2)
[45] (4, –2, 1)
12
[46] [D] 108 14


[47] (–4, 5, –2)
[48] [B] x = 3, y =  5
[49] [B] 7

5
7.27
Advanced Algebra 2 – Final Review Packet KG3 2011
[50]
Page 10 of 18
1 4 2 3 
5 7 1 4 


3
2
2
8




y


y
 x
10
y
10


10 x
–10
y

10 x
–10
[52]
3
x


axis of symmetry:
2
vertex:  23 , 43
[53] Down 
 Vertex: (2, 4); Axis: x = 2
[54]
[51]
–10
–10

[60]
vertex: 5, 0
axis of symm: x  5

Theonly x - intercept is at the vertex.
y
(–2, 3)
 x
y
10
3
–5
–3 –2
–1
–2
[55] Up
[57] [A]
[56]
–3
1

x
10 x
–10
[61]
vertex: 1.5, 1.25; axis of
symm: x  1.5;

–10
y   x 2  2x  5
y

x- intercepts at  2.6,  0.4
[62] 7.96 in.
 [63] (2x  5)(4x  5)
2
[58] y  x  2x  3
[64] 9

[65] x = 3 or x =  7

 [66] x = 4 or x =  6
[67] Base: 15 ft; Height: 10 ft
[59] [B] y  x 2  2x  2
[68] –5, 3
[69] 2

 [70] x   3

[71] 4.44 seconds

 x


y

 x




Advanced Algebra 2 – Final Review Packet KG3 2011
[72] 4.56 seconds
[73] [C] [C] max = –5
[74]  14  i 423
[75] i 3
[76] 1  2i

[77] 14 – 5i
 [78] 11  13i
Page 11 of 18
10
[89] x   3
1
[90] 2 sec


y

8
[79] 1 19 i 3

[91]
[80] 5, –7

[81] x  2  6
2
 [82] c  25; x5
2
[83] y  3x  8  9
 vertex = (–8, 9)
4
 [84] y  x  152  6
vertex = (15, –6)
[85] max = 4
 [86] max = –15
[87] y = (x  1) 2
y

–1
2
4
5
x
–2
–3
[92]
–4
y


[93]
 x

y
10
–10
10 x
[94]
[95] x  4 or x  10
[96] x <  5 or x > 9
8 

[97] ,    5, 
–10
+1
 x

[88] 1  2
(2, 2)
1
y



2
3

 x


9 
[98] between 0.87 and
Advanced Algebra 2 – Final Review Packet KG3 2011
41.58 seconds
2
f

x


2

x

9

8
[99]
[100] f x   2x 2
Page 12 of 18
y


 x
[112]

y
10
21
c
[101]
d 28
[102] 4s6 t 8 u 2


[113]
[114] The function is a cubic
–10
27
[103]  x 6
polynomial with degree 3 and
leading coefficient 2.
[115] The function is a
quartic polynomial with degree
4 and leading coefficient 1.
[116]The function is a
quadratic polynomial with
degree 2 and leading coefficient 4.
2x 3
 [104] 5
3y
x2

[105] 2y 5
 [106]
10 x
–10
4
5x 3 y 2
y
3
 [107]
2
5
y
5
–5
–5
[108]
x=5
[109] 3
 [110] 137
10
[117]
y
5
[118]
–5
10 x
[111]
–10
x
–5
[119] 11 f 5  8 f 3  3 f
[120] x 3  x 2  2x  12
[121] 4x 4 (2  7x 2 )

[122] 2x x  7x  2
 [123] 10(x 2  4)(x  2)(x  2)
[124] (x  2)(4x 2  3)

[125] (2x  3)(x 2  2)
3
–10
5
x


2
2
Advanced Algebra 2 – Final Review Packet KG3 2011
Page 13 of 18
[142]
1
 ,  3
2
10

3
 [132]

[133]
 [134]
2x 2  x  2 
4
x 5
7
x4
 [135]
[136] 1, 2, -2
[137] 4, -1, -2
[138] 3, -3
[139] x = 5,  5, and 1
[140] –5, –3, –1

[141]
y


[144] [B] [B]
1
,  5
2
 x

y = x3  x2  4 x  4
[145] f (x)  x 3  x 2  4 x  4
[146]  1,  11
5x 2  13 
2x 2  4 x  4 
–10
y = x 3  7x  6
2
30
x 2  2 
2
2x 2  x  1
x4
10 x
–10
[143]
[126] 6,  6, 3,  3
[127] 1; 2
[128] –2
[129] 3
[130] c 2  7c  49 
4
[131] 2x  2x  8x  8 + x  4
y
[147]


y




6
2
0
y 84
0
6 20 360
x
2
6

 x

[148] (2, 0), (–2, 0), (1, 0)
y


 x

[149] (x = 2.25)
Volume = 160.3
Advanced Algebra 2 – Final Review Packet KG3 2011
Page 14 of 18
length = 9.5
width = 7.5
height = 2.25
[165]  x 2  x  12
[166] 3,342,000
[150] [C] [C]
f (x )  x  4 x  15x  18
3
2
[167]

[168]
[169]

9
[151] 5 x 4
[152] 16
[153] 32
[154] 0.397
5
[155] 7
[156] 6 18
[157] 3.717
[158]
 [159]
1
5
7
x 12
v15
 [160]
w9

g(x) =
1
1
x 
3
3

[170] (7, 1), (5, 2), (3, 3),(1, 4)

[171] (5, 1), (3, 2), (1, 3),
(–1, 4)
[172] (1, 1), (2, 2), (3, 1),
(4, 4)
9 x
[173] y  11

[174]
y
x7
18
f(x )
f (x)

2
4 y 3xz
[161] 9z 3
[162] 2.041
 [163] x 2  x
2

x
 x 2
[164]

x3
5
x 2
y =
5
y =
–2
–1
–2
–3
[175]
f
2
-1
(x)
x
Advanced Algebra 2 – Final Review Packet KG3 2011
Page 15 of 18
g(x )
6
5
4
1
[181]
[182] x  3
–1
1
2
3
4
5
x
g(x )
h(x )
5
4
f(x )
2
–1
f
2
3
–4 –3 –2
-1
(x)
x
–2
[176]
[177] No
f(x )
5
2
1
–3
–1
1
3
–2
1
f (x)
3
–2
3
x
[178] Shift the graph of
y  x left 3 units, and
down 3 units.
–1
1
2
–2
x
[184]
[185] 18 weeks
[186] –59
[187] x  11
[188] [B] [B] –120
[189] y = 127
 [190] y = 64
[191] [A] [A] 19

[183]
y
10
10 x
–10
y

10
[192]
[193] $1967.15
[194] 3%; $269.86
million
[195] 4,127,000
[196] $6,300,000
[197] $16,734.32
–10
10 x
–10
[179]
–10
y


[180]
 x

1
–3
2
3
4
x
Advanced Algebra 2 – Final Review Packet KG3 2011
Page 16 of 18
[198] [C] [C] 2
x
[199] y  8
f( x )
6
4
2
–6 –4
x
[200]
y
1 
5 
4
x
[214]
 [201] 4
[202] –2
[203] 3
[204] 3
[205] 1.898

[206]
2.681
[215] x = 2, y = –1
[216] [B] [B]
x3
[217] x  1
[218]
y
10
10 x
–10
2
[207]
Domain: x | x  1; Range:
all real numbers
 [A] [A]
[208]
log a 5  log a x  3log a y  5log a z
[209] 2log 3 x  3log 3 y
13
[210] [A] [A] 9
2
x

x4
x3
 [219] x  5

[220]
–10
x7
x3
area:
18
m2
y(y  4)
1
 [221] x  4

2x
6x 2  18x  24
[222]
[223] 25

[224] distance = 122
midpoint =  15 ,  7
 2 2 
2.710
2
2
y
0.07x
x

y
 25
[211]
[225] 

 [212] f(x) = 250(1.11)x ; 421

[213] 
[B] [B] 64.13 ppm



Advanced Algebra 2 – Final Review Packet KG3 2011
Page 17 of 18
[234] ( 2 , 1)
[235] 4, 20, 48, 88

[226]
x2  y2 =

[227]
900

[239]

y


 x
[228]
2
2

x

5


y

2
 4


[229]
[230]center (–4, –3); r = 4
[231] (x  1)2 + (y  3)2 = 16
[243]


y
10
10 x
–10

–10
[232] vertices =
foci = (0,  2 5)

[233]
768, 3072, 12,288
1
16
[240]
[241] –4
[242] –3
 x


[237]
[238] –0.3
y


[236]
1
6
5
2
( 0,  6)
;
(x 
 3) 2 + (y  5) 2 = 1; The figure is a circle.
[244]

n 1
 3
an  24 
 4 
n 1
 2
an  48 
 3
n 1
 2
an  3 
 3
[245]
[246] [C] [C] 17th day;
$1997.97 total income

[247] 64
[248] 120
[249] 40,320
[250] 5! = 120
[251] 6! = 720
[252] 126
Advanced Algebra 2 – Final Review Packet KG3 2011
[253] 36
[254] 1365
[255] 22,100
[256] q3  6q2 r  12qr 2  8r 3
[257] 8s3  36s2t  54st 2  27t 3

[258] 8


7
8
[259] [A] [A]

[260] 0.48
[261] 14 C1 (.5)14  0.0009
 [A] 48.62 m
[262] [A]
[263] [B] [B] 8.33
[264] [D] [D]
a  34.20, B  60.17, C  51.83
[265] x  62.9 t and
y  210.8 t
[266] 2.65 m

Page 18 of 18