20. Jeffrey's company has been operating for 83 months. In the beginning, Jeffrey was seeing incredible profit growth every month. As more companies started to adjust their services to compete with Jeffrey's business, Jeffrey's profits leveled out and have started to fall off. Jeffrey's business life cycle can be modeled by the equation $p=1.96 m-0.02 m^{2}$ where $p$ is the profit of his company in millions, and $m$ is the number of months since his business started. Graph this situation on your calculator or on a sheet of paper. What is the reasonable domain during this company's profitability? $D=0 \leq m \leq 83$ $D=0 \leq p \leq 111$ $D=0 \leq p \leq 92$ $D=0 \leq m \leq 98$

Step 1 ：First, we need to understand the meaning of the problem. The problem is asking for the reasonable domain of the company's profitability. This means we need to find the range of months 'm' during which the company is profitable.

Step 2 ：Next, we need to find the roots of the equation $p=1.96 m-0.02 m^{2}$, which represent the points where the company breaks even (making no profit).

Step 3 ：We can rewrite the equation as $0.02 m^{2}-1.96 m+p=0$ and solve for 'm' using the quadratic formula $m=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a=0.02$, $b=-1.96$, and $c=p$.

Step 4 ：Substituting the values into the formula, we get $m=\frac{1.96\pm\sqrt{(-1.96)^{2}-4*0.02*p}}{2*0.02}$.

Step 5 ：Since the company has been operating for 83 months, the reasonable domain for 'm' is from 0 to 83. Therefore, the reasonable domain during this company's profitability is $D=0 \leq m \leq 83$.

Step 6 ：Finally, we need to check if our result meets the requirements of the problem. The problem asks for the reasonable domain during the company's profitability, and our result provides the range of months during which the company is profitable. Therefore, our result meets the requirements of the problem.