Question

ALGUIEN QUE SEPA 'PROPORCIONALIDAD NUMERICA' ME RESUELVE ESTA CUENTA?? :(

Answer 1

$\dfrac{ \sqrt[3]{- \frac{1}{27} }}{-0,\stackrel{\frown}{2}-0,\stackrel{\frown}{6}} = \dfrac{ \sqrt{1- \frac{3}{4} }}{y} \\ \\ \\ \\ ( \sqrt[3]{- \frac{1}{27} })*y= ( \sqrt{1- \frac{3}{4} })*(-0,\stackrel{\frown}{2}-0,\stackrel{\frown}{6}) \\ \\ \\ ( -\frac{1}{3} })*y= ( \sqrt{ \frac{1}{4} })*(- \frac{2}{9}- \frac{6}{9}) \\ \\ \\ ( -\frac{1}{3} })*y= ( \frac{1}{2} )*(- \frac{8}{9}) \\ \\ \\ y= - \frac{4}{9} : - \frac{1}{3} \to y= + \frac{12}{9}\to \boxed{ \boxed{y= \frac{4}{3} }}$


Espero que te sirva, salu2!!!!