Question Number 189563 by Rupesh123 last updated on 18/Mar/23
Answered by FelipeLz last updated on 19/Mar/23
$${V}\:=\:\left({r}_{\mathrm{1}} +\mathrm{2}\right)×\left({r}_{\mathrm{2}} +\mathrm{2}\right)×\left({r}_{\mathrm{3}} +\mathrm{2}\right)\:=\:{r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} +\mathrm{2}\left({r}_{\mathrm{1}} {r}_{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{3}} +{r}_{\mathrm{2}} {r}_{\mathrm{3}} \right)+\mathrm{4}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)+\mathrm{8}\: \\ $$$${r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} \:=\:\frac{\mathrm{6}}{\mathrm{10}} \\ $$$${r}_{\mathrm{1}} {r}_{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{3}} +{r}_{\mathrm{2}} {r}_{\mathrm{3}} \:=\:\frac{\mathrm{29}}{\mathrm{10}} \\ $$$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} \:=\:\frac{\mathrm{39}}{\mathrm{10}} \\ $$$${V}\:=\:\frac{\mathrm{6}}{\mathrm{10}}+\mathrm{2}×\frac{\mathrm{29}}{\mathrm{10}}+\mathrm{4}×\frac{\mathrm{39}}{\mathrm{10}}+\mathrm{8} \\ $$$${V}\:=\:\frac{\mathrm{220}}{\mathrm{10}}+\mathrm{8}\:=\:\mathrm{30}\: \\ $$
Commented by Rupesh123 last updated on 19/Mar/23
Excellent!
Answered by manxsol last updated on 19/Mar/23
$${r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} =\frac{\mathrm{6}}{\mathrm{10}}={V}_{{i}} \\ $$$$\left(\mathrm{2}{r}_{\mathrm{1}} \right)\left(\mathrm{2}{r}_{\mathrm{2}} \right)\left(\mathrm{2}{r}_{\mathrm{3}} \right)=\frac{\mathrm{48}}{\mathrm{10}}=\frac{\mathrm{24}}{\mathrm{5}}={V}_{{f}} \\ $$