To solve the problem where [tex]\( y \)[/tex] varies inversely as the square of [tex]\( x \)[/tex], and given that [tex]\( y = \frac{7}{4} \)[/tex] when [tex]\( x = 1 \)[/tex], we need to find the expression for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] and then determine [tex]\( y \)[/tex] for [tex]\( x = 3 \)[/tex].
1. Understanding the Relationship:
Since [tex]\( y \)[/tex] varies inversely as the square of [tex]\( x \)[/tex], the relationship can be written as:
[tex]\[
y = \frac{k}{x^2}
\][/tex]
where [tex]\( k \)[/tex] is a constant.
2. Finding the Constant [tex]\( k \)[/tex]:
We are given that [tex]\( y = \frac{7}{4} \)[/tex] when [tex]\( x = 1 \)[/tex]. Substitute these values into the equation to find [tex]\( k \)[/tex]:
[tex]\[
\frac{7}{4} = \frac{k}{1^2}
\][/tex]
[tex]\[
k = \frac{7}{4}
\][/tex]
3. Expression for [tex]\( y \)[/tex]:
Now that we have [tex]\( k = \frac{7}{4} \)[/tex], we can write the equation for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[
y = \frac{\frac{7}{4}}{x^2} = \frac{7}{4x^2}
\][/tex]
4. Finding [tex]\( y \)[/tex] when [tex]\( x = 3 \)[/tex]:
Substitute [tex]\( x = 3 \)[/tex] into the equation:
[tex]\[
y = \frac{7}{4 \cdot (3^2)} = \frac{7}{4 \cdot 9} = \frac{7}{36}
\][/tex]
Thus, the value of [tex]\( y \)[/tex] when [tex]\( x = 3 \)[/tex] is [tex]\( \frac{7}{36} \)[/tex], and the equation that represents the relationship is [tex]\( y = \frac{7}{4x^2} \)[/tex].
Therefore, the correct answer is:
c. [tex]\( y = \frac{7}{4 x^2} ; y(3) = \frac{7}{36} \)[/tex]
