Download ExamView - Algebra 1CP Final

Name: ________________________ Class: ___________________ Date: __________
ID: A
Algebra 1CP Final
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
____
____
____
____
____
____
ÔÏÔÔ 3x − 3y = −27
1. Tell whether the ordered pair (–4, 5) is a solution of the system ÔÔÌ
.
ÔÔ 3x + y = −8
Ó
a. no
b. yes
ÔÏÔÔ x − y = −4
Ô
2. Solve ÌÔ
by substitution. Express your answer as an ordered pair.
ÔÔ x = 2y − 12
Ó
a. (–6, 0)
c. (8, 0)
b. (0, –8)
d. (0, 8)
ÔÏÔÔ −9x + 2y = 12
Ô
3. Solve ÔÌ
by elimination. Express your answer as an ordered pair.
ÔÔ −7x + 2y = −12
Ó
a. (5, -5)
c. (–12, –48)
b. (0, 5)
d. (5, -5)
4. At the local pet store, zebra fish cost $2.20 each and neon tetras cost $1.75 each. If Sameer bought 17 fish for
a total cost of $33.80, not including tax, how many of each type of fish did he buy?
a. 9 zebra fish, 8 neon tetras
c. 6 zebra fish, 11 neon tetras
b. 8 zebra fish, 9 neon tetras
d. 11 zebra fish, 6 neon tetras
5. An airplane travels 900 miles from Houston to Miami in 6 hours against the wind. On its return trip, with the
wind, it takes only 5 hours. Find the rate of the airplane with no wind. Find the rate of the wind.
a. The airplane flies at 155 mi/h with no wind. The rate of the wind is 135 mi/h.
b. The airplane flies at 165 mi/h with no wind. The rate of the wind is 135 mi/h.
c. The airplane flies at 155 mi/h with no wind. The rate of the wind is 15 mi/h.
d. The airplane flies at 165 mi/h with no wind. The rate of the wind is 15 mi/h.
6. A paint mixer wants to mix paint that is 15% gloss with paint that is 30% gloss to make 6 gallons of paint
that is 25% gloss. How many gallons of each paint should the paint mixer mix together?
a. The paint mixer should use 4 gallons of 15% gloss paint and 2 gallons of 30% gloss
paint..
b. The paint mixer should use 3 gallons of 15% gloss paint and 3 gallons of 30% gloss
paint.
c. The paint mixer should use 2 gallons of 15% gloss paint and 4 gallons of 30% gloss
paint.
d. The paint mixer should use 1 gallon of 15% gloss paint and 5 gallons of 30% gloss paint.
7. Tell whether (3, 2) is a solution of y > 4x − 5 .
a. No, (3, 2) is not a solution of y > 4x − 5 .
b. Yes, (3, 2) is a solution of y > 4x − 5 .
1
Name: ________________________
____
8. Write an inequality to represent the graph.
a.
b.
____
ID: A
y ≥ 3x + 2
y ≥ 2x + 3
9. Simplify 4 −2 .
a. –8
b.
16
c.
d.
y > 3x + 2
y ≤ 3x + 2
c.
− 16
d.
1
16
1
____ 10. Simplify (−4) 0 .
1
a.
1
c.
−4
b.
–4
d.
0
____ 11. Find the value of the power 10 2 .
a. 10
b. 0.01
c.
d.
100
20
____ 12. Find the value of the expression 174 × 10 −3 .
a. 1.74
b. 0.0174
c.
d.
–5,220
0.174
____ 13. Simplify 2 2 ⋅ 2 4 .
a. 12
1
b. 64
c.
d.
64
Cannot simplify
c.
m81 ⋅ y 4
m18 ⋅ y 4
____ 14. Simplify m9 ⋅ y 4 ⋅ m9 .
a.
b.
m0 ⋅ y 4
(m ⋅ y) 22
____ 15. Simplify (x 5 ) −8 x 3 .
1
a.
x 37
b.
x −37
d.
c.
d.
2
1
x9
1
120
x
Name: ________________________
ID: A
65
.
6
a. Cannot simplify
b. 46,656
ÊÁ 8 ˆ˜ 2
Á 4r ˜
____ 17. Simplify ÁÁÁÁ 5 3 ˜˜˜˜ .
Ár s ˜
Ë
¯
16
a.
r5 s6
8r 10
b.
s5
____ 16. Simplify
c.
d.
c.
d.
1,296
5
16r 6
s6
8r 6
s6
4
5
____ 18. Simplify the expression 32 .
a. 16
b. 13
c.
d.
20
8
____ 19. Tell whether the number 7 is a root of 2r 2 − 11r − 63.
a. Yes
b. No
____ 20. Add or subtract.
3z 2 − z 4 + 5z 2 + 14z 4
a. 21z 6
c. 8z 4 + 13z 8
b. −2z 2 − 15z 4
d. 8z 2 + 13z 4
____ 21. Subtract.
(8b 4 − b 3 ) − (b 4 + 3b 3 − 1)
a. 8b 4 − 4b 3 − 1
b. 8b 4 + 3b 3 − 1
c. 7b 4 − 4b 3
d. 7b 4 − 4b 3 + 1
____ 22. Multiply.
Ê
ˆ
−7x y 2 ÁÁ x 3 y 4 − 4x y 3 ˜˜
Ë
¯
4 6
2 5
a. −7x y + 28x y
b. −6x 4 y 6 − 11x 2 y 5
____ 23. Multiply.
(n − 4)(n + 1)
d.
−7x 3 y 8 + 28x 1 y 6
−7x 5 y 7 − 7x 3 y 6
n2 − 4
n(n + 1) − 4(n + 1)
____ 24. Multiply.
(m − 7) 2
a. p 2 + 64
b. p 2 − 64
c.
d.
n 2 − 4n − 4
n 2 − 3n − 4
c.
m2 + 14m + 49
m2 − 14m + 49
a.
b.
c.
d.
3
Name: ________________________
ID: A
____ 25. Multiply.
(r + 6)(r − 6)
r 2 − 6r + 36
r 2 − 36
c.
d.
2r − 12
r 2 + 12
____ 27. Factor the trinomial d 2 + 2d − 48.
a. (d − 6)(d + 8)
b. (d − 8)(d − 6)
c.
d.
(d − 1)(d − 48)
(d + 1)(d − 48)
____ 28. Factor 3x 2 + 10x − 25.
a. (x − 5)(3x + 5)
b. (x + 5)(3x + 5)
c.
d.
(x − 5)(3x − 5)
(x + 5)(3x − 5)
a.
b.
____ 26. Factor the polynomial 15x 4 + 36x 3 − 6x 2 .
a. 3(5x 4 + 12x 3 − 2x 2 )
b. Cannot be factored
c. x 2 (15x 2 + 36x − 6)
d.
3x 2 (5x 2 + 12x − 2)
____ 29. Without graphing, tell whether the point (–3, 4) is on the graph of y = −3x 2 − 8 .
a. No
b. Yes
4
Name: ________________________
ID: A
____ 30. Graph y = x 2 − 3x + 4 . Find the axis of symmetry and the vertex.
a.
c.
The axis of symmetry is x = 0. The
vertex is (0, 4).
The axis of symmetry is x = 4. The
vertex is (4, 0).
d.
b.
3
3
The axis of symmetry is x = 2 . The
Ê3 7ˆ
vertex is ÁÁÁ 2 , 4 ˜˜˜ .
Ë
¯
The axis of symmetry is x = − 2 . The
Ê 3 7ˆ
vertex is ÁÁÁ − 2 , 4 ˜˜˜ .
Ë
¯
____ 31. Solve the quadratic equation b 2 − 8b + 15 = 0 by factoring.
a. 3 and 5
c. –3 and 5
b. –3 and –5
d. 3 and –5
____ 32. Solve x 2 = 64 by using square roots.
a. The solutions are 8 and –8.
b. The solutions are 4096 and –4096.
____ 33. Solve 100x2 – 121 = 0 by using square roots.
100
a. ± 121
b.
10
± 11
c.
d.
The solution is 8.
The solution is 4096.
c.
± 10
d.
No solution
11
____ 34. Complete the square for x 2 − 14x + ? to form a perfect square trinomial.
a. x 2 − 14x − 49
c. x 2 − 14x + 49
b. x 2 − 14x + 196
d. x 2 − 14x − 196
5
Name: ________________________
ID: A
____ 35. Solve 2x 2 − x − 4 = 0 by using the Quadratic Formula.
1±
33
a.
x=
b.
no solution
4
____ 36. Simplify the rational expression
a.
5t; t ≠ 2 or 0
Ê
ˆ
b. 5t ÁÁ t 2 − 2t ˜˜ ; t ≠ 2 or 0
Ë
¯
____ 37. Multiply. Simplify your answer.
x2 − x − 6
x2 + x
⋅
.
2x 2 − 6x x 2 + 4x + 4
x+1
a.
2x + 4
x
b.
2
x +4
____ 38. Divide. Simplify your answer.
1 n−7
÷
n
7n
1(7n)
a.
n (n − 7)
n−7
b.
7
____ 39. Add. Simplify your answer.
3x + 8
4
+ 2
2
x − 16 x − 16
3
a.
x+4
3 (x + 4 )
b.
x 2 − 16
____ 40. Add. Simplify your answer.
3y
3y
−
2
18y
9y
a.
b.
1
3
y+2
6y
c.
x=
d.
x=
1±
−31
4
−1 ± 33
4
5t 3 − 10t
. Identify any excluded values.
t 2 − 2t
c. 5t; t ≠ 2
d.
c.
d.
c.
d.
c.
d.
5t; no excluded values
2x 2 − 6
3x 2 − 2x + 4
1
16
7
n−7
7
n
3
x−4
3
2 (x − 4 )
c.
y−2
6y
d.
1
y
6
Name: ________________________
ID: A
____ 41. Divide. Simplify your answer.
(12x 4 − 18x 3 + 36x 2 ) ÷ 6x 3
36
a. 12x − 18 +
x
6
b. 2x − 3 +
x
c.
2x 4 − 3x 3 + 6x 2
d.
6x −
c.
2x + 9 +
2x + 21
d.
x −3+
a.
4
3
=
. Check your answer.
m − 3 5m
9
m = 17
c.
m = − 23
b.
m = − 17
d.
m=
6 5 −7
+ =
. Check your answer.
t 4
4
t=2
t=12
c.
d.
t=-12
t=-2
____ 42. Divide.
(2x 2 + 3x − 15) ÷ (x − 3 )
6
a. 2x − 3 − x − 3
b.
12 30
+
x2 x3
12
x−3
12
2x + 9
____ 43. Solve
9
9
9
23
____ 44. Solve
a.
b.
____ 45. Simplify the expression
16x 5 y 2 . All variables represent nonnegative numbers.
a.
4x 2 y
x
c.
4x 4 y 2
b.
4x 2 y
x2
d.
4 x2
____ 46. Simplify
b9
. The variable represents a nonnegative number.
16b
a.
b8
4
c.
b4
4
b.
b4
4
d.
b8
16
____ 47. Simplify
x
80
.
121
a.
20
11
c.
4 5
11
b.
5 4
11
d.
5
11
____ 48. Simplify the expression
75b + 4 12b − 2 27b .
a.
147b
c.
b.
7 3b
d.
7
ÊÁ
ˆ
ÁÁ 75 + 4 12 − 2 27 ˜˜˜
Ë
¯
12 3b
b
Name: ________________________
ID: A
____ 49. Multiply. Write the product in simplest form.
25h
10h
a.
4b
15
c.
30b
b.
2b
15
d.
b
____ 50. Simplify the quotient
5
.
15
5
a.
b.
3
c.
3
5
d.
____ 51. Solve the equation
a. b = −324
b. b = 324
____ 52. Solve
60
b = 18. Check your answer.
c.
d.
3
5
3
15
b = 36
b = −36
2y + 8 = 5. Check your answer.
9
2
a.
y=
b.
No solution.
c.
y=
2
9
d.
y=
3
2
8
ID: A
Algebra 1CP Final
Answer Section
MULTIPLE CHOICE
1. ANS: A
Substitute –4 for x and 5 for y in both equations. Since these values make the second equation false, (–4, 5)
is not a solution of the system.
Feedback
A
B
Correct!
Use substitution to check that the ordered pair satisfies both equations.
PTS:
OBJ:
STA:
KEY:
1
DIF: Basic
REF: Page 329
6-1.1 Identifying Solutions of Systems
1A9.0
TOP: 6-1 Solving Systems by Graphing
ordered pair | system of equations | solution
1
NAT: 12.5.4.g
ID: A
2. ANS: D
Step 1
x = −6y + 48
Solve the second equation for x.
Step 2
3(−6y + 48) – 2y = –16
Substitute −6y + 48 for x in the first equation.
Step 3
−18y + 144 – 2y = –16
–20y + 144 = –16
–20y = –16 – (144)
–20y = –160
y=8
Use the Distributive Property to simplify.
Collect like terms.
Subtract 144 from both sides.
Divide both sides by –20.
Step 4
x + 6y = 48
x + 6(8) = 48
x + 48 = 48
x = 48 – (48)
x=0
Write one of the original equations.
Substitute 8 for y.
(0, 8)
Write the solution as an ordered pair.
Step 5
Subtract 48 from both sides.
Feedback
A
B
C
D
After solving one equation for a variable, substitute the value into the other original
equation, not the one that has just been solved.
Check the signs.
After solving one equation for a variable, substitute the value into the other original
equation, not the one that has just been solved.
Correct!
PTS: 1
DIF: Average
REF: Page 338
OBJ: 6-2.2 Using the Distributive Property
NAT: 12.1.5.e
STA: 1A9.0
TOP: 6-2 Solving Systems by Substitution
2
ID: A
3. ANS: C
−9x + 2y = 12
7x − 2y = 12
−2x + 0y = 24
24
x=
−2
x = −12
12 + 9(−12)
2
y = −48
y=
Multiply all expressions in the second equation by −1.
Add the two equations together.
Divide both sides by –2.
Solve for x.
Substitute the value for x into one of the original equations and
solve for y.
Feedback
A
B
C
D
Multiply all terms in the second equation by -1 before combining the equations.
Multiply all terms in the second equation by -1 before combining the equations.
Correct!
Multiply all terms in the second equation by -1 before combining the equations.
PTS:
OBJ:
STA:
KEY:
1
DIF: Average
REF: Page 344
6-3.2 Elimination Using Subtraction
NAT: 12.5.4.g
1A9.0
TOP: 6-3 Solving Systems by Elimination
system of equations | elimination
3
ID: A
4. ANS: A
Let z be the number of zebra fish and let n be the number of neon tetras that Sameer bought. Then solve the
following system of equations.
Sameer spent $33.80.
2.20z + 1.75n = 33.80
Sameer bought 17 fish.
n = 17
z+
2.20z + 1.75n = 33.80
Multiply the second equation by –2.20
Add the two equations to eliminate the z term.
−2.20z − 2.20n = −37.40
−0.45n = −3.60
Solve for n.
n=8
To solve for z, substitute 8 for n in the first equation.
2.20z + 1.75 (8) = 33.80
2.20z = 19.8
z=9
Simplify.
Solve for z.
Feedback
A
B
C
D
Correct!
You switched the prices of zebra fish and neon tetras.
Write an equation expressing the total cost and a second equation expressing the total
number of fish. Solve for z and n using elimination.
Write an equation expressing the total cost and a second equation expressing the total
number of fish. Solve for z and n using elimination.
PTS: 1
NAT: 12.5.4.g
DIF: Average
STA: 7AF1.1
REF: Page 346
OBJ: 6-3.4 Application
TOP: 6-3 Solving Systems by Elimination
4
ID: A
5. ANS: D
Let p be the rate at which the airplane travels with no wind, and let w be the rate of the wind. Use a table to
set up two equations—one for the trip against the wind and one for the trip with the wind.
•
=
Rate
Time
Distance
p −w
•
6
900
With the wind
=
p +w
•
5
900
Against the wind
=
ÔÏÔ
ÔÏÔ
ÔÔÔ 6(p − w) = 900
ÔÔ 6p − 6w = 900
Solve the system ÔÌ
. First write the system as ÔÔÌ
, and then use elimination.
ÔÔÔ 5(p + w) − 900
ÔÔÔ 5p + 5w − 900
Ó
Ó
Multiply each term in the first equation by
5(6p − 6w = 900)
Step 1
5 and each term in the second equation by 6.
6(5p + 5w = 900)
30p − 30w = 4500
+30p + 30w = 5400
Step 2
Step 3
Step 4
60p = 9900
p = 165
6p − 6w = 900
6(165) − 6w = 900
990 − 6w = 900
−990
− 990
−6w = −90
w = 15
(165,15)
Add the new equations.
Simplify and solve for p.
Write one of the original equations.
Substitute 165 for p.
Subtract 990 from both sides.
Simplify and solve for w.
Write the solution as an ordered pair.
The airplane flies at 165 mi/h with no wind. The rate of the wind is 15 mi/h.
Feedback
A
B
C
D
Use a table to set up two equations—one for the trip against the wind and one for the
trip with the wind.
Check your math.
Use a table to set up two equations—one for the trip against the wind and one for the
trip with the wind.
Correct!
PTS: 1
STA: 1A15.0
DIF: Advanced
REF: Page 356
TOP: 6-5 Applying Systems
5
OBJ: 6-5.1 Solving Rate Problems
ID: A
6. ANS: C
Let x be the number of gallons of 15% gloss paint, and let y be the number of gallons of 30% gloss paint. Use
a table to set up two equations—one for the amount of gloss in the first paint, and one for the amount of gloss
in the second paint.
=
+
15% gloss
30% gloss
25% gloss
x
+
y
6
Amount of paint (gallons)
=
0.15x
0.30y
0.25(6) = 1.5
+
Amount of gloss (gallons)
=
ÔÏÔ
ÔÔ
x+y = 6
Solve the system ÔÌÔ
. Use substitution.
ÔÔÔ 0.15x + 0.30y = 1.5
Ó
Solve the first equation for x by subtracting
x+y = 6
Step 1
y from both sides.
− y −y
x
Step 2
Step 3
Step 4
Step 5
= 6−y
0.15x + 0.30y = 1.5
Substitute 6 − y for x in the second
equation.
0.15(6 − y) + 0.30y = 1.5
0.15(6) − 0.15(y) + 0.30y = 1.5
0.90 − 0.15y + 0.30y = 1.5
0.90 + 0.15y = 1.5
−0.90
−0.90
0.15y = 0.60
0.15y 0.60
=
0.15
0.15
y=4
x+y = 6
x+4= 6
− 4 −4
x= 2
(2, 4)
Distribute 0.15 to the expression in
parentheses.
Simplify. Solve for y.
Subtract 0.90 from both sides.
Divide both sides by 0 .15
Write one of the original equations.
Substitute 4 for y.
Subtract 4 from both sides.
Write the solution as an ordered pair.
The paint mixer should use 2 gallons of 15% gloss paint and 4 gallons of 30% gloss paint.
Feedback
A
B
C
D
Use a table to set up two equations—one for the amount of gloss in the first paint, and
one for the amount of gloss in the second paint.
Use a table to set up two equations—one for the amount of gloss in the first paint, and
one for the amount of gloss in the second paint.
Correct!
Use substitution to solve the system of equations.
PTS: 1
STA: 1A15.0
DIF: Advanced
REF: Page 357
TOP: 6-5 Applying Systems
6
OBJ: 6-5.2 Solving Mixture Problems
ID: A
7. ANS: A
Substitute (3, 2) for (x, y) in y > 4x − 5.
y > 4x − 5
2 > 4(3) − 5
2 > 7, false
(3, 2) is not a solution of y > 4x − 5 .
Feedback
A
B
Correct!
Substitute the values for (x, y) into the inequality to see if the ordered pair is a solution.
PTS: 1
DIF: Basic
REF: Page 364
OBJ: 6-6.1 Identifying Solutions of Inequalities
NAT: 12.5.4.a
STA: 1A6.0
TOP: 6-6 Solving Linear Inequalities
8. ANS: A
Use the graph to determine the slope and y-intercept, and then write an equation in the form y = mx + b. A
graph shaded above the line means greater than and the graph shaded below the line means less than. Use ≤
or ≥ if the line is solid; use < or > if the line is dashed.
Feedback
A
B
C
D
Correct!
Use the graph to find the slope and y-intercept. Then write an equation for the boundary
line in the form y = mx + b, where m is the slope and b is the y-intercept.
Use "greater than or equal to" or "less than or equal to" for a solid line. Use "greater
than" or "less than" for a dashed line.
Check the direction of the inequality symbol.
PTS:
OBJ:
STA:
KEY:
1
DIF: Average
REF: Page 367
6-6.4 Writing an Inequality from a Graph
1A9.0
TOP: 6-6 Solving Linear Inequalities
graph | inequality | equation of a line
7
NAT: 12.5.3.d
ID: A
9. ANS: D
1
4 −2 = 2
4
=
1
16
1
The reciprocal of 4 is 4 .
4 2 = 16.
Feedback
A
B
C
D
A nonzero number raised to a negative exponent is equal to 1 divided by that number
raised to the opposite (positive) exponent.
A nonzero number raised to a negative exponent is equal to 1 divided by that number
raised to the opposite (positive) exponent.
Check the sign of your answer. A negative exponent does not affect the sign of the
answer.
Correct!
PTS: 1
DIF: Average
REF: Page 395
OBJ: 7-1.2 Zero and Negative Exponents
NAT: 12.1.1.d
STA: 7AF2.1
TOP: 7-1 Integer Exponents
KEY: negative exponent | evaluate | power | exponent
10. ANS: A
Any nonzero base to the zero power is equal to 1.
(−4) 0 = 1
Feedback
A
B
C
D
Correct!
A nonzero number raised to the zero power is equal to 1.
A nonzero number raised to the zero power is equal to 1.
A nonzero number raised to the zero power is equal to 1.
PTS: 1
DIF: Average
REF: Page 395
OBJ: 7-1.2 Zero and Negative Exponents
NAT: 12.1.1.d
STA: 7AF2.1
TOP: 7-1 Integer Exponents
KEY: zero exponent | zero power | evaluate | power | exponent
11. ANS: C
Start with 1 and move the decimal point two places to the right.
10 2 = 100
Feedback
A
B
C
D
Count the number of places to move the decimal point.
If the exponent is positive, move the decimal point to the right. If the exponent is
negative, move the decimal point to the left.
Correct!
Start with 1 and move the decimal point.
PTS: 1
NAT: 12.1.1.f
DIF: Basic
STA: 7AF2.1
REF: Page 400
OBJ: 7-2.1 Evaluating Powers of 10
TOP: 7-2 Powers of 10 and Scientific Notation
8
ID: A
12. ANS: D
Move the decimal point 3 places to the left.
0.174
Feedback
A
B
C
D
Move the decimal point the correct number of places.
Move the decimal point the correct number of places.
For powers of 10, the exponent tells the number of places to move the decimal point.
Correct!
PTS: 1
DIF: Average
REF: Page 401
OBJ: 7-2.3 Multiplying by Powers of 10
NAT: 12.1.1.f
STA: 7NS1.1
TOP: 7-2 Powers of 10 and Scientific Notation
KEY: exponents | multiplication | power | powers of 10
13. ANS: C
To multiply powers with the same base, keep the same base and add the exponents. Then, evaluate the power.
2 2 ⋅ 2 4 = 2 6 = 64
Feedback
A
B
C
D
The exponent tells how many times to multiply the base number by itself.
Check the sign of the exponent.
Correct!
If the bases are the same, add the exponents. Then evaluate the power.
PTS: 1
DIF: Basic
REF: Page 409
OBJ: 7-3.1 Finding Products of Powers
NAT: 12.5.3.c
STA: 1A2.0
TOP: 7-3 Multiplication Properties of Exponents
KEY: evaluate | product | multiply | power | exponent
14. ANS: D
To multiply powers with the same base, keep the same base and add the exponents.
m9 ⋅ y 4 ⋅ m9 = (m9 ⋅ m9 ) ⋅ y 4 = m18 ⋅ y 4
Feedback
A
B
C
D
To multiply powers with the same base, add the exponents, not subtract.
Rewrite only powers with the same base. Do not combine powers with different bases.
To multiply powers with the same base, add the exponents, not multiply.
Correct!
PTS: 1
DIF: Average
REF: Page 409
OBJ: 7-3.1 Finding Products of Powers
NAT: 12.5.3.c
STA: 1A2.0
TOP: 7-3 Multiplication Properties of Exponents
KEY: evaluate | product | multiply | power | exponent
9
ID: A
15. ANS: A
(x 5 ) −8 x 3
x
5 (−8 )
x3
Use the Power of a Power Property.
x −40 x 3
Simplify the exponent of the first term.
x −40 + 3
Add the exponents since the powers have the same base.
x −37
1
x 37
Write with a positive exponent.
Feedback
A
B
C
D
Correct!
Continue simplifying to a fraction with a positive exponent.
Multiply the exponents when a power is raised to another power. Add the exponents
when a power is multiplied by another power.
Multiply the exponents when a power is raised to another power.
PTS: 1
DIF: Advanced
REF: Page 410
OBJ: 7-3.3 Finding Powers of Powers
NAT: 12.5.3.c
STA: 1A2.0
TOP: 7-3 Multiplication Properties of Exponents
16. ANS: C
To divide powers with the same base, keep the same base and subtract the exponents.
Feedback
A
B
C
D
The bases are the same, so the expression can be simplified.
To divide powers with the same base, subtract the exponents.
Correct!
To divide powers with the same base, subtract the exponents.
PTS: 1
DIF: Basic
REF: Page 415
OBJ: 7-4.1 Finding Quotients of Powers
STA: 1A2.0
TOP: 7-4 Division Properties of Exponents
KEY: exponent | power | division | base
10
ID: A
17. ANS:
ÊÁ 8
ÁÁ 4r
ÁÁ 5 3
ÁÁ r s
Ë
C
ˆ˜ 2
˜˜
˜˜
˜˜
¯
2
ÁÊÁ 4r 3 ˆ˜˜
Á
˜
= ÁÁÁ 3 ˜˜˜
Á s ˜
Ë
¯
ÊÁ 3 ˆ˜ 2
Á 4r ˜
Ë ¯
=
ÊÁ 3 ˆ˜ 2
Ás ˜
Ë ¯
Simplify exponents with like bases:
r8
= r3 .
r5
Use the Power of a Quotient Property.
2
=
=
=
(4) (r 3 ) 2
(s 3 ) 2
(16)(r 3 ) 2
(s 3 ) 2
16r 6
s6
Use the Power of a Product Property.
2
Simplify: (4) = 16.
Use the Power of a Power Property to simplify the
exponents.
Feedback
A
B
C
D
Check to see if the terms are correctly placed in the numerator and denominator.
Use the Power of a Power Property to raise every term in the problem to the exponent
outside the parenthesis.
Correct!
Use the Power of a Power Property to raise the constant to the power outside the
parenthesis.
PTS: 1
DIF: Average
REF: Page 417
OBJ: 7-4.4 Finding Positive Powers of Quotients
NAT: 12.5.3.c
STA: 1A2.0
TOP: 7-4 Division Properties of Exponents
11
ID: A
18. ANS: A
4
5
Ê
ˆ4
32 = ÁÁÁ 5 32 ˜˜˜
Ë
¯
ÊÁ 5 5 ˜ˆ 4
= ÁÁÁÁ 2 ˜˜˜˜
Ë
¯
= (2)
= 16
m
n
Definition of b .
4
Feedback
A
B
C
D
Correct!
A number raised to the power of m/n is equal to the nth root of the number raised to the
mth power.
Rewrite the base as a number raised to a power.
Rewrite the denominator of the power as a root.
PTS: 1
DIF: Average
REF: Page 423
OBJ: 7-5.2 Simplifying Expressions with Fractional Exponents
STA: 1A2.0
TOP: 7-5 Fractional Exponents
KEY: fractional exponent
19. ANS: B
2r 2 − 11r − 63
= 2(7) 2 − 11(7) − 63
Substitute for r.
= 98 − 77 − 63
Simplify.
= –42 No!
−42 ≠ 0, so 7 is not a root of 2r 2 − 11r − 63.
Feedback
A
B
A root of a polynomial in one variable is a value of the variable for which the
polynomial is equal to 0.
Correct!
PTS: 1
DIF: Average
REF: Page 432
OBJ: 7-6.5 Identifying Roots of Polynomials
TOP: 7-6 Polynomials
KEY: polynomial
12
STA: 1A25.0
ID: A
20. ANS: D
3z 2 − z 4 + 5z 2 + 14z 4 = 3z – z4 + 5z 2
+ 14z4
= 3z2 + 5z2 – z4
+ 14z4
= 8z 2 + 13z 4
Identify like terms.
Use the Commutative Property to
move like terms together.
Combine like terms.
Feedback
A
B
C
D
Only add or subtract coefficients on like terms.
Check your addition and subtraction.
When combining like terms, only add or subtract the coefficients. The powers stay the
same.
Correct!
PTS: 1
DIF: Advanced
REF: Page 438
OBJ: 7-7.1 Adding and Subtracting Monomials
NAT: 12.5.3.c
STA: 1A10.0
TOP: 7-7 Adding and Subtracting Polynomials
KEY: monomial
21. ANS: D
(8b 4 − b 3 ) − (b 4 + 3b 3 − 1)
Rewrite subtraction as addition of the opposite.
= (8b 4 − b 3 ) + (−b 4 − 3b 3 + 1)
= (8b 4 − b 4 ) + (−b 3 − 3b 3 ) + (1)
= 7b 4 − 4b 3 + 1
Identify like terms. Rearrange terms to get like terms
together.
Combine like terms.
Feedback
A
B
C
D
Check the coefficients and the signs.
First, rewrite the subtraction as an addition of the opposite. Then, combine the like
terms.
Check that you have included all the terms.
Correct!
PTS: 1
NAT: 12.5.3.c
KEY: polynomial
DIF: Average
STA: 1A10.0
REF: Page 439
OBJ: 7-7.3 Subtracting Polynomials
TOP: 7-7 Adding and Subtracting Polynomials
13
ID: A
22. ANS: A
Use the Distributive Property to multiply the monomial by each term inside the parentheses. Group terms to
get like bases together, and then multiply.
Feedback
A
B
C
D
Correct!
Multiply the coefficients for each term; don't add.
When multiplying like bases, add the exponents.
Don't forget to multiply the coefficients for each term.
PTS: 1
DIF: Advanced
REF: Page 446
OBJ: 7-8.2 Multiplying a Polynomial by a Monomial
NAT: 12.5.3.c
STA: 1A10.0
TOP: 7-8 Multiplying Polynomials
KEY: polynomial | multiplication
23. ANS: D
(n − 4)(n + 1)
Use FOIL.
n(n + 1) − 4(n + 1)
Distribute n and –4.
n(n) + n(+1) − 4(n) − 4(+1)
Distribute n and –4 again.
2
Multiply.
n + n − 4n − 4
2
Combine like-terms.
n − 3n − 4
Feedback
A
B
C
D
You did not multiply the inner and outer terms.
Distribute again, multiply and combine like-terms.
You did not multiply the outer terms.
Correct!
PTS: 1
DIF: Basic
NAT: 12.5.3.c
STA: 1A10.0
KEY: binomial | multiplication
24. ANS: D
(m − 7) 2
2
(a − b ) = a 2 − 2ab + b 2
(m − 7) 2 = m2 − 2(m)(7) + 7 2
m2 − 14m + 49
REF: Page 448
OBJ: 7-8.3 Multiplying Binomials
TOP: 7-8 Multiplying Polynomials
2
Use the rule for (a − b ) .
Use the FOIL method, and then combine like terms.
Simplify.
Feedback
A
B
C
D
The last term in the product should be the square of the second term in the binomial.
Check the signs.
Check the signs.
Correct!
PTS: 1
DIF: Basic
REF: Page 456
OBJ: 7-9.2 Finding Products in the Form (a – b)^2
STA: 1A10.0
TOP: 7-9 Special Products of Binomials
14
NAT: 12.5.3.c
KEY: binomial | multiplication
ID: A
25. ANS: B
(r + 6)(r − 6)
(a + b ) (a − b ) = a 2 − b 2
(r + 6)(r − 6) = r 2 − 6 2
r 2 − 36
Use the rule for (a + b ) (a − b ) .
Use the FOIL method, and then combine like terms.
Simplify.
Feedback
A
B
C
D
First, use the FOIL method. Then, combine the like terms.
Correct!
Use the FOIL method.
The terms in the product should be squares.
PTS: 1
DIF: Basic
REF: Page 457
OBJ: 7-9.3 Finding Products in the Form (a + b)(a – b)
NAT: 12.5.3.c
STA: 1A10.0
TOP: 7-9 Special Products of Binomials KEY: binomial | multiplication
26. ANS: D
15x 4 + 36x 3 − 6x 2
Find the GCF. The GCF of 15x 4 , 36x 3 , and −6x 2 is 3x 2 .
2
2
2
2
Write the terms as products using the GCF.
3x (5x ) + 3x (12x) − 3x (2)
Use the Distributive Property to factor out the GCF.
3x 2 (5x 2 + 12x − 2)
Feedback
A
B
C
D
First, find the GCF. Then, write the terms as products using the GCF.
First, find the GCF. Then, use the Distributive Property to factor out the GCF.
First, find the GCF. Then, write the terms as products using the GCF.
Correct!
PTS: 1
NAT: 12.1.5.b
27. ANS: A
d 2 + 2d − 48
(d + ? )(d + ? )
(d − 6)(d + 8)
DIF: Average
STA: 1A11.0
REF: Page 487
OBJ: 8-2.1 Factoring by Using the GCF
TOP: 8-2 Factoring by GCF
Look for the factors of –48 whose sum is 2.
The factors are –6 and 8.
Feedback
A
B
C
D
Correct!
Use the FOIL method to check your answer.
Check the signs.
Use the FOIL method to check your answer.
PTS: 1
DIF: Basic
REF: Page 498
OBJ: 8-3.3 Factoring x^2 + bx + c When c Is Negative
STA: 1A11.0
TOP: 8-3 Factoring x^2 + bx + c
15
NAT: 12.5.3.d
ID: A
28. ANS: D
Try factors of 3 for the coefficients and factors of –25 for the constant terms.
The combination that works is:
(x + 5)(3x − 5) = 3x 2 − 5x + 15x − 25 = 3x 2 + 10x − 25
Feedback
A
B
C
D
Multiply the factors to check your answer.
Check the signs.
Multiply the factors to check your answer.
Correct!
PTS: 1
DIF: Basic
REF: Page 505
OBJ: 8-4.1 Factoring ax^2 + bx + c
NAT: 12.5.3.d
STA: 1A11.0
TOP: 8-4 Factoring ax^2 + bx + c
KEY: factor | trinomial | guess and check
29. ANS: A
Substitute (–3, 4) into y = −3x 2 − 8 .
y = −3x 2 − 8
4 =? −3(−3) 2 − 8
4 =? −3 ⋅ 9 − 8
4 =? −27 − 8
4 ≠ −35 No!
No, since (–3, 4) is not a solution of y = −3x 2 − 8 , (–3, 4) is not on the graph.
Feedback
A
B
Correct!
Substitute the x-coordinate of the point into the equation and check if the solution is
equal to the corresponding y-coordinate.
PTS: 1
DIF: Average
REF: Page 544
OBJ: 9-1.1 Determining Whether a Point Is on a Graph
TOP: 9-1 Quadratic Equations and Functions
16
STA: 1A17.0
ID: A
30. ANS: B
Step 1: Find the axis of symmetry.
−(−3)
−b
3
Use x = 2a . Substitute 1 for a and −3 for b.
x= 2 =2
Step 2: Find the vertex.
y = x 2 − 3x + 4
3
Ê 3 ˆ2
Ê3ˆ
Substitute 2 for x.
y = ÁÁÁ 2 ˜˜˜ − 3 ÁÁÁ 2 ˜˜˜ + 4
Ë ¯
Ë ¯
7
7
y= 4
The y-coordinate is 4 .
Ê3 7ˆ
The vertex is ÁÁÁ 2 , 4 ˜˜˜ .
Ë
¯
Feedback
A
B
C
D
Graph the equation correctly.
Correct!
Check the vertex.
To find the axis of symmetry, substitute for a and b in –b/2a.
PTS: 1
DIF:
KEY: graph | quadratic
31. ANS: A
b 2 − 8b + 15 = 0
(b − 3) (b − 5) = 0
b − 3 = 0 or b − 5 = 0
b = 3 or b = 5
The solutions are 3 and 5.
Advanced
STA: 1A21.0
TOP: 9-3 Graphing Quadratic Functions
Factor the trinomial.
Use the Zero Product Property.
Solve each equation.
Feedback
A
B
C
D
Correct!
Check your factorization by multiplying the factors together.
Check your factorization by multiplying the factors together.
Substitute the solutions into the original equation to check your answer.
PTS: 1
DIF: Average
REF: Page 577
OBJ: 9-5.2 Solving Quadratic Equations by Factoring
NAT: 12.5.4.a
STA: 1A14.0
TOP: 9-5 Solving Quadratic Equations by Factoring
17
ID: A
32. ANS: A
Take the square root of both sides of the equation. Remember that there are both positive
and negative solutions to the square root.
Feedback
A
B
C
D
Correct!
Check your answers by substituting into the original equation.
Every positive number has two square roots, and negative numbers have no square roots
that are real numbers.
Every positive number has two square roots, and negative numbers have no square roots
that are real numbers.
PTS: 1
DIF: Basic
REF: Page 582
OBJ: 9-6.1 Using Square Roots to Solve x^2 = a
NAT: 12.5.4.a
STA: 1A2.0
TOP: 9-6 Solving Quadratic Equations by Using Square Roots
33. ANS: C
100x2 – 121 = 0
+ 121 = + 121
Add 121 to both sides.
2
121
100x
=
Divide both sides by 100.
100
100
121
Take the square root of both sides.
x2 = 100
11
x = ±
Use the plus/minus sign to show positive and negative roots.
10
11
11
The solutions are 10 and – 10 .
Feedback
A
B
C
D
If possible, take the square root to find x.
Check your solution.
Correct!
Check your positive and negative signs.
PTS: 1
DIF: Average
REF: Page 583
OBJ: 9-6.2 Using Square Roots to Solve Quadratic Equations NAT: 12.5.4.a
STA: 1A2.0
TOP: 9-6 Solving Quadratic Equations by Using Square Roots
18
ID: A
34. ANS: C
Ê − 14 ˆ 2
Divide the coefficient of the x-term by 2, and then square it to get the constant term, ÁÁÁ 2 ˜˜˜ .
Ë
¯
Add the result to the expression to form a perfect square trinomial.
Feedback
A
B
C
D
The final term of a perfect square trinomial must be added, not subtracted.
Divide the coefficient of the x-term by 2, and then square it to get the constant term.
Correct!
Divide the coefficient of the x-term by 2, and then square it to get the constant term.
PTS: 1
DIF: Basic
REF: Page 591
OBJ: 9-7.1 Completing the Square
NAT: 12.5.3.b
STA: 1A14.0
TOP: 9-7 Completing the Square
35. ANS: A
Write the equation in standard form: x 2 − 2x − 8 = 0. Substitute for a, –2 for b, and –8 for c in the Quadratic
b 2 − 4ac
. Write as two equations, one adding the square root and one subtracting it.
2a
Simplify each equation to find the values of x.
Formula, x =
−b ±
Feedback
A
B
C
D
For the numerator in the Quadratic Formula, find the square root before adding it to or
subtracting it from –b.
The denominator of the Quadratic Formula is 2 times a.
Use the opposite of b, the coefficient of the x-term in the equation.
Correct!
PTS: 1
DIF: Average
REF: Page 599
OBJ: 9-8.1 Using the Quadratic Formula
NAT: 12.5.4.a
STA: 1A20.0
TOP: 9-8 The Quadratic Formula
36. ANS: A
Factor common factors out of the numerator and/or denominator. Divide out the common factors to simplify
the expression. Finally, use the original denominator to determine excluded factors.
Feedback
A
B
C
D
Correct!
Divide out common factors.
Determine excluded values from the original denominator.
Determine excluded values from the original denominator.
PTS: 1
DIF: Basic
REF: Page 643
OBJ: 10-3.2 Simplifying Rational Expressions
NAT: 12.5.3.c
STA: 1A12.0
TOP: 10-3 Simplifying Rational Expressions
19
ID: A
37. ANS: A
x2 − x − 6
x2 + x
⋅
2x 2 − 6x x 2 + 4x + 4
x(x + 1)
(x + 2)(x − 3)
⋅
=
2x(x − 3)
(x + 2)(x + 2)
1 (x + 1)
= ⋅
2 (x + 2)
x+1
=
2x + 4
Factor the numerator and denominator.
Simplify.
Multiply the remaining factors.
Feedback
A
B
C
D
Correct!
Factor the numerator and denominator and divide out the common factors.
Factor the numerator and denominator and divide out the common factors.
Factor the numerator and denominator and divide out the common factors.
PTS: 1
DIF: Average
REF: Page 653
OBJ: 10-4.3 Multiplying Rational Expressions Containing Polynomials
NAT: 12.5.3.c
STA: 1A13.0
TOP: 10-4 Multiplying and Dividing Rational Expressions
38. ANS: C
1 n−7
÷
n
7n
Write as multiplication by the reciprocal.
1 7n
= ⋅
n n−7
Multiply the numerators and the denominators.
1(7n)
=
n (n − 7)
Divide out common factors. Simplify.
7
=
n−7
Feedback
A
B
C
D
Divide out common factors, and simplify.
Write as multiplication by the reciprocal first.
Correct!
First, write as multiplication by the reciprocal. Then, multiply the numerators and the
denominators.
PTS: 1
DIF: Basic
REF: Page 654
OBJ: 10-4.4 Dividing by Rational Expressions and Polynomials
NAT: 12.5.3.c
STA: 1A13.0
TOP: 10-4 Multiplying and Dividing Rational Expressions
20
ID: A
39. ANS: C
3x + 8
4
+ 2
2
x − 16 x − 16
3x + 8 + 4 3x + 12
= 2
= 2
x − 16
x − 16
3(x + 4)
=
(x + 4)(x − 4)
3
=
x−4
Combine like terms in the numerator.
Factor. Divide out common factors.
Simplify.
Feedback
A
B
C
D
Divide out only common factors.
Factor the denominator and continue simplifying your answer.
Correct!
The denominators are the same, so keep the common denominator.
PTS: 1
DIF: Average
REF: Page 659
OBJ: 10-5.1 Adding Rational Expressions with Like Denominators
NAT: 12.5.3.c
STA: 1A13.0
TOP: 10-5 Adding and Subtracting Rational Expressions
40. ANS: C
3y
3y
Identify the LCD, 18y 2 .
+
2
18y
9y
3y ⋅ 2
3y ⋅ y
+
= 2
18y
⋅y
9y ⋅ 2
=
=
6y
18y
2
3y 2
+
6y + 3y
18y
2
Rewrite each fraction with a denominator of 18y 2 .
2
18y 2
3y(2 + y)
=
3y(6y)
2+y
=
6y
Add.
6y and 3y 2 are not like terms, so they cannot be combined.
Factor and divide out common factors.
Simplify.
Feedback
A
B
C
D
To simplify the numerator, factor out 3y.
First find a common denominator. Then add the numerators.
Correct!
To simplify the numerator, factor out 3y.
PTS: 1
DIF: Basic
REF: Page 661
OBJ: 10-5.4 Adding and Subtracting with Unlike Denominators
NAT: 12.5.3.c
STA: 1A13.0
TOP: 10-5 Adding and Subtracting Rational Expressions
21
ID: A
41. ANS: B
(12x 4 − 18x 3 + 36x 2 ) ÷ 6x 3
12x 4 − 18x 3 + 36x 2
6x 3
12x 4 18x 3 36x 2
=
−
+
6x 3
6x 3
6x 3
2x 4 3x 3 6x 2
= 3 − 3 + 3
x
x
x
6
= 2x − 3 +
x
=
Rewrite as a rational expression.
Divide each term in the numerator by the denominator.
Simplify by dividing out common factors.
Simplify by using powers of exponents.
Feedback
A
B
C
D
Write the division of each term by the divisor as a quotient. Simplify by dividing out
common factors and using the rule of exponents.
Correct!
Write the division of each term by the divisor as a quotient. Simplify by dividing out
common factors and using the rule of exponents.
Write the division of each term by the divisor as a quotient. Simplify by dividing out
common factors and using the rule of exponents.
PTS:
OBJ:
STA:
42. ANS:
1
DIF: Average
REF: Page 667
10-6.1 Dividing a Polynomial by a Monomial
1A10.0
TOP: 10-6 Dividing Polynomials
C
2x + 9 +
NAT: 12.5.3.c
12
x−3
x − 3 2x 2 + 3x − 15
Write the problem in long division form. There will be a remainder when you finish with the long division.
Write the remainder as a rational expression using the divisor as the denominator.
Feedback
A
B
C
D
Use long division.
Write the remainder as a rational expression using the divisor as a denominator.
Correct!
Use long division.
PTS: 1
DIF: Average
REF: Page 670
OBJ: 10-6.4 Long Division with a Remainder
STA: 1A10.0
TOP: 10-6 Dividing Polynomials
22
NAT: 12.5.3.c
ID: A
43. ANS: B
4
3
=
m − 3 5m
4 (5m) = 3 (m − 3)
20m = 3m − 9
17m = −9
9
m = − 17
Use cross products.
Multiply. Distribute 3 on the right side..
Subtract 3m from both sides.
Divide both sides by 17.
Feedback
A
B
C
D
First compute the cross products. Then solve for the variable.
Correct!
First compute the cross products. Then solve for the variable.
First compute the cross products. Then solve for the variable.
PTS: 1
DIF: Average
REF: Page 674
OBJ: 10-7.1 Solving Rational Equations by Using Cross Products
NAT: 12.5.4.a
STA: 1A13.0
TOP: 10-7 Solving Rational Equations
44. ANS: D
Multiply both sides of the equation by the LCD: 3z(z – 21).
Divide out common factors, then distribute. This gives the equation:
–9z + 189 + 4z – 84 = 30z
Combine like terms and simplify. Then solve for z.
Feedback
A
B
C
D
When solving for the variable, perform the same operations on both sides of the
equation.
Find the LCD of all the rational expressions in the equation.
Keep track of signs when multiplying by the LCD and solving for the variable.
Correct!
PTS: 1
DIF: Average
REF: Page 674
OBJ: 10-7.2 Solving Rational Equations by Using the LCD
STA: 1A13.0
TOP: 10-7 Solving Rational Equations
23
NAT: 12.5.4.a
ID: A
45. ANS: A
Factor perfect squares out of the radicand. Use the Product Property of Square Roots to take the square root
of each factor separately. Simplify.
Feedback
Correct!
First, factor perfect squares out of the radicand. Then, take the square roots of the
perfect squares.
Take the square root of the factor when removing the radical.
First, factor perfect squares out of the radicand. Then, take the square roots of the
perfect squares.
A
B
C
D
PTS:
OBJ:
STA:
46. ANS:
1
DIF: Average
REF: Page 706
11-2.2 Using the Product Property of Square Roots
1A2.0
TOP: 11-2 Radical Expressions
C
NAT: 12.5.3.c
b9
16b
b8
16
=
=
Simplify the radicand.
Use the Quotient Property of Square Roots.
b4
4
always indicates a nonnegative square root.
Feedback
A
B
C
D
Simplify the numerator. Find the square root of the radicand (the expression under the
radical sign).
The radical sign always indicates a nonnegative square root.
Correct!
Find the square root of the expression.
PTS: 1
DIF: Average
REF: Page 706
OBJ: 11-2.3 Using the Quotient Property of Square Roots
STA: 1A2.0
TOP: 11-2 Radical Expressions
24
NAT: 12.5.3.c
ID: A
47. ANS: C
80
121
=
=
=
=
80
Use the Quotient Property.
121
16(5)
Find perfect square factors if possible. Write 80 as 16(5).
121
16
5
121
4 5
11
Use the Product Property.
Simplify.
Feedback
A
B
C
D
Factors that aren't perfect squares remain beneath the radical sign.
Only factors that aren't perfect squares remain beneath the radical sign.
Correct!
Check your calculations. A factor from the numerator is missing.
PTS:
OBJ:
NAT:
48. ANS:
75b
=
1
DIF: Basic
REF: Page 707
11-2.4 Using the Product and Quotient Properties Together
12.5.3.c
STA: 1A2.0
TOP: 11-2 Radical Expressions
B
+ 4 12b − 2 27b
Factor the radicands using perfect-square factors.
25(3b) + 4 4(3b) − 2 9(3b)
= 5 3b + 4(2)
3b − 2(3) 3b
Use the Product Property of Square Roots.
= 5 3b + 8 3b − 6 3b
= 7 3b
Combine like radicals.
Feedback
A
B
C
D
The answer should have the radical sign with at least the variable as the radicand. You
cannot square the radical expression without changing its value.
Correct!
Factor the radicands using perfect-square factors.
Use the Product Property of Square Roots.
PTS: 1
DIF: Average
REF: Page 712
OBJ: 11-3.2 Simplifying Before Adding or Subtracting
NAT: 12.5.3.c
STA: 1A2.0
TOP: 11-3 Adding and Subtracting Radical Expressions
25
ID: A
49. ANS: B
25h
10h =
=
250h 2
Product Property of Square Roots.
25 (10)h 2
Factor 250 using a perfect-square factor.
25
10
ÊÁ
ˆ
= ÁÁ 5 10 ˜˜˜ h
Ë
¯
=
Product Property of Square Roots.
h2
Simplify.
Feedback
A
B
C
D
Take the square root of the perfect-square factor.
Correct!
Use the Product Property of Square Roots. Then factor the number using a
perfect-square factor.
Factor the number using a perfect-square factor.
PTS: 1
NAT: 12.5.3.c
50. ANS: A
3
=
5
=
DIF: Advanced
STA: 1A2.0
Ê
3 ÁÁÁÁ
ÁÁ
5 ÁÁ
Ë
15
25
=
ˆ
5 ˜˜˜˜
˜˜
5 ˜˜
¯
REF: Page 716
OBJ: 11-4.1 Multiplying Square Roots
TOP: 11-4 Multiplying and Dividing Radical Expressions
Multiply by a form of 1 to get a perfect-square radicand
in the denominator.
Simplify the denominator.
15
5
Feedback
A
B
C
D
Correct!
First, multiply by a form of 1 to get a perfect-square radicand in the denominator. Then,
simplify the denominator.
Rationalize the denominator by finding the appropriate form of 1 to multiply by.
A quotient with a square root in the denominator is not simplified.
PTS: 1
DIF: Basic
REF: Page 718
OBJ: 11-4.4 Rationalizing the Denominator
NAT: 12.5.3.c
STA: 1A2.0
TOP: 11-4 Multiplying and Dividing Radical Expressions
26
ID: A
51. ANS: B
b = 18
2
b = 18 2
b = 324
Square both sides.
Feedback
A
B
C
D
The radical cannot be negative.
Correct!
Square both sides, not multiply.
Square both sides to isolate the variable.
PTS: 1
DIF: Basic
REF: Page 722
OBJ: 11-5.1 Solving Simple Radical Equations
STA: 1A2.0
TOP: 11-5 Solving Radical Equations
52. ANS: B
Isolate the radical on one side of the equation.
Square both sides.
Solve for y.
Check your answer in the original equation to determine if a solution is extraneous.
Feedback
A
B
C
D
Check your answer in the original equation.
Correct!
Square both sides of the equation.
Square both sides of the equation.
PTS: 1
STA: 1A17.0
DIF: Average
REF: Page 725
TOP: 11-5 Solving Radical Equations
27
OBJ: 11-5.5 Extraneous Solutions