What we know:
Football field is 100 yards in length
End zones are 10 yards each in length
Perimeter between pylons is 306 2/3 yards
What we need to find:
a. Perimeter and area of one end zone
b. Perimeter and area of without end zones
c. Perimeter and area of playing field with end zones
First we need to find the measurements of the field between pylons using the perimeter of 306 2/3 yards. We already know the length is 100 yards so we need to find width (w).
P=2l + 2w
306 2/3=2 (100) + 2w
306 2/3= 200 + 2w
300 2/3-200=200-200+2w
106 2/3=2w
106 2/3/2=2/2w
53 1/3=w
a. Perimeter=2 (10)+2 (53 1/3)=126 2/3 yards
Area=10×53 1/3=533 1/3 yd²
b. Perimeter=2 (100)+2 (53 1/3)=306 2/3 yards
Area=100×53 1/3=5333 1/3 yd²
c. Perimeter=2 (120) + 2 (53 1/3)=346 2/3 yards
Area=120×53 1/3=6400 yd²
Answer:1. 7 2. 
Step-by-step explanation:
1. 
= 7
2. ( doesn't have any simplest radical form)
The answer is the answer is C and F.
This is found by inserting the equation above into the quadratic formula or -b plus or minus the square root of b^2 - 4ac over 2a.
With that, you find that the first number is a -3, which eliminates choices B and D.
Then, solving the 4ac portion, you find that inside the square root the number is 29, which gives you the last two answer choices.
Hope this helps!
1x20
2x10
4x5
They are all dimensions of wood by wood.
