ka1leE

2021-09-30

Simplify and express the final result using positive exponents.
${\left(\frac{8{y}^{2}}{2{y}^{-1}}\right)}^{-1}$

Liyana Mansell

Given expression:
${\left(\frac{8{y}^{2}}{2{y}^{-1}}\right)}^{-1}$
Simplification:
${\left(\frac{8{y}^{2}}{2{y}^{-1}}\right)}^{-1}=\frac{1}{\frac{8{y}^{2}}{2{y}^{-1}}}$
$=\frac{2{y}^{-1}}{8{y}^{2}}$
$=\frac{{y}^{-1}}{4{y}^{2}}$
$=\frac{1}{4}{y}^{-1}×{y}^{-2}$
$=\frac{1}{4}{y}^{-3}$
$=\frac{1}{4{y}^{3}}$
Hence ${\left(\frac{8{y}^{2}}{2{y}^{-1}}\right)}^{-1}=\frac{1}{4{y}^{3}}$

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