Given:
![\begin{gathered} consant\text{ rate = }2.5\text{ gallons/min} \\ bucket\text{ size = }18\text{ gallon} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/o48hhu076aqmzta2v7defoql0qlmzig30x.png)
Let the time at which the bucket is filled be x
We can write:
At time t = 0 , Volume (V) = 0
At time t = x, Volume (V) = 18
Using the slope formula:
![\begin{gathered} \text{slope = }(V_(t=x)-V_(t=0))/(x-0) \\ (18-0)/(x-0)=\text{ 2.5} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5parpynkhts75r9o9uhsnp3wrc7ip6zjj3.png)
Solving for x:
![\begin{gathered} 2.5x\text{ = 18} \\ x\text{ = 7.2 min} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jx61vk6dg9lnd8jm4rgz8q94ing5yjqzsu.png)
Recall that the domain of the function V is the time interval that satisfies the function.
Since the bucket becomes filled at 7.2min, the domain of the function V is:
![0\text{ }\leq\text{ t }\leq\text{ 7.2 }](https://img.qammunity.org/2023/formulas/mathematics/college/qvqmu2ohn7ua615gq8p8i6qgscnso5k811.png)
Answer: Option A