MCQ
$\Big(\frac{-1}{2}\Big)^5\times\Big(\frac{-1}{2}\Big)^3$ is equal to:
  • $\Big(\frac{-1}{2}\Big)^{8}$
  • B
    $-\Big(\frac{1}{2}\Big)^8$
  • C
    $\Big(\frac{1}{4}\Big)^8$
  • D
    $\Big(-\frac{1}{2}\Big)^{15}$

Answer

Correct option: A.
$\Big(\frac{-1}{2}\Big)^{8}$
We have:
$\Big(\frac{-1}{2}\Big)^5\times\Big(\frac{-1}{2}\Big)^3$
$=\Big(\frac{-1}{2}\Big)^{5+3}$
$=\Big(\frac{-1}{2}\Big)^8$

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