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21 votes
21 votes
Evaluate
(1)/(2^(-2)x^(-3)y^(5) ) for x=2 and y=-4.

A)-16
b)-4
c)
-(1)/(32)
d)16

User DCookie
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2 Answers

18 votes
18 votes


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

User Hanzallah Afgan
by
2.8k points
9 votes
9 votes

Answer:


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

Explanation:


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


\tt x=2\\y=-4

~

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


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

First let's solve
\tt 2^(-2)* \:\:2^(-3)\left(-4\right)^5


\tt 2^(-2)* \: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-(1)/(32)* \:1024


\tt \cfrac{1}{32}* \cfrac{1024}{1}

Apply fraction rule:-


\tt -\cfrac{1* \:1024}{32* \: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

☆-------☆-------☆-------☆

User Mgttlinger
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