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We are given the function \( h(k) = \sqrt{7 - k} + \sqrt{k + 3} \) and asked to find the intervals on which the function is continuous.

### Step 1: Analyze the domain of the function
For the function to be continuous, both square roots must be defined, which requires that the expressions inside the square roots be non-negative. Specifically:
1. \( 7 - k \geq 0 \) for the first square root.
2. \( k + 3 \geq 0 \) for the second square root.

### Step 2: Solve the inequalities
1. \( 7 - k \geq 0 \) implies \( k \leq 7 \).
2. \( k + 3 \geq 0 \) implies \( k \geq -3 \).

### Step 3: Determine the domain
The function is defined for \( k \) values that satisfy both inequalities. Combining them, we get:
\[
-3 \leq k \leq 7
\]

### Step 4: Conclusion
The function \( h(k) \) is continuous on the interval \( [-3, 7] \).

Would you like more details on any specific part? Here are some related questions for further exploration:

1. What happens to \( h(k) \) at the endpoints \( k = -3 \) and \( k = 7 \)?
2. How can we determine the smoothness of \( h(k) \) within the interval?
3. How would \( h(k) \) behave outside the interval \( [-3, 7] \)?
4. Can you graph \( h(k) \) to visualize its continuity?
5. What if a different function were under the square roots? How would the domain change?

**Tip:** For square roots, always check the domain by ensuring the argument of the square root is non-negative.

Let h(k) = sqrt(7 - k) + sqrt(k + 3). Give the interval(s) on which h(k) is continuous.

Explore how to determine the intervals of continuity for the function h(k) = sqrt(7 - k) + sqrt(k + 3). The problem involves key algebraic concepts such as inequalities and square roots. Suitable for students in Grades 10-12.

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