For the half-life, the mass $m(t)$ should be half of the initial mass. What number should you set $m(t)$ equal to?

Set $120e^{-0.025t}$ equal to that half-mass value and divide both sides by 120 so the equation has the form $e^{-0.025t}=$ (some number).

Once you have an equation like $e^{-0.025t}=$ (some number), use the natural logarithm (ln) on both sides to solve for $t$.

In Desmos, enter the function as $y=120*e^{-0.025x}$ on one line, and enter $y=60$ on another line (60 is half of the initial 120 grams).

Zoom or pan until you see where the curve $y=120*e^{-0.025x}$ meets the horizontal line $y=60$, then click on the intersection point; the x-coordinate of this point is the half-life in days.

Half-life is the time it takes for a substance to decay to half of its initial amount.

From the function $m(t)=120e^{-0.025t}$, the initial mass is when $t=0$, which gives $m(0)=120$ grams.

So at the half-life, the mass should be $120/2=60$ grams.

Set the model equal to half the initial mass:

Divide both sides by 120 to isolate the exponential term:

When an exponential equation has the form $e^{\text{(expression)}}=\text{number}$, take the natural logarithm (ln) of both sides.

Taking ln of both sides of $e^{-0.025t}=1/2$ gives:

On a calculator, evaluate $\ln(0.5)$ to get a negative number.

Now solve for $t$ by dividing both sides by $-0.025$:

Using a calculator, $\ln(0.5)\approx-0.693$, so

This means the half-life is about 28 days, which matches answer choice D.