Split up the interval [0, 2] into 4 subintervals, so that
![[0,2]=\left[0,\dfrac12\right]\cup\left[\dfrac12,1\right]\cup\left[1,\dfrac32\right]\cup\left[\dfrac32,2\right]](https://tex.z-dn.net/?f=%5B0%2C2%5D%3D%5Cleft%5B0%2C%5Cdfrac12%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac12%2C1%5Cright%5D%5Ccup%5Cleft%5B1%2C%5Cdfrac32%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac32%2C2%5Cright%5D)
Each subinterval has width
. The area of the trapezoid constructed on each subinterval is
, i.e. the average of the values of
at both endpoints of the subinterval times 1/2 over each subinterval
.
So,


Answer:

Step-by-step explanation:
Hello,
Touchdown is 7 points, field goals is 3 points
Let's note a the number of touchdowns and b the number of field goals, we can write
(1) a + b = 80
(2) 7a + 3b = 516
(2) - 3*(1) gives
7a + 3b - 3(a + b) = 516 - 3*80
<=> 7a + 3b -3a - 3b = 516 - 240 = 276
<=> 4a = 276
<=> 
And then from (1) b = 80 - 69 = 11
Hope this helps
Answer:
40 cubic units
Step-by-step explanation:
I think the answer might be x=4/8