$Solve \: for \: r: \\ = > v= \pi {h}^{2}(r - \frac{h}{3} ) \\ \\ v=\pi {h}^{2}(r - \frac{h}{3} )is \: equivalent \: to \: {h}^{2}\pi(r - \frac{h}{3} ) = v: \\ = > {h}^{2}\pi(r - \frac{h}{3} ) = v \\ \\ Divide \: both \: sides \: by \: \pi {h}^{2} : \\ = > r - \frac{h}{3} = \frac{v}{\pi {h}^{2} } \\ \\ Add \: \frac{h}{3} \: to \: both \: sides: \\ = > r = \frac{h}{3} + \frac{v}{\pi {h}^{2} }$

7 0

2 years ago

What is <br><img src="https://tex.z-dn.net/?f=8.4%20-%204.2" id="TexFormula1" title="8.4 - 4.2" alt="8.4 - 4.2" align="absmiddle

STALIN [3.7K]

Answer:

4.2

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\space\space&8&.&4\\ -&4&.&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-2=2$

$\frac{\begin{matrix}\space\space&8&.&\textbf{4}\\ -&4&.&\textbf{2}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{2}\end{matrix}}$

$\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}$

$\frac{\begin{matrix}\space\space&8&\textbf{.}&4\\ -&4&\textbf{.}&2\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{.}&2\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:8-4=4$

$\frac{\begin{matrix}\space\space&\textbf{8}&.&4\\ -&\textbf{4}&.&2\end{matrix}}{\begin{matrix}\space\space&\textbf{4}&.&2\end{matrix}}$

$=4.2$

3 0

2 years ago