Respuesta :

abscisa15

[tex]\it \mathcal{V} =\dfrac{\mathcal{A}_b\cdot h}{3} \Rightarrow h = \dfrac{3\mathcal{V}}{\mathcal{A}_b}= \dfrac{3\cdot3072}{\mathcal{A}_b}\ \ \ \ \ (*) \\\;\\ \\\;\\ \mathcal{A}_b = \dfrac{\ell^2\sqrt3}{4} = \dfrac{16^2\sqrt3}{4} =\dfrac{16\cdot16\sqrt3}{4} =4\cdot16\sqrt3=64\sqrt3\ \ \ \ (**) \\\;\\ \\\;\\ (*),\ (**) \Rightarrow h=\dfrac{3\cdot3072}{64\sqrt3} = \dfrac{3\cdot48}{\sqrt3} = \dfrac{\sqrt3\cdot\sqrt3\cdot48}{\sqrt3} = 48\sqrt3\ inches[/tex]


Otras preguntas