Exam Review
1. Determine the GCF of each set of numbers.
a) 120, 160, 180
120
160
180
2
60
2
80
2
90
2 30
2
40
2
45
2 15
2 20
3 15
3 5
2
10
3 5
2
5
2 ∙ 2 ∙ 2 ∙ 3 ∙ 5 = 2 ∙ 3 ∙ 5
2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 5 = 2
∙ 5
2 ∙ 2 ∙ 3 ∙ 3 ∙ 5 = 2 ∙ 3 ∙ 5
∴ ℎℎℎℎ: 2 ∙ 2 ∙ 5 = 20
b) 245, 280, 385
245
5
49
7 7
280
2
385
140
2
5
70
77
7
11
2
35
5
7
5∙7∙7= 5∙7
2 ∙ 2 ∙ 2 ∙ 5 ∙ 7 = 2 ∙ 5 ∙ 7
5 ∙ 7 ∙ 11
∴ ℎℎℎℎ: 5 ∙ 7 = 35
c) 176, 320, 368
176
320
368
2
88
2
160
2
184
2 44
2
80
2
92
2 22
2 40
2 46
2 11
2
20
2 23
2
10
2 5
2 ∙ 2 ∙ 2 ∙ 2 ∙ 11 = 2" ∙ 11
2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 5 = 2# ∙ 5
2 ∙ 2 ∙ 2 ∙ 2 ∙ 23 = 2" ∙ 23
∴ ℎℎℎℎ: 2" = 16
d) 484, 496, 884
484
2
242
2 121
11 11
496
2
884
248
2
442
2
124
2
221
62
13 17
2
31
2 ∙ 2 ∙ 11 ∙ 11 = 2 ∙ 11
2 ∙ 2 ∙ 2 ∙ 2 ∙ 31 = 2" ∙ 31
2 ∙ 2 ∙ 13 ∙ 17 = 2 ∙ 13 ∙ 17
∴ ℎℎℎℎ: 2 = 4
2
2. Determine the LCM of each set of numbers.
a) 70, 90, 140
70
90
2
35
2
45
5 7
3
15
3 5
140
2
70
2
35
5
7
2∙5∙7
2 ∙ 3 ∙ 3 ∙ 5 = 2 ∙ 3 ∙ 5
2 ∙ 2 ∙ 5 ∙ 7 = 2 ∙ 5 ∙ 7
∴ ℎ%&ℎ'(ℎ)*ℎ+,+-: 2 ∙ 3 ∙ 5 ∙ 7 = 1260
b) 120, 130, 309
120
130
309
2
60
2
65
3
103
2 30
5
13
2 15
3 5
2 ∙ 2 ∙ 2 ∙ 3 ∙ 5 = 2 ∙ 3 ∙ 5
2 ∙ 5 ∙ 13
3 ∙ 103
∴ ℎ%&ℎ'(ℎ)*ℎ+,+-: 2 ∙ 3 ∙ 5 ∙ 13 ∙ 103 = 160680
c) 200, 250, 500
200
250
500
2
100
2
125
2
250
2 50
5
25
2
125
2 25
5 5
5
25
5 5
5 5
2 ∙ 2 ∙ 2 ∙ 5 ∙ 5 = 2 ∙ 5
2 ∙ 5 ∙ 5 ∙ 5 = 2 ∙ 5
2 ∙ 2 ∙ 5 ∙ 5 ∙ 5 = 2 ∙ 5
∴ ℎ%&ℎ'(ℎ)*ℎ+,+-: 2 ∙ 5 = 1000
d) 180, 240, 340
180
240
340
2
90
2
120
2
170
2 45
2
60
2
85
3 15
2 30
5
17
3 5
2 15
3 5
2∙2∙3∙3∙5= 2 ∙3 ∙5
2 ∙ 2 ∙ 2 ∙ 2 ∙ 3 ∙ 5 = 2" ∙ 3 ∙ 5
2 ∙ 2 ∙ 5 ∙ 17 = 2 ∙ 5 ∙ 17
∴ ℎ%&ℎ'(ℎ)*ℎ+,+-: 2" ∙ 3 ∙ 5 ∙ 17 = 12240
3. Is each number a perfect square, a perfect cube, or neither? Determine the square root of each perfect
square and the cube root of each perfect cube.
a)
256
2
128
c)
729
e)
1936
2 64
3
243
2
968
2 32
3 81
2 484
2 16
3 27
2 242
2 8
3 9
2 121
2 4
3 3
11 11
2 2
3∙3∙3∙3∙3∙3
2 ∙ 2 ∙ 2 ∙ 2 ∙ 11 ∙ 11
2∙2∙2∙2∙2∙2∙2∙2
729).
1936).
(2 even groups)
256).
+)
(2 even groups)
(2 or 3 even groups)
f)
9261
b)
324
3
3087
d) 1298
2
162
3 1029
2
649
2 81
3 343
11 59
3 27
7 49
3 9
7 7
2 ∙ 11 ∙ 59
3 3
Neither (can’t make
even groups)
3∙3∙3∙7∙7∙7
2∙2∙3∙3∙3∙3
9261)
(3 even groups)
324).
(2 even groups)
Area = 3969 cm2
|
|
|
4. Determine the side length of this square.
To determine the length of a side of a square given the area take the
Square root of the area
Making 2 even groups
3969
3∙3∙7
3 1323
3∙3∙7
3 441
Therefore the square root is 63
3 147
The side length of this square is 63 cm
3 49
7 7
|
5. A square has an area of 18 225 square feet. What is the perimeter of the square?.
To determine the length of a side of a square given the area take the
Square root of the area
Making 2 even groups
18225
5∙3∙3∙3
5 3645
5∙3∙3∙3
5 729
Therefore the square root is 135
3 243
The side length of this square is 135 ft
3 81
The perimeter is the sum of all the sides
3 27
of the square: 135+135+135+135= 540 ft
3 9
3 3
6. A cube has a surface area of 11 616 cm3. What is the edge length of the cube?
To determine the length of a side of a cube given the surface area divide by 6 because a cube has six
faces, then take the Square root of that answer which is the area of one face.
11616 ÷ 6 = 1936
1936
Making 2 even groups
2 968
2 ∙ 2 ∙ 11
2 484
2 ∙ 2 ∙ 11
2 242
Therefore the square root is 44
2 121
The edge length of the cube is 44 cm
11 11
7. Write each radical in simplest form.
c) √112
a) √150
2 75
2 56
3 25
2 28
5 5
2 14
2 7
√5 ∙ 5 ∙ 2 ∙ 3
√2 ∙ 2 ∙ 2 ∙ 2 ∙ 7
√5 ∙ 5√2 ∙ 3
√2 ∙ 2 ∙ 2 ∙ 2√7
5√6
2 ∙ 2√7
2
4√7
b) √135
3
5 27
d) √162
3 9
2 81
3 3
3 27
2
3 9
√3 ∙ 3 ∙ 3 ∙ 5
2
2
3 3
√3 ∙ 3 ∙ 3 √5
3
2
√3 ∙ 3 ∙ 3 ∙ 3 ∙ 2
3√5
3
3
√3 ∙ 3 ∙ 3 ∙ 3 √2
3
3√2
8. Write each mixed radical as an entire radical.
b) 3√14
a) 6√5
√6 ∙ 6 ∙ 5
√3 ∙ 3 ∙ 14
√126
√180
c) 4√3
2
√4 ∙ 4 ∙ 4 ∙ 3
2
√192
9. Express each power as a radical.
a) 123
3
√12
d) 7 )2 4
=
?
b) 7−50) 2
2
:−50
7 √−50)
2
c) 1.2= √1.2
4
C
1.4
3
3
@A B 8
9
2
10. Express each radical as a power.
a) √1.4
b)
d) 2√3
3
√2 ∙ 2 ∙ 2 ∙ 2 ∙ 3
3
√48
2
c) (√2.5)4
√13
9
2
d) DE
F
3
"
2.5
13
4
3
G
H
11. Using the following formula, solve for q when M = 475.
2
q= 70& 3
. = 707475)3
2
. = 70I √475J
. = 707101.75)
. = 7122.3
Simplify the following:
a)
73 )
3C∙ C∙ ∙
3 #
3
b) (16a b )
K4
2 4
=
LC
LC
LC
16C∙ ∙ "∙ LC
16 LC L
LC
LC L
I√16J 4LC LC L
1
4 c)
MN 3
N
OP "LC
OP d) GQ K= RK= H
Q 2 RK4
C
L7L) LCL7L)
e) (a3b)(a-1b4)
S7LC) CS"
f) T
M=N
4
U
M=
C
L
O P
C
O CL
" C
O L
O
g)
V2
V9
L
S7L)
. a-3
L
1
h) GO = PH GO = P L H
4
C 2
O S P CS7L)
"
O P LC
O
P
12. Evaluate the following.
a)
2
4
=
=
G H G H
C
S
3 .
c)
T U
2
"
3 T U
2
3 T U
2
9
4
b)
T U
−5
#
12 T U
−5
12 T U
−5
144
25
2
W.C#3
4
W.C#3
#
d) E ?C
3
C
0.16"L"
√256
3
0.16"
C
0.16
√81
4
3
3
√0.16
0.4
13. Which of these numbers is rational?
a) E
b) ]
"
C#Z
4 #
C 2
XG H Y
L
C#
12 ∙C
c) √8 = 2
2
= Rational
[
\
d)
Irrational
14. Is the cube root of 250 rational or irrational?
Justify your answer.
2
√250 = 6.299605246 Irrational number does not end or repeat.
Rational
=2.84604……..Irrational
Rational numbers end or repeat and can
be written as a fraction, irrational
numbers never end or repeat.
15. Order these numbers from least to greatest:
,
,
,
,
2
2
2
√75 = 4.21716 … √14 = 3.74165 … √100 = 4.641588 … √17 = 4.12310 … √30 = 3.10723 …
√30
2
√14
√17 √75
√100
2
2
16. Expand and simplify the following:
a. 7O + 3)7O + 5)
O + 5O + 3O + 15
O + 8O + 15
b. 7O– 5)73O + 2)
3O + 2O − 15O − 10
3O − 13O − 10
c. 75– 2)
75 − 2)75 − 2)
25 − 10 − 10 + 4
25 − 20 + 4
d. 2x (x + 2)
2O + 4O
e. 2(x2 + 3x – 2) + 3(4x – 7 – 6x2)
2O + 6O − 4 + 12O − 21 − 18O −16O + 18O − 25
f. I3– 5J –I– 9J
73 − 5)73 − 5) − 7 − 9)7 − 9)
9 − 15 − 15 + 25 − 7 − 9 − 9 + 81)
9 − 30 + 25 − 7 − 18 + 81)
8 − 12 − 56
17. Fully factor the following:
a. 2a– 8ab
271 − 4)
b. 3x – 48
37O − 16)
37O − 4)7O + 4)
c. x – 16x + 28
c = −167−2) + 7−14)
& = 287−2) × 7−14)
7O − 2)7O − 14)
d. 9m – 1
73 − 1)73 + 1)
e. 2x – 2y 27O − P )
27O − P)7O + P)
f. 3x + 10x + 7
c = 1073) + 77)
& = 2173) × 77)
3O + 3O + 7O + 7
3O7O + 1) + 77O + 1)
7O + 1)73O + 7)
g. 2x + 3x + 1
c = 372) + 71)
& = 272) × 71)
2O + 2O + 1O + 1
2O7O + 1) + 17O + 1)
7O + 1)72O + 1)
h. 8x + 14x– 15
c = 147−6) + 720)
& = −1207−6) × 720)
8O − 6O + 20O − 15
2O74O − 3) + 574O − 3)
74O − 3)72O + 5)
i. x + 12xy + 27y c = 1279) + 73)
& = 2779) × 73)
7O + 9P)7O + 3P)
j. x – 11x + 30
c = −117−5) + 7−6)
& = 307−5) × 7−6)
7O − 5)7O − 6)
k. y – 3y – 18y
P7P − 3P − 18)
c = −373) + 7−6)
& = −1873) × 7−6)
P7P + 3)7P − 6)
l. 6O – 7O + 1
c = −77−6) + 7−1)
& = 67−6) × 7−1)
6O − 6O − 1O + 1
6O7O − 1) − 17O − 1)
7O − 1)76O − 1)
18. Which of the following are examples of functions?
a.
b.
Relation
c.
Function
Function
a. {(-2,7), (-1,5), (0,3), (1,1), (2,1)}
m. 3P + 12P − 15
c = 12715) + 7−3)
& = −45715) × 7−3)
3P + 15P − 3P − 15
3P7P + 5) − 37P + 5)
7P + 5)73P − 3)
n. 6P + 19P + 15
c = 1979) + 710)
& = 9079) × 710)
6P + 9P + 10P + 15
3P72P + 3) + 572P + 3)
72P + 3)73P + 5)
o. 8O + 10O + 3
c = 1074) + 76)
& = 2474) × 76)
8O + 4O + 6O + 3
4O72O + 1) + 372O + 1)
72O + 1)74O + 3)
p. 2O – 7O − 15
c = −77−10) + 73)
& = −307−10) × 73)
2O − 10O + 3O − 15
2O7O − 5) + 37O − 5)
7O − 5)72O + 3)
Function
e. {(-7,20), (3,5), (0,5), (-2,0), (6,-4), (-6,-9), (4,4)} Function
b. {(4,8), (-3,-2), (9,6), (2,-1), (-4,-5), (2,7), (-8,0)} Relation
19. If ℎ7O) = O ,7O) = 2O + 5 and (7O) = 1 + 3O, find
a. ℎ72)
ℎ72) = 2
ℎ72) = 4
b. (7−3)
(7−3) = 1 + 37−3)
(7−3) = 1 − 9
(7−3) = −8
20. This is a graph of the function
.
Determine the range value when the domain
value is 2.
(72) = −272) + 3
(72) = −4 + 3
(72) = −1
21. This is a graph of the function
.
Determine the domain value when the range
value is –4.
y
−4 = −3O + 2
−4 − 2 = −3O + 2 − 2
4
−6 = −3O
L#
LM
L = L
2
g(x) = –3x + 2
2=O
y
4
2
–4
–2
0
g(x) = –2x + 3
2
4
c. 70)
70) = 270) + 5
70) = 0 + 5
70) = 5
d. ℎ70)
ℎ70) = 0
ℎ70) = 0
–4
x
0
–2
–2
–2
–4
–4
2
4
x
22. The equation
represents the total cost, C dollars, for a sports banquet when g people attend.
a) Write the function in function notation.
7() = 11( + 250
b)
Determine C(46). What does this number (your answer) represent?
746) = 11746) + 250
746) = 506 + 250
746) = 756
The answer represents the cost of a sports banquet when 46 people attend.
c) Determine the value of g when C(g) = 1581. What does this number (your answer) represent?
1581 = 11( + 250
1581 − 250 = 11( − 250
1331 = 11(
1331 11(
=
11
11
121 = (
The answer represents the number of people who attended the banquet if it costs $1581
Volum e of Water in a Bottle
2.5
23. This graph shows the volume of water in Katherine’s water bottle as she cycles
around town. Describe what is happening for line segments FG, HI, KL and LM. 2
A
B
I
J
FG
HI
KL
LM
She stops and drinks most of her water
She refills her water bottle
She cycles without drinking any water
She stops and empties her water bottle
Volume (L)
D
1.5
L
C
K
F
1
E
0.5
G
H
M
10
20
30
40
50
Distance from hom e (km )
24. Identify the domain and range of these relations.
D:{x|5,6,8,9}
a)
R:{y|-11,-8,7,10}
b)
1
3
6
8
9
–2
0
3
5
D:{x|1,3,6,8,9}
R:{y-2,0,3,5}
25. Determine the domain and range of the following graphs.
A)D:{x|−6 g O g 6, O ∈ j}
R:{y|0 g P g 6, P ∈ j}
B)D:{x|−7 g O k 5, O ∈ j}
R:{y|−3 g P k 1, P ∈ j}
C)D:{x|5}
R:{y|−2 k P g 6, P ∈ j}
D)D:{x|-4g O g 4, O ∈ j}
R:{y|-4g P g 4, P ∈ j}
E)D:{x|0g O, O ∈ j}
R:{y|P ∈ j}
A)D:{|0k O, O ∈ j}
R:{y|4}
Slope : =
QlRm
Qno
= ∆M = M=LM4
∆N
P = O + 26. Calculate the slope of each line.
a)
−2
-) =
=
= −1
2
b) going between (-1, 4) and (3, -2)
P − PC
−2 − 4
−6 −3
=
=
=
=
O − OC 3 − 7−1)
4
2
LC
"
x+5
−1
=
4
c) y =
N LN
=
Slope y-int
d) O + P = 5
O − O + P = −O + 5
P = −O + 5
∴ = −1
e) 3O– 5P = 12
−3O + 3O − 5P = −3O + 12
−5P = −3O + 12
−5P −3O 12
=
+
−5
−5 −5
3O 12
P=
−
5
5
3
=
5
f) a horizontal line through (-2, 3)
Horizontal line has a slope of zero.
4
27. What is the equation of each line? Express answers in the P = O + b)
a)
f) through (4, 2) and (0, -3)
3
P − PC
=
=
=
1
O − OC
P − = = 2
−3 − 2
P = O + =
0−4
P = 3O + 2
−5 5
=
=
−4 4
Point (0,-3) is the y-int
∴ = −3
c) m= ½ and a y-intercept of -5
5
P = O + P
=
O−3
1
4
P= O−5
g) through (1, -5) and (-7, 3)
2
P − PC
d) with a slope of 2 going through (3, =
O − OC
1)
3
− 7−5)
P = O + =
−7 − 1
−1 = 72)73) + 8
−1 = 6 + =
= −1
−8
−1 − 6 = 6 + − 6
P = O + −7 = −5 = 7−1)71) + P = 2O − 7
−5 = −1 + e) m = -3, through (1, 6)
−5 + 1 = −1 + + 1
P = O + −4 = 6 = 7−3)71) + P = −1O − 4
6 = −3 + 6 + 3 = −3 + + 3
To determine if a point lies on a line
9=
substitute the x and y values if they
P = −3O + 9
satisfy the equation (both side equal)
then the point lies on the line.
28. Which point(s) are on the line given by 2x - 4y + 8 = 0?
a) (2, 2)
e) (-4, 0)
272) − 472) + 8 = 0
27−4) − 470) + 8 = 0
4−8+8= 0
−8 − 0 + 8 = 0
4q0
0=0
Not on the line
Lies on the line
b) (4, 0)
f) (-12, -4)
274) − 470) + 8 = 0
27−12) − 47−4) + 8 = 0
8−0+8= 0
−24 + 16 + 8 = 0
0=0
0=0
Lies on the line
Lies on the line
c) (-4, -2)
g) (0, -2)
27−4) − 47−2) + 8 = 0
270) − 47−2) + 8 = 0
7−8)
−8 −
+8=0
0+8+8= 0
8q0
16 q 0
Not on the line
Not on the line
d) (8, 4)
278) − 474) + 8 = 0
16 − 16 + 8 = 0
8q0
Not on the line
29. Graph each of the following equations.
C
a) y = x- 2
b) 2y - 3x = 4
c) x-intercept of 4, y-intercept of -3
2P − 3O + 3O = 3O + 4
2P = 3O + 4
2P 3
4
= O+
2
2
2
3
P= O+2
2
30. The Civic Center is planning a Fall Banquet. The Center charges a fixed cost of $200 plus $5 per guest.
a) Write an equation to represent this relationship.
= 5( + 200
b) What does the slope represent?
The slope represents cost per guest.
c) What does the y-intercept represent?
The y-intercept represents the fixed cost to rent the hall.
d) How much will the Fall Banquet cost if 79 guests attend?
= 5( + 200
= 5779) + 200
= 595
It will cost $595 if 79 guests attend the fall banquet.
e) Suppose your teacher wanted to pay for everyone’s meal at the banquet. How many people
could attend if she had $1500 to spend?
= 5( + 200
1500 = 5( + 200
1500 − 200 = 5( + 200 − 200
1300 = 5(
1300 5(
=
5
5
260 = (
260 guests could attend.
f) What would the equation become if the cost per person was raised by $2?
= 7( + 200
31. What is the distance between the points (3, 5) and (10, 15)?
+ = :7O − OC ) + 7P − PC )
+ = :710 − 3) + 715 − 5)
+ = :77) + 710)
+ = √49 + 100
+ = √149 = 12.2
32. What is the midpoint between the points S (−3,6) and T (5,7) ?
+) = G
,
M4 SM= N4 SN=
LS
#Sr
+) = G
C
,
+) = G , H
C
+) = G1, H
H
H
33. Solve by graphing
P = O−2
a) s
P = O+3
LC
"
Intersection point: (-4,0)
b) s
P = O − 3
P=
C
LC
O−1
Intersection point: (2,-2)
34. Solve the following systems of equations using the substitution method.
P = 3O + 2
O−P =3
a) t
b) t
P = −2O + 12
3O − 2P = 11
This is a good example of comparison
O−P+P =3+P
method,
O = 3+P
set both equations equal to each other
373 + P) − 2P = 11
3O + 2 = −2O + 12
9 + 3P − 2P = 11
3O + 2O + 2 − 2 = −2O + 2O + 12 − 2
9 + P = 11
5O = 10
−9 + 9 + P = 11 − 9
5O 10
P=2
=
O = 3+P
5
5
O=2
O = 3+2
P = 372) + 2
O=5
P=8
75,2)
72,8)
35. Solve the following systems of equations using the elimination method.
−2O + P = −5
3O − 5P = 13
a) t
b) t
4O − P = 7
O − 2P = 5
7−3)O − 2P7−3) = 57−3)
2O = 2
2O 2
−3O + 6P = −15
=
3O − 5P = 13
2
2
O=1
1P = 2
−2O + P = −5
O − 2P = 5
−271) + P = −5
O − 272) = 5
−2 + P = −5
O−4=5
−2 + 2 + P = −5 + 2
O−4+4= 5+4
P = −3
O=9
71, −3)
79,2)