
Math Equations, Inequalities and Functions Quiz
Authored by Wilhelmina Ramelb
Mathematics
8th Grade
CCSS covered

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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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