Unit Test — Unit test Answers

23 verified answers
3

Which graph represents the function f(x) = |x| – 4?

A
On a coordinate plane, an absolute value graph has a vertex at (0, 4).
Option A
B
On a coordinate plane, an absolute value graph has a vertex at (negative 4, 0).
Option B
C
On a coordinate plane, an absolute value graph has a vertex at (0, negative 4).
Option C
D
On a coordinate plane, an absolute value graph has a vertex at (4, 0).
Option D
6

What is the simplified form of the following expression?

Question illustration
A
6 StartRoot 2 EndRoot
Option A
B
18 StartRoot 2 EndRoot
Option B
C
30 StartRoot 2 EndRoot
Option C
D
36 StartRoot 2 EndRoot
Option D
7

Which number line represents the solutions to –2|x| = –6?

A
A number line from negative 8 to 8 in increments of 1. Two points, one at negative 6 and one at negative 2.
Option A
B
A number line from negative 8 to 8 in increments of 1. Two points, one at negative 8 and one at negative 4.
Option B
C
A number line from negative 8 to 8 in increments of 1. Two points, one at negative 4 and one at 4.
Option C
D
A number line from negative 8 to 8 in increments of 1. Two points, one at negative 3 and one at 3.
Option D
8

What is the following sum?

Question illustration
A
7 x (RootIndex 6 StartRoot x squared y EndRoot)
Option A
B
7 x squared (RootIndex 6 StartRoot x y squared EndRoot)
Option B
C
7 x squared (RootIndex 3 StartRoot x y squared EndRoot)
Option C
D
7 x (RootIndex 3 StartRoot x squared y EndRoot)
Option D
9

What is the solution to the inequality ?

Question illustration
A
–3 > n > –2
B
2 < n < 3
C
n –2
D
n 3
10

What is the simplest form of ?

Question illustration
A
3 x squared (RootIndex 4 StartRoot y Superscript 5 EndRoot
Option A
B
3 x squared y (RootIndex 4 StartRoot y EndRoot)
Option B
C
9 x squared y (RootIndex 4 StartRoot y EndRoot)
Option C
D
9 x Superscript 4 Baseline y squared (RootIndex 4 StartRoot y EndRoot)
Option D
12

Johan found that the equation –2|8 – x| – 6 = –12 had two possible solutions: x = 5 and x = –11. Which explains whether his solutions are correct?

A
He is correct because both solutions satisfy the equation.
B
He is not correct because he made a sign error.
C
He is not correct because there are no solutions.
D
He is not correct because there is only one solution: x = 5.
14

What is the following product? Assume

Question illustration
A
80 x Superscript 4 Baseline + 8 x Superscript 4 Baseline StartRoot 30 x EndRoot + 24 x Superscript 4
Option A
B
80 x Superscript 6 Baseline + 8 x Superscript 5 Baseline + 8 x Superscript 5 Baseline StartRoot 30 EndRoot + 24 x Superscript 4
Option B
C
104 x Superscript 4
Option C
D
104 x Superscript 4 + 16 x Superscript 4 Baseline StartRoot 30 EndRoot
Option D
16

What is the following quotient?

Question illustration
A
StartFraction StartRoot 3 EndRoot Over 4 EndFraction
Option A
B
StartFraction 1 + StartRoot 3 EndRoot Over 4 EndFraction
Option B
C
StartFraction 1 minus StartRoot 3 EndRoot Over 4 EndFraction
Option C
D
StartFraction negative 1 + StartRoot 3 EndRoot Over 2 EndFraction
Option D
17

What is the following simplified product? Assume

Question illustration
A
10 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot + x squared StartRoot 15 x EndRoot
Option A
B
11 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus x Superscript 4 Baseline StartRoot 75 EndRoot + x squared StartRoot 15 EndRoot
Option B
C
10 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot minus x squared StartRoot 15 EndRoot
Option C
D
11 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot minus x cubed StartRoot 15 x EndRoot
Option D
18

Which expression is equivalent to ?

Question illustration
A
StartFraction RootIndex 12 StartRoot 27 EndRoot Over 2 EndFraction
Option A
B
StartFraction RootIndex 4 StartRoot 24 EndRoot Over 2 EndFraction
Option B
C
StartFraction RootIndex 12 StartRoot 55296 EndRoot Over 2 EndFraction
Option C
D
StartFraction RootIndex 12 StartRoot 177147 EndRoot Over 3 EndFraction
Option D
20

What is the following product?

Question illustration
A
x squared (RootIndex 3 StartRoot 28 x squared EndRoot)
Option A
B
x Superscript 5 Baseline (RootIndex 3 StartRoot 28 x EndRoot)
Option B
C
4 x squared (RootIndex 3 StartRoot 3 x squared EndRoot)
Option C
D
4 x Superscript 5 Baseline (RootIndex 3 StartRoot 3 x EndRoot)
Option D

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