Which of the following is an example of using the Division Property of Equality to rearrange the equation PV=nRT ?(1 point)

Question

Responses

R=nTPV
upper R equals Start Fraction n upper T over upper P upper V End Fraction

n=PVRT
n equals Start Fraction upper P upper V over upper R upper T End Fraction

T=PVnR
upper T equals upper P upper V n upper R

R=PV−nT
upper R equals upper P upper V minus n upper T

Answer 1

The Division Property of Equality states that if you divide both sides of an equation by the same non-zero number, the two sides remain equal.

In the equation \( PV = nRT \), if we want to solve for \( R \), we can divide both sides by \( nT \).

Doing this gives us:

\[ R = \frac{PV}{nT} \]

So, the correct response that shows the use of the Division Property of Equality to rearrange the equation \( PV = nRT \) is:

R = \(\frac{nTPV}\).

Thus, the answer option that corresponds to this is:

upper R equals Start Fraction n upper T over upper P upper V End Fraction.

Answer 2

Given the formula for the circumference of a circle, C=2πr , how would you rearrange the formula if you were interested in finding the radius of the circle? (1 point)
Responses

r=Cπ
r equals Start Fraction upper C over pi End Fraction

r=2πC
r equals 2 pi upper C

r=C2π
r equals Start Fraction upper C over 2 pi End Fraction

r=Cπ2

Answer 3

To rearrange the formula for the circumference of a circle, \( C = 2\pi r \), to solve for the radius \( r \), you would divide both sides of the equation by \( 2\pi \).

Starting with:

\[ C = 2\pi r \]

Dividing both sides by \( 2\pi \) gives:

\[ r = \frac{C}{2\pi} \]

So the correct response is:

r = \(\frac{C}{2\pi}\), which corresponds to:

r = C2π.