Question

[3^-1/(-2)^-2]^-2 simplify it ​

Answer 1

Answer:

$-\frac{9}{16}$

Step-by-step explanation:

We can start by separating the numbers that have exponents.

$3^{-1} = \frac{1}{3}$

$-2^{-2} = -\frac{1}{4}$

These both need to be put back together

$\frac{1}{3} /-\frac{1}{4}$

Which can be simplified to:

-4/3

Then the last $x^{-2}$

Since it's a negative, we swap the fraction over

$-\frac{3^{2} }{4^{2}}$

Normal squares

$-\frac{9}{16}$

Answer 2

$\displaystyle{\sf{(\frac{3^-^1}{(-2)^-^2})^2 }}$

$\displaystyle{\sf{=(\frac{(-2)^2}{3^1})^2 }}$

$\displaystyle{\sf{=(\frac{4}{3} )^2}}$

$\underline{\boxed{\displaystyle{\sf{=\frac{16}{9} }}}}$