Given:
Consider the given function is:
![f(x)=5^x+1](https://tex.z-dn.net/?f=f%28x%29%3D5%5Ex%2B1)
To find:
The average rate of change between x = 0 and x = 4.
Solution:
The average rate of change of a function f(x) over the interval [a,b] is:
![m=\dfrac{f(b)-f(a)}{b-a}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D)
We have,
![f(x)=5^x+1](https://tex.z-dn.net/?f=f%28x%29%3D5%5Ex%2B1)
At
,
![f(0)=5^0+1](https://tex.z-dn.net/?f=f%280%29%3D5%5E0%2B1)
![f(0)=1+1](https://tex.z-dn.net/?f=f%280%29%3D1%2B1)
![f(0)=2](https://tex.z-dn.net/?f=f%280%29%3D2)
At
,
![f(4)=5^4+1](https://tex.z-dn.net/?f=f%284%29%3D5%5E4%2B1)
![f(4)=625+1](https://tex.z-dn.net/?f=f%284%29%3D625%2B1)
![f(4)=626](https://tex.z-dn.net/?f=f%284%29%3D626)
Now, the average rate of change between x = 0 and x = 4 is:
![m=\dfrac{f(4)-f(0)}{4-0}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7Bf%284%29-f%280%29%7D%7B4-0%7D)
![m=\dfrac{626-2}{4}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B626-2%7D%7B4%7D)
![m=\dfrac{624}{4}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B624%7D%7B4%7D)
![m=156](https://tex.z-dn.net/?f=m%3D156)
Hence, the average rate of change between x = 0 and x = 4 is 156.
Answer: 8,000 Liters
Step-by-step explanation:
There are 1,000 liters in 1 cubic meter. So you would just do 8 times 1,000, which is 8,000 liters.
2 ounces each sammich made
x=number of sammiches made
y=amount left
amount left=full-gotten
y=40-2 times number of sammiches
y=40-2x
or
y=-2x+40
D is da answer
The sample space is:
(1, 1); (1, 2) - sum of 3; (1, 3); (1, 4); (1, 5) - sum of 6; (1, 6);
(2, 1) - sum of 3; (2, 2); (2, 3); (2, 4) - sum of 6; (2, 5); (2, 6);
(3, 1); (3, 2); (3, 3) - sum of 6; (3, 4); (3, 5); (3, 6) - sum of 9;
(4, 1); (4, 2) - sum of 6; (4, 3); (4, 4); (4, 5) - sum of 9; (4, 6);
(5, 1) - sum of 6; (5, 2); (5, 3); (5, 4) - sum of 9; (5, 5); (5, 6);
(6, 1): (6, 2); (6, 3) - sum of 9; (6, 4); (6, 5); (6, 6)