Literal Equations Answers

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1
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The formula for the nth term of an arithmetic sequence can be found using the formula . Which of the following is equivalent to this equation?

Question illustration
A
n = a Subscript n Baseline + a Subscript 1
Option A
B
n = StartFraction a Subscript n Baseline + a Subscript 1 Baseline minus 3 Over d EndFraction
Option B
C
n = a Subscript n Baseline minus a Subscript 1
Option C
D
n = a Subscript n Baseline minus a Subscript 1 Baseline + d Over d EndFraction
Option D
2
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If , which equation is solved for t?

Question illustration
A
I – pr = t
B
StartFraction uppercase I minus p Over r EndFraction = t
Option B
C
StartFraction uppercase I Over p r EndFraction = t
Option C
D
I + pr = t
3

Which equation is equivalent to

Question illustration
A
s = t minus 2
Option A
B
s = StartFraction 4 Over t + 2 EndFraction
Option B
C
s = StartFraction t + 2 Over 4 EndFraction
Option C
D
s = t + 6
Option D
4

Paolo wrote the following equation for the perimeter of a rectangle. Which equation is equivalent to the equation Paolo wrote?

Question illustration
A
w = uppercase P minus 2 l
Option A
B
2 = uppercase P minus l
Option B
C
w = StartFraction uppercase P minus 2 l Over 2 EndFraction
Option C
D
w = StartFraction uppercase P + 2 l Over 2 EndFraction
Option D
5

The equation is the slope-intercept form of the equation of a line. What is the equation solved for b?

Question illustration
A
y minus m = b
Option A
B
y minus m x = b
Option B
C
StartFraction y Over m x EndFraction = b
Option C
D
StartFraction y Over m EndFraction minus x = b
Option D
6

A student solves the equation using the steps in the table. Original equationCross multiplicationDistributive propertySubtraction property of equality5 = xWhich method of solving for the variable could be used instead of cross multiplication?

Question illustration
A
distributing x + 3 and then 3x + 5 to both sides of the equation
B
distributing x – 3 and then 3x – 5 to both sides of the equation
C
using the multiplication property of equality to multiply both sides of the equation by 10
D
using the multiplication property of equality to multiply both sides of the equation by
Option D
7

The final velocity, V, of an object under constant acceleration can be found using the formula , where v is the initial velocity (in meters per second), a is acceleration (in meters per second), and s is the distance (in meters). What is the formula solved for a?

Question illustration
A
uppercase V squared minus v squared minus 2 x = a
Option A
B
uppercase V squared minus v squared + 2 s = a
Option B
C
StartFraction uppercase V squared minus v squared Over 2 s EndFraction = a
Option C
D
StartFraction uppercase V squared plus v squared Over 2 s EndFraction = a
Option D
8

A student solves the equation for v. Start with the original equation.Use the division property of equality.Use the multiplication property of equality.Use the square root property of equality.Simplify.Which statement explains how to correct the error that was made?

Question illustration
A
The subtraction property of equality should have been applied to move m to the other side of the equation.
B
The multiplication property of equality should have been applied in the last step.
C
The division property of equality should have been used to divide by k instead of m.
D
The square root property should have been applied to both complete sides of the equation instead of to select variables.
9

Gavin wrote the equation to represent p, the profit he makes from s sales in his lawn-mowing business. Which equation is solved for s?

Question illustration
A
s = StartFraction p minus 100 Over 3 EndFraction
Option A
B
s = StartFraction 4 p minus 300 Over 3 EndFraction
Option B
C
s = StartFraction 4 p Over 300 EndFraction
Option C
D
s = StartFraction 400 p Over 3 EndFraction
Option D
10

When is solved for c, one equation is Which of the following is an equivalent equation to find c?

Question illustration
A
c = 10 b minus 10 minus 5
Option A
B
c = (10 b minus 10 minus 5) squared
Option B
C
c = StartFraction (10 b minus 2) squared Over 25 EndFraction
Option C
D
c = StartFraction (10b minus 10) squared Over 25 EndFraction
Option D

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