The quantity of y varies directly with the cube of x,lf y = 25 when x = 2/find y when x is
6?

The relationship between \( y \) and the cube of \( x \) is a direct variation, which can be written as:

\[
y = kx^3
\]

Where \( k \) is the constant of proportionality. To find \( k \), we use the given values \( y = 25 \) when \( x = 2 \):

\[
25 = k \cdot 2^3
\]
\[
25 = k \cdot 8
\]
\[
k = \frac{25}{8}
\]

Now, to find \( y \) when \( x = 6 \), substitute \( x = 6 \) into the equation:

\[
y = \frac{25}{8} \cdot 6^3
\]
\[
y = \frac{25}{8} \cdot 216
\]
\[
y = \frac{25 \cdot 216}{8}
\]
\[
y = 25 \cdot 27
\]
\[
y = 675
\]

Thus, \( y = 675 \) when \( x = 6 \).

Would you like further details or have any questions?

Here are 5 related questions for practice:
1. If \( y \) varies directly with \( x^4 \), and \( y = 32 \) when \( x = 2 \), find \( y \) when \( x = 4 \).
2. Given \( y \) varies inversely with \( x \), and \( y = 10 \) when \( x = 5 \), find \( y \) when \( x = 2 \).
3. If \( y \) varies directly with the square of \( x \), and \( y = 9 \) when \( x = 3 \), find \( y \) when \( x = 5 \).
4. Solve for \( y \) when \( y \) varies directly with \( x^2 \), given \( y = 50 \) and \( x = 5 \), and \( x = 10 \).
5. If \( y \) varies inversely with the square of \( x \), and \( y = 16 \) when \( x = 2 \), find \( y \) when \( x = 4 \).

**Tip:** Always solve for the constant of variation first before solving for other values of \( x \).

The quantity of y varies directly with the cube of x, if y = 25 when x = 2, find y when x is 6?

Learn how to solve direct variation problems where y varies with the cube of x. In this example, if y = 25 when x = 2, find y when x = 6. This solution covers key concepts such as direct variation, exponents, and includes a step-by-step guide to finding the constant of proportionality.

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