Answer:
The average value of
over the interval
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
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:
10=y/11-13
We move all terms to the left:
-10-(y/11-13)=0
-y/11+13-10=0
We multiply all the terms by the denominator
-y+13*11-10*11=0
We add all the numbers together, and all the variables
-1y+33=0
We move all terms containing y to the left, all other terms to the right
-y=-33
y=-33/-1
y=+33
The location of the vertex tells you the horizontal and vertical shift. (The parent function f(x) = x² has its vertex at the origin, (0, 0). The vertical distance of the point 1 unit left or right of the vertex in relation to the vertex tells you the vertical scale factor (stretch).
g(x) = f(x +3) -3
horizontal shift left 3
vertical shift down 3
h(x) = -3f(x)
reflection across the x-axis
vertical stretch of 3
d(x) = f(x -3) -3
horizontal shift right 3
vertical shift down 3
Lets get all the information we need first.
A week has 7 days, you practice 30 min each day, so in a week you will practice a total time t of 30 min for 7 times, that is:
t = 7day(30 min/day)
t = 210 min
So in a week you will practice a total of 210 minutes.
Number 1 is Rio Grande and number 2 is St. Lawrence River
Step-by-step explanation: