PLS HELP!!!

6 months ago@Vocaloid

6 months ago@mhchen

6 months agoSubstitute the x and y that were given

6 months agoy=2, x=-4

6 months agoAlso \(x^{-n}=\frac{1}{x^n}\) n being any number Likewise \(\frac{1}{x^{-n}}=>x^n\)

6 months agoYes

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

6 months agoThat 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

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

6 months agoAlso quick note, if you want to copy the latex, right-click the math and Show Math As > Tex Commands When pasting include `\( \)` in between

6 months ago@dude thanks how do I get the final answer

6 months ago\(\large \frac{1}{2^{-2}2^{-3}-4^5}\) After flipping the numbers to the numerator =\(\large \frac{2^{2}2^{3}}{-4^5}\) \(2^2=4\) \(2^3=8\) and \(-4^5=-1024\)

6 months ago\(\large \frac{4\cdot 8}{-1024}\) =\(\large -\frac{32}{1024}=>\boxed{\frac{1}{32}}\)

6 months ago@dude pardon but the result is minus 1/32

6 months ago@ThisGirlPretty do you agree it ?

6 months agoRight, thank you \(\large -{\frac{1}{32}}\)

6 months ago