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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

1 answer

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