Certainly! Let's break down the solution step-by-step for each part.
### Part (a): Solve the formula for [tex]\( w \)[/tex]
We start with the formula for the perimeter of the soccer field:
[tex]\[ P = 2 \ell + 2 w \][/tex]
To solve for [tex]\( w \)[/tex], follow these steps:
1. Subtract [tex]\( 2\ell \)[/tex] from both sides:
[tex]\[ P - 2 \ell = 2 w \][/tex]
2. Divide both sides by 2:
[tex]\[ w = \frac{P - 2 \ell}{2} \][/tex]
Thus, the width [tex]\( w \)[/tex] in terms of the perimeter [tex]\( P \)[/tex] and the length [tex]\( \ell \)[/tex] is:
[tex]\[ w = \frac{P - 2 \ell}{2} \][/tex]
### Part (b): Find the width of the field
Using the specific values given, where the perimeter [tex]\( P = 240 \)[/tex] yards and the length [tex]\( \ell = 80 \)[/tex] yards, we can use the formula derived in Part (a) to find the width [tex]\( w \)[/tex]:
[tex]\[ w = \frac{240 - 2 \cdot 80}{2} \][/tex]
Calculate the terms inside the parentheses first:
[tex]\[ 240 - 2 \cdot 80 = 240 - 160 = 80 \][/tex]
Now, divide by 2:
[tex]\[ w = \frac{80}{2} = 40 \][/tex]
The width [tex]\( w \)[/tex] of the field is [tex]\( 40 \)[/tex] yards.
### Part (c): About what percent of the field is inside the circle?
Assume a circle with a radius equal to the width [tex]\( w \)[/tex].
1. Calculate the area of the circle:
The formula for the area of a circle is:
[tex]\[ \text{Area of the circle} = \pi r^2 \][/tex]
where [tex]\( r = w = 40 \)[/tex] yards. Therefore:
[tex]\[ \text{Area of the circle} = \pi \cdot (40)^2 \][/tex]
The calculated area of the circle is approximately:
[tex]\[ \text{Area of the circle} \approx 5026.55 \text{ square yards} \][/tex]
2. Calculate the area of the field:
The area of the rectangular field is given by the product of its length and width:
[tex]\[ \text{Area of the field} = \ell \cdot w = 80 \cdot 40 = 3200 \text{ square yards} \][/tex]
3. Calculate the percent of the field inside the circle:
The percent is found by dividing the area of the circle by the area of the field and then multiplying by 100:
[tex]\[ \text{Percent inside the circle} = \left( \frac{\text{Area of the circle}}{\text{Area of the field}} \right) \times 100 \][/tex]
[tex]\[ \text{Percent inside the circle} = \left( \frac{5026.55}{3200} \right) \times 100 \approx 157.08\% \][/tex]
About 157.08% of the field is inside the circle.
Since this percentage exceeds 100%, it might indicate that the circle with radius equal to the width extends well beyond the boundaries of the field.
