How do you solve #((x^3y)/(2z))^2#?

Question

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

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Answer 1 · 2018-05-18T07:42:34

#256/25#

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

#(((2)^3(-4))/(2(-5)))^2 = ((-32)/-10)^2=(16/5)^2=256/25.#