Answer :

[tex] \bf \textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}~~
\begin{cases}
r=radius\\[-0.5em]
\hrulefill\\
r=3\cdot 10^2
\end{cases}\implies V=\cfrac{4\pi (3\cdot 10^2)^3}{3}
\implies
V=\cfrac{4\pi (3^3\cdot 10^{2\cdot 3})}{3}
\\\\\\
V=4\pi (3^3\cdot 10^6)3^{-1}\implies V=4\pi 3^33^{-1}\cdot 10^6\implies V=4\pi 3^{3-1}10^6
\\\\\\
V=4\cdot \pi \cdot 3^2\cdot 10^6 [/tex]

Other Questions