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(1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)
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03 Jun 2020, 05:39
03 Jun 2020, 05:39
Expert Reply
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Question Stats:
83% (00:51) correct
16% (00:26) wrong based on 43 sessions
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\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)
(A) \((\frac{1}{2})^{(-48)}\)
(B) \((\frac{1}{2})^{(-11)}\)
(C) \((\frac{1}{2})^{(-6)}\)
(D) \((\frac{1}{8})^{(-11)}\)
(E) \((\frac{1}{8})^{(-6)}\)
(A) \((\frac{1}{2})^{(-48)}\)
(B) \((\frac{1}{2})^{(-11)}\)
(C) \((\frac{1}{2})^{(-6)}\)
(D) \((\frac{1}{8})^{(-11)}\)
(E) \((\frac{1}{8})^{(-6)}\)
Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)
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03 Jun 2020, 05:39
03 Jun 2020, 05:39
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)
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03 Jun 2020, 05:50
03 Jun 2020, 05:50
1
Carcass wrote:
\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)
(A) \((\frac{1}{2})^{(-48)}\)
(B) \((\frac{1}{2})^{(-11)}\)
(C) \((\frac{1}{2})^{(-6)}\)
(D) \((\frac{1}{8})^{(-11)}\)
(E) \((\frac{1}{8})^{(-6)}\)
(A) \((\frac{1}{2})^{(-48)}\)
(B) \((\frac{1}{2})^{(-11)}\)
(C) \((\frac{1}{2})^{(-6)}\)
(D) \((\frac{1}{8})^{(-11)}\)
(E) \((\frac{1}{8})^{(-6)}\)
Nice rule: \((\frac{a}{b})^{-n} = (\frac{b}{a})^{n}\)
\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1} = (\frac{2}{1})^{3}(\frac{4}{1})^{2}(\frac{16}{1})^{1}\)
\(= (2^{3})(4^{2})(16^{1})\)
\(= (2^{3})(2^{4})(2^{4})\)
\(= 2^{11}\)
\(= (\frac{2}{1})^{11}\)
\(= (\frac{1}{2})^{-11}\)
Answer: B
