Default
Question
Simplify: ${\frac{(6x^3y^{-4})^{-2}}{(3x^2y^5)^{-3}}}$
Solution
The correct answer is 64/9
Explanation
$7\dfrac{1}{9}$ = $7$ + $\dfrac{1}{9}$ = $\dfrac{7 * 9}{9}$ + $\dfrac{1}{9}$ = $\dfrac{64}{9}$
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