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Answer :

Answer:

[tex]\bf C)\: -\cfrac{1}{32}[/tex]

Step-by-step explanation:

[tex]\tt \cfrac{1}{2^{-2}x^{-3}y^5}[/tex]

[tex]\tt x=2\\y=-4[/tex]

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Substitute ''x'' with '2' & 'y' with "-4":-

[tex]\tt \cfrac{1}{2^{-2}\times \:2^{-3}\left(-4\right)^5}[/tex]

First let's solve [tex]\tt 2^{-2}\times \:\:2^{-3}\left(-4\right)^5[/tex]

[tex]\tt 2^{-2}\times \:2^{-3}=\boxed{\cfrac{1}{32} }[/tex]

[tex]\tt \cfrac{1}{32}\left(-4\right)^5[/tex]

Calculate exponents:- -4^5= -1024

[tex]\tt \cfrac{1}{32}\left(-1024\right)[/tex]

Remove parentheses, apply rule:- [tex]\tt a\left(-b\right)=-ab[/tex]

[tex]\tt-\frac{1}{32}\times \:1024[/tex]

[tex]\tt \cfrac{1}{32}\times \cfrac{1024}{1}[/tex]

Apply fraction rule:-

[tex]\tt -\cfrac{1\times \:1024}{32\times \:1}[/tex]

[tex]\tt \cfrac{1024}{32}[/tex]

Divide numbers:-

[tex]\tt -32[/tex]

Now that we're done factoring the denominator let's bring down the numerator which is ' 1 ':-

[tex]\tt -\cfrac{1}{32}\;\; \bf Answer[/tex]

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Loneguygo

[tex]\dfrac 1{2^{-2}\cdot x^{-3}\cdot y^5}\\\\=\dfrac 1{2^{-2} \cdot2^{-3} \cdot (-4)^5}\\\\=-\dfrac 1{2^{-2} \cdot 2^{-3} \cdot (2^2)^5}\\\\=-\dfrac{1}{2^{-2} \cdot 2^{-3} \cdot 2^{10}}\\\\=-\dfrac{1}{2^{-2-3+10}}\\\\=-\dfrac 1{2^{5}}\\\\=-\dfrac 1{32}[/tex]

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