AnswersOnline Algebra Function Number AInequalities in One Variable

Inequalities in One Variable Answers

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Which did you include in your response? Check all that apply.

A
Multiplying or dividing by the same value on both sides of an equation does not change the solution set.
B
Multiplying or dividing by the same positive value on both sides of an inequality does not change the solution set.
C
Multiplying or dividing by the same negative value on both sides of an inequality changes the sense of the inequality.
3

What is the solution to the given inequality?

Question illustration
A
x –1
Option A
B
x ≤ –1
C
x ≥ 3
D
x ≤ 3
4

Which graph shows the correct representation of {x | x ≤ 5}?

A
[Image: A number line going from 4 to 6. An open circle ap...]
B
[Image: A number line going from 4 to 6. An open circle ap...]
C
[Image: A number line going from 4 to 6. A solid circle ap...]
D
[Image: A number line going from 4 to 6. A solid circle ap...]
5

The solution to an inequality is (–∞, 6.5]. Is 6.5 a solution to the inequality? Explain your answer.

A
The solution includes numbers from negative infinity to 6.5. Because a bracket is used, the solution does not include 6.5.
B
The solution includes numbers from 6.5 to infinity. Because a parenthesis is used, the solution does not include 6.5.
C
The solution is the point (–∞, 6.5). Because a bracket is used, 6.5 is a solution to the inequality.
D
The solution includes numbers from negative infinity to 6.5. Because a bracket is used, the solution includes 6.5.
6

Let x represent the number of canvases. Which inequality can be used to find x?

A
350 ≥ 125 + 15x
B
350 > 125 – 15x
C
350 ≤ 125 – 15x
D
350 < 125 + 15x
7

The club can buy up to 14 canvases.The club can buy at least 14 canvases. ✔ The club can buy up to 15 canvases.The club can buy at least 15 canvases.

A
The club can buy up to 14 canvases.
B
The club can buy at least 14 canvases.
C
✔ The club can buy up to 15 canvases.
D
The club can buy at least 15 canvases.
11

Which inequality is true for all real numbers? Check all that apply.a – b d, then a– c < a – dIf a < b, then a < b + c

A
a – b < a + b
B
ac ≥ bc
C
If a ≥ b, then a + c ≥ b + c
D
If c > d, then a– c < a – d
E
If a < b, then a < b + c

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