What is ((4x^{-3}y^4)/(8x^2y^{-2)})^{-2} ?

The solution to ((4x^{-3}y^4)/(8x^2y^{-2)})^{-2} is (4x^{10})/(y^{12)}

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simplify ((4x^{-3}y^4)/(8x^2y^{-2)})^{-2}

simplify

(\frac{4 x ^{- 3} y ^{4}}{8 x ^{2} y ^{- 2}})^{- 2}

\frac{4 x ^{10}}{y ^{12}}

Solution steps

(\frac{4 x ^{- 3} y ^{4}}{8 x ^{2} y ^{- 2}})^{- 2}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

(\frac{4 x ^{- 3} y ^{4}}{8 x ^{2} y ^{- 2}})^{- 2} = \frac{1}{( \frac{4 x ^{- 3} y ^{4}}{8 x ^{2} y ^{- 2}} ) ^{2}}

= \frac{1}{( \frac{4 x ^{- 3} y ^{4}}{8 x ^{2} y ^{- 2}} ) ^{2}}

Simplify

(\frac{4 x ^{- 3} y ^{4}}{8 x ^{2} y ^{- 2}})^{2} : \frac{y ^{12}}{4 x ^{10}}

= \frac{1}{\frac{y ^{12}}{4 x ^{10}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{4 x ^{10}}{y ^{12}}

- 2 x^{2} + 98 = 0

5 ∣ x + 9 ∣ + 2 < 17

(2 + i) (2 - 2 i)

3 x + 7 \leq - 2 and 2 x \leq 0

\frac{3 x + 1}{7} - \frac{2 - 4 x}{3} = \frac{5 x - 4}{14} + \frac{7 x}{6}

What is ((4x^{-3}y^4)/(8x^2y^{-2)})^{-2} ?
The solution to ((4x^{-3}y^4)/(8x^2y^{-2)})^{-2} is (4x^{10})/(y^{12)}