To solve the problem given that [tex]\( y \)[/tex] varies directly as [tex]\( x \)[/tex], we need to follow the concept of direct variation, which means [tex]\( y = kx \)[/tex] for some constant [tex]\( k \)[/tex].
Here’s the step-by-step approach to solving the problem:
1. Determine the constant of variation [tex]\( k \)[/tex]:
- We are given that [tex]\( y = 6 \)[/tex] when [tex]\( x = 72 \)[/tex].
- Since [tex]\( y \)[/tex] varies directly as [tex]\( x \)[/tex], we can express this relationship as:
[tex]\[
y = kx
\][/tex]
- Substitute the given values into the equation to solve for [tex]\( k \)[/tex]:
[tex]\[
6 = k \cdot 72
\][/tex]
- Solving for [tex]\( k \)[/tex]:
[tex]\[
k = \frac{6}{72} = \frac{1}{12}
\][/tex]
2. Use the constant [tex]\( k \)[/tex] to find [tex]\( y \)[/tex] when [tex]\( x = 8 \)[/tex]:
- Now that we have [tex]\( k = \frac{1}{12} \)[/tex], we use this to find the new value of [tex]\( y \)[/tex] for [tex]\( x = 8 \)[/tex].
- Substitute [tex]\( k \)[/tex] and [tex]\( x = 8 \)[/tex] back into the direct variation equation [tex]\( y = kx \)[/tex]:
[tex]\[
y = \frac{1}{12} \cdot 8
\][/tex]
- Simplify the expression:
[tex]\[
y = \frac{8}{12} = \frac{2}{3}
\][/tex]
So the value of [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 8 is [tex]\( \frac{2}{3} \)[/tex].
Therefore, the correct answer is:
[tex]\[
\boxed{\frac{2}{3}}
\][/tex]
