Literal Equations Review

Literal Equations Review

8th - 9th Grade

12 Qs

quiz-placeholder

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Literal Equations Review

Literal Equations Review

Assessment

Quiz

Mathematics

8th - 9th Grade

Practice Problem

Hard

CCSS
HSA.CED.A.4, 6.EE.C.9

Standards-aligned

Used 16+ times

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12 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

Solve the literal equation for the given variable.

2Ah\frac{2A}{h}

2h\frac{2}{h}

2Ah

Ah2\frac{Ah}{2}

2.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

h=V3πr2h=\frac{V}{3\pi r^2}

h=3Vπr2h=\frac{3V}{\pi r^2}

h=Vr23πh=\frac{Vr^2}{3\pi}

h=3πVr2h=\frac{3\pi}{Vr^2}

3.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

s = -Dn + C

s = Dn - C

s=DnCs=\frac{Dn}{C}

s=CDns=\frac{C}{Dn}

4.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

 F=GMmr2F=\frac{GMm}{r^2}  

(solve for G) 

 G=Fr2MnG=\frac{Fr^2}{Mn}  

 G=FMmr2G=FMmr^2  

 Fr2=GMmFr^2=GMm  

 G=Fr2M2G=Fr^2M^2  

5.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

T = PV/K
T = PK/V
T=KV/P
Greenland

Tags

CCSS.HSA.CED.A.4

6.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

The volume of a sphere can be expressed as  V=43πr3V=\frac{4}{3}\pi r^3 , where 𝒓 is the radius of the sphere.
Which formula can be used to determine the radius of the sphere? 

 r=3V4πr=\frac{3V}{4\pi}  

 r=4V3πr=\frac{4V}{3\pi}  

 r= 33V4πr=\ ^3\sqrt{\frac{3V}{4\pi}}  

 r= 33V4πr=\ ^3\sqrt{\frac{3V}{4\pi}}  

7.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

 ax5 = b; for aax-5\ =\ b;\ for\ a  

 x(b+5)x\left(b+5\right)  

 (b5)x\frac{\left(b-5\right)}{x}  

 (b+5)x\frac{\left(b+5\right)}{x}  

 x(b5)x\left(b-5\right)  

Tags

CCSS.HSA.CED.A.4

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