Answer:
The average value of 
over the interval ![[-2,5]](https://tex.z-dn.net/?f=%5B-2%2C5%5D)
is 
.
Step-by-step explanation:
Let suppose that function is continuous and integrable in the given intervals, by integral definition of average we have that:
(1)
(2)
By Fundamental Theorems of Calculus we expand both expressions:
(1b)
(2b)
We obtain the average value of over the interval ![[-2, 5]](https://tex.z-dn.net/?f=%5B-2%2C%205%5D)
by algebraic handling:
![F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)](https://tex.z-dn.net/?f=F%285%29%20-%20F%283%29%20%2B%5BF%283%29-F%28-2%29%5D%20%3D%2040%20%2B%20%28-30%29)
The average value of over the interval is .
Answer is 11w-6
Add all the like terms. So all with the w’s which is how I got 11 and then 6 stays the same hence the -6
Answer:
Tan θ = y/z
Step-by-step explanation:
Reference angle = θ
Opposite side to θ = y
Adjacent side = z
Thus:
Tan θ = opposite/adjacent
Tan θ = y/z
Ok, I'm not giving you the answer. Instead, I'm going to give you the way to solve it. Then, you can do it yourself, message it to me, and I will check it.
Do .11 x 25. Remember to add the decimal back in.
