How do you simplify #((16x^2y^-1)/(4x^0y^-4z))^-2#?

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Answer 1 · 2017-04-14T17:29:44

#(16z^2)/(256x^4y^8)#

Recall: #(a/b)^-2 = (b/a)^2#

(Flip the fraction and the index changes sign.)

#((16x^2y^-1)/(4x^0y^-4z))^-2 = ((4x^0y^-4z)/(16x^2y^-1))^2#

Now simplify the fraction inside the bracket.

#((4x^0y^-4z)/(16x^2y^-1))^2 = ((4(1)yz)/(16x^2y^4))^2#

#((4z)/(16x^2y^3))^2" "larr# now square all the factors.

#=(16z^2)/(256x^4y^8)#