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We are given a piecewise function \( h(x) \) defined as follows:

\[
h(x) =
\begin{cases}
-2x - 16 & \text{for } x < -4 \\
5 & \text{for } -4 \leq x < 5 \\
x + 7 & \text{for } x \geq 5
\end{cases}
\]

We need to evaluate the function at the given points: \( h(-5) \), \( h(0) \), \( h(5) \), and \( h(9) \).

### Step-by-step evaluation:

1. **For \( h(-5) \):**
   Since \( -5 < -4 \), we use the first part of the piecewise function: \( h(x) = -2x - 16 \).
   \[
   h(-5) = -2(-5) - 16 = 10 - 16 = -6
   \]

2. **For \( h(0) \):**
   Since \( -4 \leq 0 < 5 \), we use the second part of the piecewise function: \( h(x) = 5 \).
   \[
   h(0) = 5
   \]

3. **For \( h(5) \):**
   Since \( 5 \geq 5 \), we use the third part of the piecewise function: \( h(x) = x + 7 \).
   \[
   h(5) = 5 + 7 = 12
   \]

4. **For \( h(9) \):**
   Since \( 9 \geq 5 \), we use the third part of the piecewise function: \( h(x) = x + 7 \).
   \[
   h(9) = 9 + 7 = 16
   \]

### Final answers:
- \( h(-5) = -6 \)
- \( h(0) = 5 \)
- \( h(5) = 12 \)
- \( h(9) = 16 \)

Let me know if you need more details or have any questions! Here are 5 related questions for further practice:

1. What is \( h(-4) \)?
2. What is the domain of the function \( h(x) \)?
3. Determine if the function is continuous at \( x = -4 \).
4. Evaluate \( h(-10) \).
5. Solve for \( x \) if \( h(x) = 0 \).

**Tip:** When working with piecewise functions, always carefully check which piece applies to the given input value based on the conditions provided.

For the piecewise function, find the values h(-5), h(0), h(5), and h(9):

h(x) =
  -2x - 16 for x < -4
  5 for -4 ≤ x < 5
  x + 7 for x ≥ 5

Learn how to evaluate the given piecewise function at specific values: h(-5), h(0), h(5), and h(9). This problem demonstrates key algebraic concepts and steps for evaluating a function piece by piece based on input conditions.

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