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

[Vincen]

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

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

B. \(xy\)

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

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

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

Answer: E

Source: Veritas Prep