Question

911!!! please answer this!!!

Answer 1

$\left(\dfrac{3a^{-3}b^2}{2a^{-1}b^0}\right)^2$

First some simplifying:

$\left(\dfrac{3a^1b^2}{2a^3b^0}\right)^2=\left(\dfrac{3b^2}{2a^2}\right)^2=\dfrac{9b^4}{4a^4}$

Then with $a=-2$ and $b=-3$, we have $a^4=(-2)^4=16=4^2$ and $b^4=(-3)^4=81=9^2$, so

$\dfrac{9b^4}{4a^4}=\dfrac{9\cdot9^2}{4\cdot4^2}=\dfrac{9^3}{4^3}=\dfrac{729}{64}$

Answer 2

Answer:

Step-by-step explanation:

Fourth option