Answer:
hmm maybe 1?
Step-by-step explanation:
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
X=-9
There is no slope in either equation. Therefore, x=-11 and x=-9 will be two vertical and parallel lines at (-11,0) and (-9,0)
Answer:
Other needs =25,2 degrees
insurance = 100,8 degrees
Taxes= 7,2
Housing = 57,6
food= 57,6
childcare = 64,8
transportation = 46,8
Step-by-step explanation:
Its % of 360 degrees so you have to calculate the percentage of each categorie on 360 degrees to find the angles. Exemple, for the Other needs, you take 7% out of 360 degrees and you find that it is equal to 25,2. (7% x360 degrees /100)
