Answer:
-1 by x^2/2
Step-by-step explanation:
I am so sorry
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: 7a^2+2a+7
Step-by-step explanation:Distribute the Negative Sign:
=9a2+7a+8+−1(2a2+5a+1)
=9a2+7a+8+−1(2a2)+−1(5a)+(−1)(1)
=9a2+7a+8+−2a2+−5a+−1
Combine Like Terms:
=9a2+7a+8+−2a2+−5a+−1
=(9a2+−2a2)+(7a+−5a)+(8+−1)
=7a2+2a+7
Answer:
equal to
Step-by-step explanation:
Butter pecan: 25
Vanilla: 35
Chocolate: 50
The dotted line between 20 and 30 (butter pecan) will be nothing more than anything including 5. In this case, it is 25 counting up to 30.
The solid lines directly leading to a number will be that number, for instance, chocolate is 50.
