Given the relationships that \( y \) is directly proportional to \( w^2 \) and \( x \) is inversely proportional to \( w \):
1. We start with the equations:
[tex]\[
y = k \cdot w^2
\][/tex]
and
[tex]\[
x = \frac{c}{w}
\][/tex]
2. When \( w = 10 \), \( y = 5 \), and \( x = 0.4 \):
- Substitute these values into the first equation to find \( k \):
[tex]\[
5 = k \cdot (10^2)
\][/tex]
[tex]\[
5 = k \cdot 100
\][/tex]
[tex]\[
k = \frac{5}{100} = 0.05
\][/tex]
- Substitute these values into the second equation to find \( c \):
[tex]\[
0.4 = \frac{c}{10}
\][/tex]
[tex]\[
c = 0.4 \times 10 = 4
\][/tex]
3. Our goal is to express \( y \) in terms of \( x \).
- From the equation \( x = \frac{c}{w} \), solve for \( w \):
[tex]\[
w = \frac{c}{x}
\][/tex]
- Substitute \( w = \frac{c}{x} \) into the equation \( y = k \cdot w^2 \):
[tex]\[
y = k \cdot \left( \frac{c}{x} \right)^2
\][/tex]
[tex]\[
y = k \cdot \frac{c^2}{x^2}
\][/tex]
4. Substitute the values of \( k \) and \( c \):
[tex]\[
y = 0.05 \cdot \frac{4^2}{x^2}
\][/tex]
[tex]\[
y = 0.05 \cdot \frac{16}{x^2}
\][/tex]
[tex]\[
y = \frac{0.8}{x^2}
\][/tex]
So, the relationship between \( y \) and \( x \) in its simplest form is:
[tex]\[
y = \frac{0.8}{x^2}
\][/tex]
