# Simplify and express the final result using positive exponents. (8y22y−1)−1

ka1leE

## Answered question

2021-09-30

Simplify and express the final result using positive exponents.

$\left(\frac{8{y}^{2}}{2{y}^{-1}}\right)}^{-1$

### Answer & Explanation

Given expression:

$\left(\frac{8{y}^{2}}{2{y}^{-1}}\right)}^{-1$

Simplification:

$\left(\frac{8{y}^{2}}{2{y}^{-1}}\right)}^{-1}=\frac{1}{\frac{8{y}^{2}}{2{y}^{-1}}$

$=\frac{2{y}^{-1}}{8{y}^{2}}$

$=\frac{{y}^{-1}}{4{y}^{2}}$

$=\frac{1}{4}{y}^{-1}\times {y}^{-2}$

$=\frac{1}{4}{y}^{-3}$

$=\frac{1}{4{y}^{3}}$

Hence $\left(\frac{8{y}^{2}}{2{y}^{-1}}\right)}^{-1}=\frac{1}{4{y}^{3}$

Do you have a similar question?

Recalculate according to your conditions!