Question

ayudaa pliis doy corona y lo sigo pero necesito ayuda ​

Answer 1

Respuesta:

$\boxed{pregunta \: a } \\ \\ {2}^{ - 3} + 0.4 - \sqrt{ \frac{9}{9} } \\ \\ \frac{1}{ {2}^{3} } + \frac{4}{10} - \sqrt{1} \\ \\ \frac{1}{8} + \frac{2}{5} - 1 \\ \\ \frac{5 \times 1 + 8 \times 2}{40} - 1 \\ \\ \frac{21}{40} - 1 \\ \\ \frac{21 - 40}{40} \\ \\ - \frac{19}{40}$

$\boxed{pregunta \: b} \\ \\ ( \sqrt{ (\frac{71 - 7}{9} } ))^{ - 1} + \frac{625}{100} \\ \\ \sqrt{ \frac{64}{9} } ^{ - 1} + \frac{25}{4} \\ \\ ( \frac{8}{3} )^{ - 1} + \frac{25}{4} \\ \\ \frac{3}{8} + \frac{25}{4} \\ \\ \frac{3}{8} + \frac{50}{8} \\ \\ \frac{53}{8}$

$\boxed{pregunta \: c} \\ \\ \sqrt{ \frac{27 - 2}{9} } + ( \frac{2}{3} )^{2} - \frac{1}{3} - (\frac{4}{3}) ^{2} \\ \\ \sqrt{ \frac{25}{9} } + \frac{4}{9} - \frac{1}{3} - \frac{16}{9} \\ \\ \frac{5}{3} - \frac{1}{3} + \frac{4}{9} - \frac{16}{9} \\ \\ \frac{4}{3} - \frac{12}{9} \\ \\ \frac{4}{3} - \frac{4}{3} \\ \\ 0$

$\boxed{pregunta \: d} \\ \\ {4}^{ - 1} - 0.125 + 1 \frac{1}{4} - 1.125 \\ \\ \frac{1}{4} - 0.125 + 1 + \frac{1}{4} - 1.125 \\ \\ \frac{1}{4} + \frac{1}{4} + 1 - 0.125 - 1.125 \\ \\ \frac{2}{4} - 0.25 \\ \\ \frac{2}{4} - \frac{25}{100} \\ \\ \frac{2}{4} - \frac{1}{4} \\ \\ \frac{1}{4}$