Question

Simplify: ((12x^-3)/(y^4))*(((y^-2x^2)^-1)/(15x^-2))

Answer 1

\[\frac{12x^{-3}}{y^4}\times \frac{(y^{-2}x^2)^{-1}}{15x^{-2}}\]

Answer 2

for the term in the brackets that its to the power of negative one,

\[(y^{-2}x^2)^{-1}\]

distribute the index, like this

\[(a^nb^n)^p=a^{n\times p}b^{m\times p}\]

Answer 3

can you do that?

Answer 4

y^2x^-2

Answer 5

I got 4/(3xy^2)

Answer 6

well you got this bit right \((y^{−2}x^2)^{−1}=y^2x^{-2}\)\(\checkmark\)

Answer 7

your final result is close but not quite right

Answer 8

...\[\frac{12x^{-3}}{y^4}\times \frac{y^{2}x^{-2}}{15x^{-2}}\]...