What is ((2h^2k)^4)/((7h^{-1)k^2)^2} ?

The solution to ((2h^2k)^4)/((7h^{-1)k^2)^2} is (16h^{10})/(49)

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simplify ((2h^2k)^4)/((7h^{-1)k^2)^2}

simplify

\frac{( 2 h ^{2} k ) ^{4}}{( 7 h ^{- 1} k ^{2} ) ^{2}}

\frac{16 h ^{10}}{49}

Solution steps

\frac{( 2 h ^{2} k ) ^{4}}{( 7 h ^{- 1} k ^{2} ) ^{2}}

Simplify

(2 h^{2} k)^{4} : 16 h^{8} k^{4}

= \frac{16 h ^{8} k ^{4}}{( 7 h ^{- 1} k ^{2} ) ^{2}}

Simplify

(7 h^{- 1} k^{2})^{2} : \frac{49 k ^{4}}{h ^{2}}

= \frac{16 h ^{8} k ^{4}}{\frac{49 k ^{4}}{h ^{2}}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{16 h ^{8} k ^{4} h ^{2}}{49 k ^{4}}

Cancel

\frac{16 h ^{8} k ^{4} h ^{2}}{49 k ^{4}} : \frac{16 h ^{10}}{49}

= \frac{16 h ^{10}}{49}

\frac{\frac{1}{x} - \frac{1}{y}}{\frac{1}{x} + \frac{1}{y}}

2 x^{2} - x^{3}

3 m^{5} n^{2} - 6 m^{4} n + m^{3}

\frac{1}{3 - 2}

x = 2^{2}

What is ((2h^2k)^4)/((7h^{-1)k^2)^2} ?
The solution to ((2h^2k)^4)/((7h^{-1)k^2)^2} is (16h^{10})/(49)