Let's solve each of the equations step by step.

(a) $V = \frac{1}{5} \pi r^2 h$ for $r$

We want to solve for $r$.

1. Start by multiplying both sides by 5 to eliminate the fraction: $5V = \pi r^2 h$

2. Divide both sides by $\pi h$: $\frac{5V}{\pi h} = r^2$

3. Take the square root of both sides: $r = \sqrt{\frac{5V}{\pi h}}$

(b) $V = s^3$ for $s$

Here, we are solving for $s$:

1. Take the cube root of both sides to solve for $s$: $s = \sqrt[3]{V}$

(c) $c^2 = a^2 + b^2$ for $a$

We are solving for $a$:

1. Subtract $b^2$ from both sides: $a^2 = c^2 - b^2$

2. Take the square root of both sides: $a = \sqrt{c^2 - b^2}$

Let me know if you need more details or have any questions!

Here are 5 related questions you could explore:

1. What is the geometric significance of the formula in part (a)?
2. How do you handle cube roots for negative volumes (if applicable)?
3. How does the Pythagorean theorem relate to part (c)?
4. Can you solve for $h$ in equation (a)?
5. What are the conditions for non-real solutions in these equations?

Tip: Always ensure to check the domain of your variables, especially when dealing with square roots or cube roots. For example, $r$, $s$, and $a$ should be non-negative in this case.