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.