Let's solve this problem step by step.

We are tasked with finding the domain of three functions in interval notation.

Part (a) $y(t) = \sqrt{36 - t}$:

• For a square root to be defined, the argument inside the square root must be non-negative: $36 - t \geq 0$ Solve for $t$: $t \leq 36$
• Therefore, the domain of $y(t)$ is: $(-\infty, 36]$

Part (b) $w(t) = \sqrt{t - 36}$:

• Again, the argument inside the square root must be non-negative: $t - 36 \geq 0$ Solve for $t$: $t \geq 36$
• Therefore, the domain of $w(t)$ is: $[36, \infty)$

Part (c) $z(t) = \frac{1}{\sqrt{36 - t}}$:

• For $\sqrt{36 - t}$ to be defined, the argument inside the square root must be positive (since division by zero is undefined): $36 - t > 0$ Solve for $t$: $t < 36$
• Therefore, the domain of $z(t)$ is: $(-\infty, 36)$

Summary of Domains:

1. $y(t) = \sqrt{36 - t}$: $(- \infty, 36]$
2. $w(t) = \sqrt{t - 36}$: $[36, \infty)$
3. $z(t) = \frac{1}{\sqrt{36 - t}}$: $(- \infty, 36)$

Let me know if you would like further clarification or a detailed explanation!

Related Questions:

1. Why does a square root require its argument to be non-negative?
2. What happens to the function if the argument inside the square root becomes negative?
3. How would the domains change if the functions were defined differently (e.g., $y(t) = \sqrt{|36 - t|}$)?
4. Can these domains ever overlap for different functions?
5. How do you find the domain of a rational function with a square root in the denominator?

Tip: Always check for conditions like square roots and denominators when determining the domain of a function!