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:
25. a = 60
26. a = 17
Step-by-step explanation:
In these two problems we are going to use a property of parallelograms that says: opposite angles have equal measure
so we have
25.
and for 26 we have
Remember:
a^m * a^n=a^(m+n)
a^-m=1/a^m
8³ x 8⁻⁵=8³⁻⁵=8⁻²=1/8²=1/64<1
Answer: the student is not correct, because 1/64<1, and 8³ x 8⁻⁵=1/64.
Yy^2 I'm pretty sure is the answer!
4.0
.40
.040
Sorry if im wrong but im sure the second one is correct
