Respuesta :
[tex]\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\[/tex]
[tex]\bf \cfrac{small}{large}\qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\implies \cfrac{2}{7}=\cfrac{\sqrt[3]{32}}{\sqrt[3]{v}}\implies \cfrac{2}{7}=\sqrt[3]{\cfrac{32}{v}}\implies \left( \cfrac{2}{7} \right)^3=\cfrac{32}{v} \\\\\\ \cfrac{2^3}{7^3}=\cfrac{32}{v}\implies v=\cfrac{7^3\cdot 32}{2^3}[/tex]
[tex]\bf \cfrac{small}{large}\qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\implies \cfrac{2}{7}=\cfrac{\sqrt[3]{32}}{\sqrt[3]{v}}\implies \cfrac{2}{7}=\sqrt[3]{\cfrac{32}{v}}\implies \left( \cfrac{2}{7} \right)^3=\cfrac{32}{v} \\\\\\ \cfrac{2^3}{7^3}=\cfrac{32}{v}\implies v=\cfrac{7^3\cdot 32}{2^3}[/tex]
Answer:
it is not u3 becuase 32- the sides of the two cubes is -u24
Step-by-step explanation:
