Question

Potenciação

Sendo $y= \frac{4^{10} . 8^{-3} . 16^{-2} }{32}$
a metade do valor de y vale ?

a)[tex2^{-3} [/tex]
b)$2^{-4}$
c)$2^{-5}$
d)$2^{-6}$

Answer 1

$y= \frac{4^{10}.8^{-3}.16^{-2}}{32}= \frac{4^{10}}{32.8^{3}.16^{2}}=0,25 \\ \frac{y}{2}= \frac{0,25}{2}=0,125$