Use the two points to find the slope: $\text{slope} = \dfrac{\text{change in volume}}{\text{change in time}}$. Be careful with the order of subtraction and the sign, since the tank is draining.

Once you know the slope and the starting amount, use the form $V = mt + b$, where $b$ is the volume at $t = 0$. Plug in what you found for $m$ and $b$.

In Desmos, type each equation using $x$ instead of $t$, for example: y = 500 - (75/2)x, y = 500 - (3/2)x, y = 500 - (2/5)x, and y = 500 - (5/2)x. All four lines will appear on the same graph.

For each equation, create a table (click the gear icon next to the expression and choose "Table") and include $x = 30$. Look at the corresponding $y$-value for each line; the correct equation is the one that gives $y = 425$ when $x = 30$.

Use time $t$ as the input (x-value) and volume $V$ as the output (y-value).

• At the start: $t = 0$, $V = 500$ gives the point $(0, 500)$.
• After 30 minutes: $t = 30$, $V = 425$ gives the point $(30, 425)$.

These two points lie on the line that models $V$ as a function of $t$.

The tank is draining at a constant rate, so the graph is a line, and the slope is

This means the volume decreases by $\dfrac{5}{2}$ gallons each minute (negative sign shows it is decreasing).

A linear model with time $t$ and volume $V$ has the form $V = mt + b$, where $m$ is the slope and $b$ is the starting value (the y-intercept).

• The starting volume is 500 gallons, so $b = 500$.
• The slope is $m = -\dfrac{5}{2}$.

Substitute these into the linear form:

This matches answer choice D.