Question

PLS HELP!!!

Answer 1

@Vocaloid

Answer 2

@mhchen

Answer 3

Substitute the x and y that were given

Answer 4

y=2, x=-4

Answer 5

Also
\(x^{-n}=\frac{1}{x^n}\)
n being any number

Likewise
\(\frac{1}{x^{-n}}=>x^n\)

Answer 6

Yes

Answer 7

\[\frac{ 1 }{ 4 }\]-\[\frac{ 1 }{ 64 }\],
32

Answer 8

That is confusing to look at
\(\large \frac{1}{2^{-2}x^{-3}y^5}\), \( x=2, y=-4\)

\(\large \frac{1}{2^{-2}2^{-3}-4^5}\)

\(\large \frac{1}{\color{red}{2^{-2}2^{-3}}-4^5}\) The red will flip to go on the numerator and will lose the - exponent

Answer 9

\[\frac{ 1 }{ 4 }\],\[\frac{ 1 }{ 8 }\],-1024

Answer 10

Also quick note, if you want to copy the latex, right-click the math and Show Math As > Tex Commands

When pasting include `\( \)` in between

Answer 11

@dude thanks
how do I get the final answer

Answer 12

\(\large \frac{1}{2^{-2}2^{-3}-4^5}\)

After flipping the numbers to the numerator
=\(\large \frac{2^{2}2^{3}}{-4^5}\)


\(2^2=4\)
\(2^3=8\)
and
\(-4^5=-1024\)

Answer 13

\(\large \frac{4\cdot 8}{-1024}\)

=\(\large -\frac{32}{1024}=>\boxed{\frac{1}{32}}\)

Answer 14

@dude pardon but the result is minus 1/32

Answer 15

@ThisGirlPretty do you agree it ?

Answer 16

Right, thank you
\(\large -{\frac{1}{32}}\)