Answer:
y= 2000 * x + 175000
Step-by-step explanation:
Given is Population is 175,000
Increase in rate = 2000/year
Let x be the population in 1995 which is C= 175,000
y= Population after per year
We know the formula for slope intercept form is
y= mx + C ............... (i)
Now from the suppose values and given
m= 2000
Now putting values in equation (i) gives
y= 2000 * x + 175000
Here x is the No of years
You could use perturbation method to calculate this sum. Let's start from:

On the other hand, we have:

So from (1) and (2) we have:

Now, let's try to calculate sum

, but this time we use perturbation method.

but:
![S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\= \sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\= \sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\ \boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}](https://tex.z-dn.net/?f=S_%7Bn%2B1%7D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%2B1%7Dk%5Ccdot%20k%21%3D0%5Ccdot0%21%2B%5Csum%5Climits_%7Bk%3D1%7D%5E%7Bn%2B1%7Dk%5Ccdot%20k%21%3D0%2B%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%2B1%29%28k%2B1%29%21%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%2B1%29%28k%2B1%29k%21%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B2k%2B1%29k%21%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%5Cleft%5B%28k%5E2%2B1%29k%21%2B2k%5Ccdot%20k%21%5Cright%5D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B%5Csum%5Climits_%7Bk%3D0%7D%5En2k%5Ccdot%20k%21%3D%5C%5C%5C%5C%5C%5C%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2%5Csum%5Climits_%7Bk%3D0%7D%5Enk%5Ccdot%20k%21%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BS_%7Bn%2B1%7D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%7D)
When we join both equation there will be:
![\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\ S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\ \sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\= (n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\= n(n+1)!+1](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7DS_%7Bn%2B1%7D%3DS_n%2B%28n%2B1%29%28n%2B1%29%21%5C%5C%5C%5CS_%7Bn%2B1%7D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%0AS_n%2B%28n%2B1%29%28n%2B1%29%21%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%3DS_n-2S_n%2B%28n%2B1%29%28n%2B1%29%21%3D%28n%2B1%29%28n%2B1%29%21-S_n%3D%5C%5C%5C%5C%5C%5C%3D%0A%28n%2B1%29%28n%2B1%29%21-%5Csum%5Climits_%7Bk%3D0%7D%5Enk%5Ccdot%20k%21%5Cstackrel%7B%28%5Cstar%29%7D%7B%3D%7D%28n%2B1%29%28n%2B1%29%21-%5B%28n%2B1%29%21-1%5D%3D%5C%5C%5C%5C%5C%5C%3D%28n%2B1%29%28n%2B1%29%21-%28n%2B1%29%21%2B1%3D%28n%2B1%29%21%5Ccdot%5Bn%2B1-1%5D%2B1%3D%5C%5C%5C%5C%5C%5C%3D%0An%28n%2B1%29%21%2B1)
So the answer is:

Sorry for my bad english, but i hope it won't be a big problem :)
Answer:
To write a polynomial in standard form, simplify and then arrange the terms in descending order.
y=ax²+bx+c Simplify each term.
y=5x²−20x+20+7
Add 20 and 7.
y=5x²−20x+27
by-step explanation:
Answer:
17.5ml- of 10 percent solution, 17.5ml- of 20 percent solution
Step-by-step explanation:
35:100*15=5.25- ml of alkali in the base solution
Suppose we need x ml of 10 percents solution and 35-x - of 20 percents.
Then The quantity of alkali in the first one (10 percents) is x/100*10=0.1x
when in the second one we have (35-x)/100*20= 7-0.2x of alkali
0.1x+7-0.2x=5.25
7-0.1x= 5.25
0.1x=1.75
x=17.5- 0f 10 percents
35-17.5=17.5 - of 20 percents