https://file.aiquickdraw.com/mathbot/file/1728865600583uha6tzfn.jpg

The problem in the image states:

- \( y \) varies directly as the cube of \( x \), meaning \( y = kx^3 \), where \( k \) is a constant.
- Given: \( y = 17 \) when \( x = 2 \).
- We need to find \( y \) when \( x = 10 \).

### Step 1: Find the constant of variation \( k \)
Using the given values \( y = 17 \) and \( x = 2 \):

\[
y = kx^3
\]
\[
17 = k(2^3)
\]
\[
17 = k(8)
\]
\[
k = \frac{17}{8} = 2.125
\]

### Step 2: Use \( k \) to find \( y \) when \( x = 10 \)
Now that we know \( k = 2.125 \), we can find \( y \) when \( x = 10 \):

\[
y = 2.125 \times (10^3)
\]
\[
y = 2.125 \times 1000
\]
\[
y = 2125
\]

Thus, \( y = 2125 \) when \( x = 10 \).

Would you like further details or explanations? 

Here are some related questions you might explore:
1. What is the formula for direct variation?
2. How is the cube function \( x^3 \) different from the square function \( x^2 \)?
3. What happens to \( y \) if \( x \) is doubled in this problem?
4. How does the constant of variation \( k \) affect the relationship between \( x \) and \( y \)?
5. Can direct variation also involve fractional powers of \( x \)?

**Tip:** When solving direct variation problems, always start by determining the constant of variation \( k \) using the initial conditions provided.

Suppose that y varies directly as the cube of x, and that y = 17 when x = 2. What is y when x = 10? Round your answer to two decimal places if necessary.

Learn how to solve a direct variation problem where y varies directly as the cube of x. Given that y = 17 when x = 2, calculate y when x = 10. This problem involves solving for the constant of variation and applying it to find y for a new x value. Suitable for students in Grades 9-11.

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