Question

Sendo y , a metade do valor de y vale....​

Answer 1

Explicação passo-a-passo:

Veja que:

• $\sf 4^{10}=(2^2)^{10}=2^{2\cdot10}=2^{20}$

• $\sf 8^{-3}=(2^3)^{-3}=2^{3\cdot(-3)}=2^{-9}$

• $\sf 16^{-2}=(2^4)^{-2}=2^{4\cdot(-2)}=2^{-8}$

• $\sf 32=2^5$

Assim:

$\sf y=\dfrac{4^{10}\cdot8^{-3}\cdot16^{-2}}{32}$

$\sf y=\dfrac{2^{20}\cdot2^{-9}\cdot2^{-8}}{2^5}$

$\sf y=\dfrac{2^{20-9-8}}{2^5}$

$\sf y=\dfrac{2^3}{2^5}$

$\sf y=2^{3-5}$

$\sf \red{y=2^{-2}}$

A metade de y vale:

$\sf \dfrac{y}{2}=\dfrac{2^{-2}}{2}$

$\sf \dfrac{y}{2}=2^{-2-1}$

$\sf \red{\dfrac{y}{2}=2^{-3}}$

Letra A