Question

Evaluate frac{1}{2^{-2}x^{-3}y^{5} }" alt="\frac{1}{2^{-2}x^{-3}y^{5} }" align="absmiddle" class="latex-formula"> for x=2 and y=-4.
A)-16
b)-4
c)$-\frac{1}{32}$
d)16

Answer 1

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

Answer 2

Answer:

$\bf C)\: -\cfrac{1}{32}$

Step-by-step explanation:

$\tt \cfrac{1}{2^{-2}x^{-3}y^5}$

$\tt x=2\\y=-4$

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

$\tt \cfrac{1}{2^{-2}\times \:2^{-3}\left(-4\right)^5}$

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

$\tt 2^{-2}\times \:2^{-3}=\boxed{\cfrac{1}{32} }$

$\tt \cfrac{1}{32}\left(-4\right)^5$

Calculate exponents:- -4^5= -1024

$\tt \cfrac{1}{32}\left(-1024\right)$

Remove parentheses, apply rule:- $\tt a\left(-b\right)=-ab$

$\tt-\frac{1}{32}\times \:1024$

$\tt \cfrac{1}{32}\times \cfrac{1024}{1}$

Apply fraction rule:-

$\tt -\cfrac{1\times \:1024}{32\times \:1}$

$\tt \cfrac{1024}{32}$

Divide numbers:-

$\tt -32$

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

$\tt -\cfrac{1}{32}\;\; \bf Answer$

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