Question

evaluate [4x^3y^-2] [3x^-2y^4] for x=-2 and y=-1.

Answer 1

Just plug in values for x and y.

Answer 2

by the way it written, the number is raised to multiple poweres, maybe reducing then plug in? :)

Answer 3

\[(\frac{4x^3}{y^2})(\frac{3y^4}{x^2})=(\frac{4(-2)^3}{(-1)^2})(\frac{3(-1)^4}{(-2)^2})=\frac{-32}{1}\times\frac{3(1)}{4}=-24\]