$$  $$$$ \text{State the restriction for }\sqrt{\left(x-9\right)^5} $$/$$ ^{} $$$$ \sqrt[3]{y} $$

To determine the restriction for $\sqrt{(x-9)^5} \cdot \sqrt[3]{y}$, we need to examine each part individually:

### 1. $\sqrt{(x-9)^5}$ (Square Root Expression):
- A square root is defined only for non-negative values. Therefore, the term $(x - 9)^5$ must be non-negative.
- Since the exponent $5$ is odd, $(x - 9)^5$ will take the sign of $x - 9$. For the expression inside the square root to be valid, $x - 9 \geq 0$.
- This implies:
  \[
  x - 9 \geq 0 \quad \Rightarrow \quad x \geq 9
  \]
  
### 2. $\sqrt[3]{y}$ (Cube Root Expression):
- A cube root is defined for all real numbers, meaning $y$ can be any real number. There are no restrictions on $y$.

### Final Restriction:
The only restriction comes from the square root part of the expression. Therefore, the restriction is:
\[
x \geq 9
\]
There are no restrictions on $y$.

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Would you like more details or have any questions?

Here are 5 related questions to expand on this:

1. What are the restrictions on $x$ for $\sqrt{(x - 5)^2}$?
2. What would be the domain of $\sqrt[4]{x}$?
3. Can you explain why cube roots do not have the same restrictions as square roots?
4. How does the exponent of a term inside a square root affect the restrictions?
5. What is the domain of $\sqrt[5]{x^2 - 4}$?

**Tip**: Always check the radicand (the expression under the root) when dealing with square roots to ensure it is non-negative.

State the restriction for $\sqrt{(x-9)^5}$ / $\sqrt[3]{y}$

Learn how to determine restrictions for expressions involving square roots and cube roots, such as $\sqrt{(x-9)^5}$ and $\sqrt[3]{y}$. This explanation covers key concepts in algebra, including the conditions for square roots and the unrestricted nature of cube roots. Suitable for high school students in Grades 9-12.

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