Question

Which expression shows the simplified form of (8 r Superscript negative 5 Baseline) Superscript negative 3? 8 r Superscript 15 StartFraction 8 Over r Superscript 15 Baseline EndFraction 512 r Superscript 15 StartFraction r Superscript 15 Baseline Over 512 EndFraction

Answer 1

Answer:

free pointzzz

Step-by-step explanation:

Answer 2

Answer:

$\frac{r^{15}}{512}$

Step-by-step explanation:

Given

$(8r^{-5})^{-3}$

Required

Simplify

This can be simplified using the following law of indices;

$(ab)^n = a^{n}b^{n}$

The equation becomes

$(8^{-3})(r^{-5})^{-3}$

Express $8^{-3}$ as a fraction

$(\frac{1}{8^{3}})(r^{-5})^{-3}$

Simplify $8^3$

$(\frac{1}{8*8*8})(r^{-5})^{-3}$

$(\frac{1}{512})(r^{-5})^{-3}$

The expression can further be simplified using the following law of indices;

$(a^m)^n = a^{mn}$

$(\frac{1}{512})(r^{-5})^{-3}$ becomes

$(\frac{1}{512})(r^{-5*-3})$

$(\frac{1}{512})(r^{15})$

$\frac{r^{15}}{512}$

Hence, the solution to $(8r^{-5})^{-3}$ is $\frac{r^{15}}{512}$