Question

If r and s are positive real numbers, which expression is equivalent to StartFraction r Superscript StartFraction 2 over 3 EndFraction Baseline s Superscript StartFraction 1 over 2 EndFraction Baseline over r s EndFraction?

Answer options with 5 options
A.
r Superscript 3 Baseline s Superscript 2 Baseline

B.
r Superscript StartFraction 1 over 3 EndFraction Baseline s Superscript StartFraction 1 over 2 EndFraction Baseline

C.
r Superscript StartFraction 2 over 3 EndFraction Baseline s Superscript StartFraction 1 over 2 EndFraction Baseline

D.
StartFraction 1 over r Superscript StartFraction 2 over 3 EndFraction Baseline s Superscript StartFraction 1 over 2 EndFraction Baseline EndFraction

E.
StartFraction 1 over r Superscript StartFraction 1 over 3 EndFraction Baseline s Superscript StartFraction 1 over 2 EndFraction Baseline EndFraction

Answer 1

We start with the expression:

\[
\frac{r^{\frac{2}{3}} s^{\frac{1}{2}}}{r s}
\]

To simplify, we can break it down into two parts:

1. The exponent for \(r\):
- We have \(r^{\frac{2}{3}}\) in the numerator and \(r\) (which is equivalent to \(r^1\)) in the denominator.
- Using the properties of exponents, we get:
\[
r^{\frac{2}{3} - 1} = r^{\frac{2}{3} - \frac{3}{3}} = r^{\frac{2 - 3}{3}} = r^{-\frac{1}{3}}
\]

2. The exponent for \(s\):
- We have \(s^{\frac{1}{2}}\) in the numerator and \(s\) (equivalent to \(s^1\)) in the denominator.
- Using the properties of exponents, we get:
\[
s^{\frac{1}{2} - 1} = s^{\frac{1}{2} - \frac{2}{2}} = s^{\frac{1 - 2}{2}} = s^{-\frac{1}{2}}
\]

Combining these results, we get:
\[
\frac{r^{\frac{2}{3}} s^{\frac{1}{2}}}{r s} = r^{-\frac{1}{3}} s^{-\frac{1}{2}}
\]

This can be rewritten as:
\[
\frac{1}{r^{\frac{1}{3}} s^{\frac{1}{2}}}
\]

Thus, the expression \( \frac{r^{\frac{2}{3}} s^{\frac{1}{2}}}{r s} \) simplifies to:
\[
\frac{1}{r^{\frac{1}{3}} s^{\frac{1}{2}}}
\]

Looking at the answer options, the matching choice is:

**E. \(\frac{1}{r^{\frac{1}{3}} s^{\frac{1}{2}}}\)**.