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:
do u got a picture by chance
Step-by-step explanation:
(a) The "average value" of a function over an interval [a,b] is defined to be
(1/(b-a)) times the integral of f from the limits x= a to x = b.
Now S = 200(5 - 9/(2+t))
The average value of S during the first year (from t = 0 months to t = 12 months) is then:
(1/12) times the integral of 200(5 - 9/(2+t)) from t = 0 to t = 12
or 200/12 times the integral of (5 - 9/(2+t)) from t= 0 to t = 12
This equals 200/12 * (5t -9ln(2+t))
Evaluating this with the limits t= 0 to t = 12 gives:
708.113 units., which is the average value of S(t) during the first year.
(b). We need to find S'(t), and then equate this with the average value.
Now S'(t) = 1800/(t+2)^2
So you're left with solving 1800/(t+2)^2 = 708.113
<span>I'll leave that to you</span>
Answer:
Step-by-step explanation:
(3m-1)(6-a)-(m+3)(6-a)
=3m(6-a)-1(6-a)-m(6-a)+3(6-a)
=18m-3ma-6+a-6m+ma+18-3a
=12m-2ma+12-2a
=2(6m-ma+6-a)
Answer:
98.9
Step-by-step explanation:
115 - 20% = 92
92 + 7.5% = 98.9
