<h2>Length = 11</h2><h2>Width = 6</h2>
The area of a rectangle is 66 ft^2:
L * W = 66
Length of the rectangle is 7 feet less than three times the width:
L = 3W-7
Substitute L in terms of W:
(3W-7) * W = 66
Factorise the equation:
3W^2 -7W = 66
3W^2 - 7W - 66 = 0
Factors of 66 =
1 66, 2 33, 3 22, 6 11
(3W + 11) (W - 6) = 0
Solve for W:
3W + 11 = 0
3W = 11
W = 3/11
W - 6 = 0
W = 0 + 6
W = 6
Using the original equation, find L:
L = 3W-7
L = 3(6)-7
L = 18-7
L = 11
L * W = 66
11 * 6 = 66
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:y=negative 1 over 3x +2
Step-by-step explanation:
Answer:
0.75 or 3/4
Step-by-step explanation:
f(-2) = 3•2^-2
f(-2) = 3•0.25
f(-2) = 0.75 or 3/4
Hope this helps! :)
