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
160 = 2 x 2 x 2 x 2 x 2 x 5 = 2^5 x 5
243 = 3 x 3 x 3 x 3 x 3 = 3^5
So
(160 * 243)^1/5
= 5th root of (160 * 243)
= 5th root of (2^5 * 5 * 3^5)
= 2 * 3 * (5th root of 5)
= 6 * (5th root of 5)
Answer is D. 6 * (5th root of 5)
Answer:
1 cubic centimeter
Step-by-step explanation:
V=l³ And l=1
so V=1³ so
V=1
since we cubed something it is in cubic centimeters so
1 cubic centimeter
The vertex would be at (3, 6).
Answer:
d. 8 units
Step-by-step explanation:
If AC = 16 units then the length of CB is 8 units because DB is perpendicular bisector from the centre of circle.
