Question

A community swimming pool starts with 250 gallons of water. Julian starts filling the pool at a rate of 500 gallons per hour.
Next, record the initial number of gallons of water in the pool.
t
g
0
1
2
3
4
5
6
7
8
9
10
0
250
500
750
1,000
1,250
1,500
1,750
2,000
2,250
2,500
Time in hours
Gallons of water in the pool
Rate of change: 500
Initial amount: 250

Gallons of water in the pool = Initial number of gallons of water + Change in gallons
of water
g =
+
⋅
Change in gallons per hour Number of hours

Answer 1

To express the total number of gallons of water in the pool as a function of time \( t \) (in hours), we can use the formula you provided.

### Given:
- **Initial number of gallons in the pool**: 250 gallons
- **Rate of change (filling rate)**: 500 gallons per hour
- **Time in hours**: \( t \)

### Formula:
The formula to find the number of gallons \( g \) in the pool after \( t \) hours is:

\[
g = \text{Initial gallons} + \text{(Rate of change)} \times t
\]

### Substituting the values:
- Initial gallons = 250
- Rate of change = 500
- Time = \( t \)

### Therefore, the equation becomes:
\[
g = 250 + 500t
\]

### Values for \( t \) from 0 to 10:
Now, let's calculate the value of \( g \) for each \( t \) from 0 to 10:

- When \( t = 0 \):
\[
g = 250 + 500 \times 0 = 250
\]

- When \( t = 1 \):
\[
g = 250 + 500 \times 1 = 750
\]

- When \( t = 2 \):
\[
g = 250 + 500 \times 2 = 1,250
\]

- When \( t = 3 \):
\[
g = 250 + 500 \times 3 = 1,750
\]

- When \( t = 4 \):
\[
g = 250 + 500 \times 4 = 2,250
\]

- When \( t = 5 \):
\[
g = 250 + 500 \times 5 = 2,750
\]

- When \( t = 6 \):
\[
g = 250 + 500 \times 6 = 3,250
\]

- When \( t = 7 \):
\[
g = 250 + 500 \times 7 = 3,750
\]

- When \( t = 8 \):
\[
g = 250 + 500 \times 8 = 4,250
\]

- When \( t = 9 \):
\[
g = 250 + 500 \times 9 = 4,750
\]

- When \( t = 10 \):
\[
g = 250 + 500 \times 10 = 5,250
\]

### Summary:
Here is a summary of the values in a table format:

| Time (hours) | Gallons of Water |
|--------------|------------------|
| 0 | 250 |
| 1 | 750 |
| 2 | 1,250 |
| 3 | 1,750 |
| 4 | 2,250 |
| 5 | 2,750 |
| 6 | 3,250 |
| 7 | 3,750 |
| 8 | 4,250 |
| 9 | 4,750 |
| 10 | 5,250 |

Thus, the gallons of water in the pool can be calculated using the equation \( g = 250 + 500t \) for values of \( t \) from 0 to 10 hours.