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Math Equations, Inequalities and Functions Quiz

Authored by Wilhelmina Ramelb

Mathematics

8th Grade

CCSS covered

Math Equations, Inequalities and Functions Quiz
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19 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

A teacher is buying bags of chips for a grade-level picnic. The teacher wants to spend at most $36 on chips. Each bag of chips costs $2.95. Which inequality represents the number of bags of chips, x, the teacher can purchase?

2.95x > 36

2.95 ≥ 36

2.95 < 36

2.95x ≤ 36

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The cost of renting a moving van is $25 a day, plus $0.35 per mile driven. Which function can be used to find the total cost C(x) in dollars of renting a van for a day and driving x miles?

C(x) = 25x - 0.35

C(x) = 25x + 0.35

C(x) = 0.35x - 25

C(x) = 25 + 0.35x

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The lateral area of a cylinder is L = 2πrh. What is this equation solved in terms of r?

r = 2πL/h

r = Lh/2π

r = 2πh/L

r = L/2πh

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The volume of a cylinder can be expressed as $ V = \pi r^2 h $, where $ r $ is the radius, and $ h $ is the height of the cylinder. Which formula can be used to determine the radius of the cylinder?

$ r = \frac{V}{\pi h} $

$ r = \frac{2V}{\pi h} $

$ r = \left( \frac{V}{\pi h} \right)^2 $

$ r = \sqrt{\frac{V}{\pi h}} $

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Solve for $ h $. $ f = 8g + 4h $

$ h = 2g + \frac{f}{4} $

$ h = 2g - \frac{f}{4} $

$ h = -2g + \frac{f}{4} $

$ h = -2g - \frac{f}{4} $

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The equation $ (y - y_1) = m (x - x_1) $ represents the point-slope form of a linear equation. Which equation can be used to find $ y_1 $?

$ y_1 = m(x - x_1) + y $

$ y_1 = -mx + mx_1 - y $

$ y_1 = y - mx + mx_1 $

$ y_1 = -mx - mx_1 + y $

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which could be the first step in solving the equation \( \frac{1}{2} (2 + 4x) = x + 5 \)? Select ALL that apply.

Multiply both sides of the equation by 2, resulting in \( 2 + 4x = x + 5 \).

Multiply both sides of the equation by 2, resulting in \( 2 + 4x = 2x + 10 \).

Distribute the \( \frac{1}{2} \) to remove the parentheses, resulting in \( 1 + 2x = x + 5 \).

Subtract \( x \) from both sides of the equation, resulting in \( \frac{1}{2} (2 + 3x) = 5 \).

Subtract \( 4x \) from both sides of the equation, resulting in \( \frac{1}{2} (2) = x + 5 \).

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