What is ((3m^{-5}n^2)/(4m^{-2)n^0})^2*((mn^4)/(9n))^2 ?

The solution to ((3m^{-5}n^2)/(4m^{-2)n^0})^2*((mn^4)/(9n))^2 is (n^{10})/(144m^4)

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simplify ((3m^{-5}n^2)/(4m^{-2)n^0})^2*((mn^4)/(9n))^2

simplify

(\frac{3 m ^{- 5} n ^{2}}{4 m ^{- 2} n ^{0}})^{2} \cdot (\frac{m n ^{4}}{9 n})^{2}

\frac{n ^{10}}{144 m ^{4}}

Solution steps

(\frac{3 m ^{- 5} n ^{2}}{4 m ^{- 2} n ^{0}})^{2} (\frac{m n ^{4}}{9 n})^{2}

Simplify

(\frac{3 m ^{- 5} n ^{2}}{4 m ^{- 2} n ^{0}})^{2} : \frac{9 n ^{4}}{16 m ^{6}}

= \frac{9 n ^{4}}{16 m ^{6}} (\frac{m n ^{4}}{9 n})^{2}

Simplify

(\frac{m n ^{4}}{9 n})^{2} : \frac{m ^{2} n ^{6}}{81}

= \frac{9 n ^{4}}{16 m ^{6}} \cdot \frac{m ^{2} n ^{6}}{81}

Apply the fraction rule:

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

= \frac{9 n ^{4} m ^{2} n ^{6}}{16 m ^{6} \cdot 81}

Factor the number:

81 = 9 \cdot 9

= \frac{9 n ^{4} m ^{2} n ^{6}}{16 m ^{6} \cdot 9 \cdot 9}

Cancel the common factor:

9

= \frac{n ^{4} m ^{2} n ^{6}}{16 m ^{6} \cdot 9}

Simplify

\frac{m ^{2}}{m ^{6}} : \frac{1}{m ^{4}}

= \frac{n ^{4} n ^{6}}{16 m ^{4} \cdot 9}

Simplify

n^{4} n^{6} : n^{10}

= \frac{n ^{10}}{16 m ^{4} \cdot 9}

Multiply the numbers:

16 \cdot 9 = 144

= \frac{n ^{10}}{144 m ^{4}}

x + 3 = - 1

m^{2} - 9 = 0

2 x^{2} y - 8 x y^{2}

t^{2} - 10 t + 24

40 - x \geq 8 x - 5

What is ((3m^{-5}n^2)/(4m^{-2)n^0})^2*((mn^4)/(9n))^2 ?
The solution to ((3m^{-5}n^2)/(4m^{-2)n^0})^2*((mn^4)/(9n))^2 is (n^{10})/(144m^4)