Question

HALLAR: (1/5)^0 (−0,6)^−1 +2/5 x 4/9 (-1/2)- 2^0

Answer 1

Pasos:
$\left(\frac{1}{5}\right)^0\left(-0.6\right)^{-1}+\frac{2}{5}\cdot \frac{4}{9}\left(-\frac{1}{2}\right)-2^0$
$1\cdot \left(-0.6\right)^{-1}+\frac{2}{5}\cdot \frac{4}{9}\left(-\frac{1}{2}\right)-1$
$=-\frac{1}{0.6}-\frac{2}{5}\cdot \frac{4}{9}\cdot \frac{1}{2}-1$
$=-\frac{1}{0.6}-\frac{4}{45}-1$
$=-\frac{1}{1}-\frac{1}{0.6}-\frac{4}{45}$
$=-\frac{27}{27}-\frac{45}{27}-\frac{4\cdot \:0.6}{27}$
$=\frac{-27-45-4\cdot \:0.6}{27}$
$=\frac{-74.4}{27}$
$=-\frac{74.4}{27}$ --- Resultado.

Saludos!!!