Use the two points to compute the slope $m$ with the formula $m=\dfrac{C_2-C_1}{t_2-t_1}$. Be careful with the order of subtraction so the sign is correct.

Once you know the slope, write $C = mt + b$ and use one of the points to find $b$. Then set $C=20$ and solve the resulting equation for $t$.

In Desmos, type (28-52)/(11-3) on a line and press Enter. The output is the slope (rate of change) of the concentration with respect to time.

Using the slope from step 1, write the equation in the form y = mx + b. You can find b by solving with one of the given points (for example, $t=3$, $C=52$). Then type that equation into Desmos, such as y = -3x + 61, to graph the concentration as a function of time.

On a new line, type y = 20 to graph the horizontal line for a concentration of 20 mg/L. Use the intersection tool (or tap on the point where the two graphs meet) and read off the $x$-coordinate of the intersection; that $x$-value is the time $t$ when the concentration reaches 20 mg/L.

Because the concentration is modeled by a linear function of time, we can think of it as a line with:

• Time $t$ (in hours) on the horizontal axis
• Concentration $C$ (in mg/L) on the vertical axis

The problem gives two points on this line:

• After 3 hours: $(t,C)=(3,52)$
• After 11 hours: $(t,C)=(11,28)$

Find the slope $m$ (rate of change) using

Compute it:

So the concentration is decreasing at $3$ mg/L per hour.

Use the slope-intercept form $C = mt + b$ with $m=-3$.

So we have

Use one of the given points to find $b$. Using $(3,52)$:

So the equation relating concentration and time is

We want to know when the concentration reaches $20$ mg/L, so set $C=20$ in the equation:

Now solve for $t$:

So, according to the model, the concentration reaches $20$ mg/L after $\dfrac{41}{3}$ hours.