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\(\Big(\dfrac{1}{2}\Big)^{-3}\Big(\dfrac{1}{4}\Big)^{-2}\Big(\dfrac{1}{16}\Big)^{-1}=\)

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by VJesus12 » Sat Sep 25, 2021 2:49 am

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A

B

C

D

E

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\(\Big(\dfrac{1}{2}\Big)^{−3}\Big(\dfrac{1}{4}\Big)^{−2}\Big(\dfrac{1}{16}\Big)^{−1}=\)


\((A) \quad\Big(\dfrac{1}{2}\Big)^{-48}\)

\((B) \quad\Big(\dfrac{1}{2}\Big)^{-11}\)

\((C) \quad\Big(\dfrac{1}{2}\Big)^{-6}\)

\((D) \quad\Big(\dfrac{1}{8}\Big)^{-11}\)

\((E) \quad\Big(\dfrac{1}{8}\Big)^{-6}\)

Answer: B

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