How do you solve ${(\frac{x^{3} y}{2 z})}^{2}$ $((x^3y)/(2z))^2$ ?

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How do you solve ${(\frac{x^{3} y}{2 z})}^{2}$ $((x^3y)/(2z))^2$ ?

x=2, y=-4, and z=-5

1 Answer

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Answer 1 · 2018-05-18T07:42:34

$\frac{256}{25}$ $256/25$

Explanation:

By plugging in the values $x = 2$ $x=2$ , $y = - 4$ $y=-4$ and $z = - 5$ $z=-5$ , we obtain:

${(\frac{{(2)}^{3} (- 4)}{2 (- 5)})}^{2} = {(\frac{- 32}{- 10})}^{2} = {(\frac{16}{5})}^{2} = \frac{256}{25} .$ $(((2)^3(-4))/(2(-5)))^2 = ((-32)/-10)^2=(16/5)^2=256/25.$

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How do you solve ${(\frac{x^{3} y}{2 z})}^{2}$ $((x^3y)/(2z))^2$ ?

x=2, y=-4, and z=-5

1 Answer

Explanation:

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How do you solve ${(\frac{x^{3} y}{2 z})}^{2}$ $((x^3y)/(2z))^2$ ?