the quantity y varies directly with the square of x. if y=16when x=2, find y when x is 6

Given that the quantity \( y \) varies directly with the square of \( x \), the relationship can be expressed as:

\[
y = kx^2
\]

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

### Step 1: Find the constant \( k \)
We are given that \( y = 16 \) when \( x = 2 \). Substituting these values into the equation:

\[
16 = k(2^2)
\]

\[
16 = k(4)
\]

\[
k = \frac{16}{4} = 4
\]

### Step 2: Use the constant \( k \) to find \( y \) when \( x = 6 \)
Now, substitute \( k = 4 \) and \( x = 6 \) into the equation \( y = kx^2 \):

\[
y = 4(6^2)
\]

\[
y = 4(36) = 144
\]

Thus, \( y = 144 \) when \( x = 6 \).

Would you like further details or explanations on any step?

Here are 5 related questions for practice:
1. What happens to \( y \) if \( x \) triples in this type of relationship?
2. How would the equation change if \( y \) varied inversely with \( x^2 \)?
3. If \( y \) is 64 when \( x = 4 \), what would \( y \) be when \( x = 8 \)?
4. How would you solve for \( x \) if \( y = 100 \) and \( k \) is known?
5. How do you derive the general equation for direct variation from experimental data?

**Tip:** In direct variation, as the independent variable increases, the dependent variable increases proportionally, but inversely if the variation is inverse.

The quantity y varies directly with the square of x. If y = 16 when x = 2, find y when x is 6.

Solve a direct variation problem where y varies with the square of x. Given y = 16 when x = 2, find y when x = 6. This example uses the direct variation formula y = kx^2 to find the solution. Suitable for students in Grades 8-10.

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