Respuesta :
Let x = the width and x + 30 = the length.
x(x + 3) = 8800
x^2 + 3x = 8800
x^2 + 3x - 8800 = 0
Using the Quadratic Formula, I get x ≈ 92.320 yd and x + 30 ≈ 122.320 yd.
2x + 2x + 60 = 600
4x + 60 = 600
4x = 540
x = 135
x + 30 = 165
The area would be 22275 yd^2.
x(x + 3) = 8800
x^2 + 3x = 8800
x^2 + 3x - 8800 = 0
Using the Quadratic Formula, I get x ≈ 92.320 yd and x + 30 ≈ 122.320 yd.
2x + 2x + 60 = 600
4x + 60 = 600
4x = 540
x = 135
x + 30 = 165
The area would be 22275 yd^2.
Answer:
110 yards and 80yards
Step-by-step explanation:
Given: Stanford University's soccer field has an area of [tex]8800[/tex] square yards. Its length is [tex]30[/tex] yards longer than its width.
To Find: Write an equation and solve for the dimensions of the soccer field.
Solution:
Total area of Stanford University's soccer field[tex]=8800[/tex] [tex]\text{square yards}[/tex]
let the length of soccer field is[tex]=\text{l}[/tex]
let the width of soccer field is[tex]=\text{b}[/tex]
Now,
as given
[tex]\text{l}=\text{b}+30[/tex]
We know that
area of soccer field[tex]=\text{length}\times\text{width}[/tex]
putting values
[tex]\text{l}\times\text{b}[/tex][tex]=8800[/tex]
[tex]\text{b}(30+\text{b})=8800[/tex]
equation for width of soccer field
[tex]\text{b}^{2}+30\text{b}-8800=0[/tex]
[tex](\text{b+110})(\text{b}-80)=0[/tex]
as [tex]\text{b}[/tex] can not be negative
[tex]\text{b}=80[/tex] [tex]\text{yards}[/tex]
[tex]\text{l}=\text{b}+30[/tex]
[tex]\text{l}=110[/tex] [tex]\text{yards}[/tex]
the dimensions of soccer field are [tex]110[/tex] [tex]\text{yards}[/tex] and [tex]80[/tex] [tex]\text{yards}[/tex]