Step 1
We are given with volume of water remaining in the tank with time t.

Step 2: (a)
The equation of slope of any secant line joining points P($x_1,y_1$) and Q($x_2,y_2$) on the curve is given by:\[{S_{PQ}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]We are given that the point P is given by:
P=(15,250)

Q=(5,694)

\[\begin{array}{l}{S_1} = \dfrac{{694 - 250}}{{5 - 15}}\\{S_1} = \dfrac{{444}}{{ - 10}}\\{S_1} = - 44.4\end{array}\]Q=(10,444)

\[\begin{array}{l}{S_2} = \dfrac{{444 - 250}}{{10 - 15}}\\{S_2} = \dfrac{{194}}{{ - 5}}\\{S_2} = - 38.8\end{array}\]Q=(20,111)

\[\begin{array}{l}{S_3} = \dfrac{{111 - 250}}{{20 - 15}}\\{S_3} = \dfrac{{-139}}{{ 5}}\\{S_3} = - 27.8\end{array}\]Q=(25,28)

\[\begin{array}{l}{S_4} = \dfrac{{28 - 250}}{{25 - 15}}\\{S_4} = \dfrac{{-222}}{{ 10}}\\{S_4} = - 22.2\end{array}\]Q=(30,0)

\[\begin{array}{l}{S_5} = \dfrac{{0 - 250}}{{30 - 15}}\\{S_5} = \dfrac{{-250}}{{ 14}}\\{S_5} = - 16.67\end{array}\]

Step 3: (b)
To find the slope of tangent line at P, we will find the average of slopes of two secant lines. Take nearby points having $t=10$ and $t=20$ to find slopes of two secant lines.

The slopes corresponding to $t=10$ is $S_2$ and that corresponding to $t=20$ is $S_3$

Average Slope\[\begin{array}{l}{S_P} = \dfrac{{{S_2} + {S_3}}}{2}\\ = \dfrac{{ - 38.8 - 27.8}}{2}\\ = - 33.3\end{array}\]Therefore,
\[{S_P} = - 33.3\]

Step 4: (c)
Draw the graph of the functionhttps://imgur.com/7MUfqg4By drawing a tangent at point P, we find that: \[\begin{array}{l}{S_P} = - \dfrac{{300}}{9}\\ = - 33.333\end{array}\]Therefore,
\[{S_P} = - 33.333\]