Question

Sendo y = 4^(10) . 8^(-3) . 16 ^(-2)/32 a matar do y vale

Answer 1

$y = 4^1^0 . 8^-^3 . 16^-^2/32$
$y = (2^2)^1^0 . (2^3)^-^3 . (2^4)^-^2/32$
$y = 2^2^0 . 2^-^9 . 2^-^8/2^5$
$y = 2^-^2$
$y = 1/4$

Resposta final: y=1/4