Question

ayuda por favor es urgente es para hoy ​

Answer 1

Respuesta:

$\frac{ {k}^{3} - \frac{1}{19} }{ {k}^{3} + \frac{1}{19} } = \frac{s}{c} \\ \frac{ {k}^{3} - \frac{1}{19} }{ {k}^{3} + \frac{1}{19} } = \frac{9k}{10k} \\ 10 {k}^{4} - \frac{10k}{19} = 9 {k}^{4} + \frac{9k}{19} \\ 1 {k}^{4} = \frac{19k}{19} \\ {k}^{3} = 1 \\ k = \sqrt[3]{1} \\ k = 1$

Reemplazar:

$s = {k}^{3} - \frac{1}{19} \\ s = {1}^{3} - \frac{1}{19} \\ s = \frac{18}{19}$

Hallar R

$r = \frac{s\pi}{180} \\ r =( \frac{18}{19} ) \times \frac{\pi}{180} \\ r = \frac{\pi}{190}$

El ángulo en radianes vale π/190