Basically what is happening is:
You start out with 15. That 1st week you have 22% more than 15, or in other words 15*1.22. The following week you have 22% more than 22% more of 15, which is 15*1.22*1.22.
Now we can write a function that models this situation:
f(n): number of views
n: number of weeks since you started
f(n) = 15(1.22^n)
We want to find out how many views there are after four weeks, so plug 4 in for n.
f(4) = 15(1.22^4)
f(4) = 33.23
This means after 4 weeks you can expect the video to have 33 views.
Answer:1002
Step-by-step explanation:
and 
and 
as 
Applying this we get
![\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csum_%7B1%7D%5E%7B1002%7D%5Cleft%20%5B%20%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7Bk%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%2B%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7B%282005-k%29%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%5Cright%20%5D)
every 
there exist 
such that 
therefore
![\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csum_%7B1%7D%5E%7B1002%7D%5Cleft%20%5B%20%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7Bk%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%2B%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7B%282005-k%29%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%5Cright%20%5D%3D1002)
-26n -22= n - 90
-26 times a number: -26n
minus 22: -22
equal: =
90 less than a number: n - 90
Solving the Equation:
Move n to the left side of the equal sign (which becomes -n) in order to combine like terms:
-26n-n-22= -90
Move -22 to the right side of the equal sign which becomes +22
-26n-n=-90+22
solve both sides; -26n-n= -27n and -90+ 22= -78
-27n=-78
n= 2.888888
Answer:
Not accomplishable.
Step-by-step explanation:
Could you please stucture this question more simply. This partly doesn't make sense, unless I am wrong.
