Answer
$11\text{mph}$.
Work Step by Step
Translate- according to the table there are two equations.
Using the formula $t=\frac{d}{r}$, we find:
$t=\frac{24}{r-4}$
$5-t=\frac{24}{r+4}$.
Carry out- substitute the value$\frac{24}{r-4}$ for $\left( t \right)$ in the second equation$5-t=\frac{24}{r+4}$.
Solving for$\left( r \right)$,
$\begin{align}
& 5-t=\frac{24}{r+4} \\
& 5-\frac{24}{r-4}=\frac{24}{r+4}
\end{align}$
Multiply by $\left( r+4 \right)\left( r-4 \right)$ on both sides of the equation,
$\begin{align}
& 5-\frac{24}{r-4}=\frac{24}{r+4} \\
& \left( r+4 \right)\left( r-4 \right)\left( 5-\frac{24}{r-4} \right)=\frac{24}{r+4}\left( r+4 \right)\left( r-4 \right) \\
& 5\left( r-4 \right)\left( r+4 \right)-24\left( r+4 \right)=24\left( r-4 \right) \\
& 5{{r}^{2}}-80-24r-96=24r-96
\end{align}$
Add $96$ on both sides,
$\begin{align}
& 5{{r}^{2}}-80-24r-96=24r-96 \\
& 5{{r}^{2}}-80-24r-96+96=24r-96+96 \\
& 5{{r}^{2}}-80-24r=24r
\end{align}$
Subtract $24r$ on both sides,
$\begin{align}
& 5{{r}^{2}}-80-24r=24r \\
& 5{{r}^{2}}-80-24r-24r=24r-24r \\
& 5{{r}^{2}}-80-48r=0
\end{align}$
Rearrange the equation,
$5{{r}^{2}}-48r-80=0$
Using the quadratic formula,
Substitute the values $b=-48,\,c=-80,\,a=5$,
$\begin{align}
& r=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& =\frac{-\left( -48 \right)\pm \sqrt{{{\left( -48 \right)}^{2}}-4\left( 5 \right)\left( -80 \right)}}{2\left( 5 \right)} \\
& =\frac{4\left( 6\pm \sqrt{61} \right)}{5}
\end{align}$
Thus, the value of $r$ is $11,\left( -1.5 \right)$.
$\begin{align}
& t=\frac{24}{r-4} \\
& =\frac{24}{11-4} \\
& =\frac{24}{7} \\
& =3.4
\end{align}$
Now, solve for the speed down river,
If $r=11$ then the speed down river is,
$\begin{align}
& r=11+4 \\
& =15
\end{align}$
Therefore, the speed down river is $15\text{mph}$.
Thus, the speed of the barge in still water is about $11\text{mph}$.
