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\(\dfrac{(x^4y^3)^{-2}}{(x^3y^2)^{-4}}=\)

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\(\dfrac{(x^4y^3)^{-2}}{(x^3y^2)^{-4}}=\)

by Vincen » Mon Nov 09, 2020 7:48 am

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Your Answer

A

B

C

D

E

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\(\dfrac{(x^4y^3)^{-2}}{(x^3y^2)^{-4}}=\)

A. \(x^{-4}y^{-2}\)

B. \(xy\)

C. \(x^3y^3\)

D. \(x^2y^2\)

E. \(x^4y^2\)

Answer: E

Source: Veritas Prep
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