Answer:
![13,571.68\text{ ft}^3\approx 384307.18\text{ liters}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/uk1ixwj586urhjj44bm2d52gy9hwfdhgs1.png)
Explanation:
We are asked to find the amount of water that the well can hold.
We will use volume of cylinder formula to solve our given problem.
, where,
V = Volume of cylinder,
r = Radius,
h = Height.
![d=2r](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3wdsyybu0qcb8ivwd5ne98555h8ik3l11d.png)
![24=2r](https://img.qammunity.org/2020/formulas/mathematics/middle-school/inw217rpxsqycwyon18b9640kihukmku51.png)
![12=r](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pgyspk9l8zi4087sro0jac9uosgzdd2xk8.png)
![V=\pi (12\text{ ft})^2* 30\text{ ft}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ln3vw010iqppxm3hwyihrpdg1b1ugjhpvu.png)
![V=\pi*144 \text{ ft}^2* 30\text{ ft}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/v1r4iujcv5nr4b2a65b6ot8s729oza5mhy.png)
![V=4,320\pi \text{ ft}^3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qf46s73msyjo3o1d89tcxcdwyja49kif3k.png)
![V=13,571.68026\text{ ft}^3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/a4ddhekhk1rm4x123s3mxixx50hgxp6vmw.png)
![V\approx 13,571.68\text{ ft}^3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/fo5xv4xietu7eptdg62445spmf4qjos7zq.png)
Since our given volume is in cubic feet, so we can convert it in liters by multiplying 13,571.68 by 28.3168.
![\text{Water in well}=13,571.68\text{ ft}^3*\frac{28.3168\text{ liter}}{ \text{ ft}^3}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tp942fp7z6eahmt4o16x9cttm43chy1v0f.png)
![\text{Water in well}=384,307.180556\text{ liter}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ybtru1jh5xlkwhoshnvwp9h1qajog8xw43.png)
Therefore, the well could hold approximately 384307.18 liters of water.