How do you solve ((x^3y)/(2z))^2(x3y2z)2?

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

1 Answer
May 18, 2018

256/2525625

Explanation:

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

(((2)^3(-4))/(2(-5)))^2 = ((-32)/-10)^2=(16/5)^2=256/25.((2)3(4)2(5))2=(3210)2=(165)2=25625.