https://file.aiquickdraw.com/mathbot/file/1727385610713faghbnwt.jpg

The equation given in the image is:

\[
36y^2 = (x^2 - 4)^3, \quad 5 \leq x \leq 6, \quad y \geq 0
\]

### Steps to solve:

1. **Solve for \(y\):**
   - Divide both sides of the equation by 36:
     \[
     y^2 = \frac{(x^2 - 4)^3}{36}
     \]
   - Take the square root of both sides, considering \(y \geq 0\):
     \[
     y = \frac{\sqrt{(x^2 - 4)^3}}{6}
     \]

2. **Evaluate for the given range of \(x\):**
   - We know that \(5 \leq x \leq 6\), so we can substitute the values of \(x\) in that range to determine the corresponding values of \(y\).
   
     - At \(x = 5\):
       \[
       y = \frac{\sqrt{(5^2 - 4)^3}}{6} = \frac{\sqrt{(25 - 4)^3}}{6} = \frac{\sqrt{21^3}}{6} = \frac{\sqrt{9261}}{6} = \frac{96.3}{6} \approx 16.05
       \]

     - At \(x = 6\):
       \[
       y = \frac{\sqrt{(6^2 - 4)^3}}{6} = \frac{\sqrt{(36 - 4)^3}}{6} = \frac{\sqrt{32^3}}{6} = \frac{\sqrt{32768}}{6} \approx \frac{181.02}{6} \approx 30.17
       \]

3. **Conclusion:**
   - For \(5 \leq x \leq 6\), the corresponding \(y\)-values range from approximately 16.05 to 30.17.

Would you like more details or have any specific questions?

Here are 5 related questions:
1. What happens to the value of \(y\) as \(x\) moves outside the given range?
2. How would the solution change if \(y\) could be negative?
3. What is the behavior of the equation for larger intervals of \(x\)?
4. How can this equation be visualized graphically?
5. How do cube roots and square roots affect the range of \(y\)?

**Tip:** Always consider the domain restrictions on \(x\) and \(y\) when solving equations with square roots and cube terms.

Solve the equation 36y^2 = (x^2 - 4)^3 with 5 ≤ x ≤ 6 and y ≥ 0.

Learn how to solve the equation 36y^2 = (x^2 - 4)^3 for x in the range of 5 to 6 and y ≥ 0. This solution covers algebraic concepts including square roots and cube powers, providing a step-by-step solution. Suitable for high school students in Grades 9-12.

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