https://file.aiquickdraw.com/mathbot/file/1729978453863171oiovu.jpg

To solve this problem, let's go through it step-by-step.

### Step 1: Set Up the Inverse Variation Equation
Since \( z \) varies inversely as \( w^2 \), we can write the equation as:
\[
z = \frac{k}{w^2}
\]
where \( k \) is a constant.

### Step 2: Find the Value of \( k \)
We know that \( z = 10 \) when \( w = 3 \). Substitute these values into the equation to solve for \( k \):
\[
10 = \frac{k}{3^2}
\]
\[
10 = \frac{k}{9}
\]
Multiply both sides by 9 to isolate \( k \):
\[
k = 90
\]

### Step 3: Substitute \( k \) and Find \( z \) When \( w = 6 \)
Now we have the equation:
\[
z = \frac{90}{w^2}
\]
Substitute \( w = 6 \):
\[
z = \frac{90}{6^2}
\]
\[
z = \frac{90}{36}
\]
\[
z = 2.5
\]

### Final Answer
\[
z = 2.5
\]

Would you like more details on any of these steps or have any questions?

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Here are some additional questions related to inverse variation:
1. What happens to \( z \) as \( w \) increases?
2. How would the answer change if \( z \) varied inversely as \( w^3 \) instead of \( w^2 \)?
3. Can we solve for \( w \) given a specific value of \( z \) if \( z = \frac{k}{w^2} \)?
4. How would you interpret the constant \( k \) in real-world applications?
5. If \( z \) varies directly as \( w^2 \) instead of inversely, how would the equation change?

**Tip:** In inverse variation, as the independent variable increases, the dependent variable decreases proportionally based on the inverse square (or other power) relationship.

If z varies inversely as w^2, and z = 10 when w = 3, find z when w = 6.

Learn how to solve an inverse variation problem where z varies inversely as w^2. Given z = 10 when w = 3, find the value of z when w = 6 with a clear, step-by-step solution. Suitable for Grades 8-10.

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