If x*x* and y*y* are in direct proportion and y*y* is 10 when x*x* is 2, find y*y* when x*x* is 6.

If \( x^2 \) and \( y^2 \) are in direct proportion, we can express the relationship as:

\[
x^2 \propto y^2 \quad \text{or} \quad \frac{x^2}{y^2} = k
\]

where \( k \) is the constant of proportionality.

We are given that \( y^2 = 10 \) when \( x^2 = 2 \). Substituting these values into the equation to find \( k \):

\[
\frac{2}{10} = k \quad \Rightarrow \quad k = \frac{1}{5}
\]

Now, we are asked to find \( y^2 \) when \( x^2 = 6 \). Using the same relationship:

\[
\frac{6}{y^2} = \frac{1}{5}
\]

Solving for \( y^2 \):

\[
y^2 = 6 \times 5 = 30
\]

Thus, when \( x^2 = 6 \), \( y^2 = 30 \).

Would you like more details or further explanations?

Here are some related questions for further exploration:

1. What is the constant of proportionality in general terms for this kind of problem?
2. How does the solution change if the relationship were inverse instead of direct?
3. Can you determine \( y \) instead of \( y^2 \) given the same conditions?
4. How would the equation look if we introduced a constant multiplier in \( x^2 \)?
5. What happens to \( y^2 \) if \( x^2 \) is doubled?

**Tip:** In problems involving direct proportion, once you know the proportionality constant, you can solve for unknowns by multiplying or dividing appropriately.

If x^2 and y^2 are in direct proportion and y^2 is 10 when x^2 is 2, find y^2 when x^2 is 6.

Learn how to solve a direct proportion problem involving x^2 and y^2. This step-by-step solution explains how to calculate the value of y^2 when x^2 is given using direct proportionality. Suitable for students in Grades 6-8.

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