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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:
Demetrius's account is $84 higher after the two transactions
Step-by-step explanation:
Let
x -----> original amount in Demetrius's account
y ----> amount in Demetrius's account after the deposit and the withdraws
we know that
The amount in Demetrius's account after the two transactions must be equal to the original amount in Demetrius's account plus the deposit minus the withdraws
so


therefore
Demetrius's account is $84 higher after the two transactions
If you can't see the picture the answer is: (y+4)(y+2).
Hope this helps, and May the Force Be With You!
-Jabba