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To solve the problem given in the image, we need to find a piecewise-defined function that represents the volume of water in the pool as a function of the height \( h \) of the water above the deepest point (9 feet).

### (a) Expressing the Volume of Water as a Function of Height \( h \):

The pool has three sections:
1. **Section 1 (Shallow End to Start of Slope)**: The height ranges from 0 to 3 feet.
2. **Section 2 (Sloped Middle Part)**: The height ranges from 3 feet to 9 feet.
3. **Section 3 (Deep End)**: The height ranges from 9 feet onwards.

The width of the pool is constant at 20 feet, and the total length is 40 feet, split into sections of 6 feet, 12 feet, 16 feet, and 6 feet respectively.

**Section 1 (0 ≤ h ≤ 3):**
- In this section, the pool has a constant depth of \( h \).
- Volume, \( V \), can be calculated as:
  \[
  V = \text{Width} \times \text{Length} \times \text{Height} = 20 \times 12 \times h = 240h \, \text{cubic feet}.
  \]

**Section 2 (3 < h ≤ 9):**
- This section is sloped, and the volume must be calculated by integrating the cross-sectional area along the sloped part.
- The slope starts at 3 feet deep and increases linearly to 9 feet deep over a 16-foot length.
- The cross-sectional shape is a trapezoid where the depth varies linearly with \( h \). 
- For a given height \( h \) in this range, the additional length from the shallow end can be calculated using the proportion:
  \[
  x = 16 \times \frac{h - 3}{9 - 3} = \frac{8}{3} (h - 3).
  \]
- The volume in this section from 3 to \( h \) feet:
  \[
  V = 20 \times \left(\frac{3 + h}{2}\right) \times \left(12 + \frac{8}{3} (h - 3)\right).
  \]

**Section 3 (h > 9):**
- The depth here is constant at 9 feet.
- Volume of this section when \( h = 9 \):
  \[
  V = 20 \times 12 \times 9 + 20 \times \left(\frac{3 + 9}{2}\right) \times 16 = 240 \times 9 + 20 \times 6 \times 16.
  \]
- For any height \( h \) greater than 9 feet, add the extra volume of water:
  \[
  V = \text{Base Volume} + 20 \times 12 \times (h - 9).
  \]

### (b) Domain and Range of the Function:

**Domain:** The height \( h \) of the water can range from 0 feet (empty pool) to the maximum water height which is the pool's depth, which is 9 feet. Hence, the domain is:
\[
0 \leq h \leq 9.
\]

**Range:** The range is the corresponding volume values from 0 to the maximum volume. Calculating the maximum volume at \( h = 9 \):
- Total volume at maximum depth (9 feet):
  \[
  V = 240 \times 9 + 20 \times \left(\frac{3 + 9}{2}\right) \times 16 = 2160 + 1920 = 4080 \, \text{cubic feet}.
  \]
So, the range is:
\[
0 \leq V \leq 4080 \, \text{cubic feet}.
\]

Would you like further details or have any questions?

Here are some related questions:
1. How would you calculate the volume of a pool with a different slope profile?
2. What changes in volume calculation if the pool has an irregular shape?
3. How does the width of the pool affect the overall volume function?
4. How can the piecewise function be plotted graphically?
5. What if the deepest end is more than 9 feet? How would the function change?

**Tip:** Always identify the different sections of a structure and how each contributes to the total when working with piecewise functions!

A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 feet deep at its deepest point. A cross-section is shown in the figure.
(a) Express the volume of the water in the pool as a function of height h of the water above the deepest point. 
(b) Determine the domain and range of the function found in part (a).

Learn how to express the volume of water in a swimming pool as a function of height using piecewise functions. This problem involves calculating the volume across three sections: shallow, sloped, and deep. Suitable for students in Grades 11-12.

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