Answer:
6.79
Step-by-step explanation:
First of all, you will have to know how much is 3% of 7 in order to decrease it by that amount.
7 x 0.03 = 0.21
Now that you know the amount, you can subtract it from seven and therefore you have decreased 7 by 3%.
7 - 0.21 = 6.79
Answer:
C(x) = 8*x + 8* (1000)/x + 80
C(x) Domain x > 0
Step-by-step explanation:
enclosed area 1000 ft²
let x be side perpendicular to the river
and y parallel to the river
Then:
A = x*y y = A / x y = 1000/x
Cost of one side (x) 4*x $ then two sides cost = 4*2*x cost = 8*x
Cost of one side (y) 8*y $ 8* (1000/x)
Cost of four cornes posts 4*20 = 80 $
Total cost C(x)
C(x) = 8*x + 8* (1000)/x + 80
R { x > 0}
Taking derivatives both sides of the equation
C´(x) = 8 - 8000/x² C´(x) = 0 8 - 8000/x² = 0
8x² -8000 = 0 x² = 1000
x = 100
Answer:
It's 40 degrees. xxxxxxxxxx'
![\displaystyle\lim_{n\to\infty}\left(k!+\frac{(k+1)!}{1!}+\cdots+\frac{(k+n)!}{n!}\right)=\lim_{n\to\infty}\dfrac{\displaystyle\sum_{i=0}^n\frac{(k+i)!}{i!}}{n^{k+1}}=\lim_{n\to\infty}\frac{a_n}{b_n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cleft%28k%21%2B%5Cfrac%7B%28k%2B1%29%21%7D%7B1%21%7D%2B%5Ccdots%2B%5Cfrac%7B%28k%2Bn%29%21%7D%7Bn%21%7D%5Cright%29%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cdfrac%7B%5Cdisplaystyle%5Csum_%7Bi%3D0%7D%5En%5Cfrac%7B%28k%2Bi%29%21%7D%7Bi%21%7D%7D%7Bn%5E%7Bk%2B1%7D%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Ba_n%7D%7Bb_n%7D)
By the Stolz-Cesaro theorem, this limit exists if
![\displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Ba_%7Bn%2B1%7D-a_n%7D%7Bb_%7Bn%2B1%7D-b_n%7D)
also exists, and the limits would be equal. The theorem requires that
![b_n](https://tex.z-dn.net/?f=b_n)
be strictly monotone and divergent, which is the case since
![k\in\mathbb N](https://tex.z-dn.net/?f=k%5Cin%5Cmathbb%20N)
.
You have
![a_{n+1}-a_n=\displaystyle\sum_{i=0}^{n+1}\frac{(k+i)!}{i!}-\sum_{i=0}^n\frac{(k+i)!}{i!}=\frac{(k+n+1)!}{(n+1)!}](https://tex.z-dn.net/?f=a_%7Bn%2B1%7D-a_n%3D%5Cdisplaystyle%5Csum_%7Bi%3D0%7D%5E%7Bn%2B1%7D%5Cfrac%7B%28k%2Bi%29%21%7D%7Bi%21%7D-%5Csum_%7Bi%3D0%7D%5En%5Cfrac%7B%28k%2Bi%29%21%7D%7Bi%21%7D%3D%5Cfrac%7B%28k%2Bn%2B1%29%21%7D%7B%28n%2B1%29%21%7D)
so we're left with computing
![\displaystyle\lim_{n\to\infty}\frac{(k+n+1)!}{(n+1)!\left((n+1)^{k+1}-n^{k+1}\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%28k%2Bn%2B1%29%21%7D%7B%28n%2B1%29%21%5Cleft%28%28n%2B1%29%5E%7Bk%2B1%7D-n%5E%7Bk%2B1%7D%5Cright%29%7D)
This can be done with the help of Stirling's approximation, which says that for large
![n](https://tex.z-dn.net/?f=n)
,
![n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n](https://tex.z-dn.net/?f=n%21%5Csim%5Csqrt%7B2%5Cpi%20n%7D%5Cleft%28%5Cdfrac%20ne%5Cright%29%5En)
. By this reasoning our limit is
![\displaystyle\lim_{n\to\infty}\frac{\sqrt{2\pi(k+n+1)}\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\sqrt{2\pi(n+1)}\left(\dfrac{n+1}e\right)^{n+1}\left((n+1)^{k+1}-n^{k+1}\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%5Csqrt%7B2%5Cpi%28k%2Bn%2B1%29%7D%5Cleft%28%5Cdfrac%7Bk%2Bn%2B1%7De%5Cright%29%5E%7Bk%2Bn%2B1%7D%7D%7B%5Csqrt%7B2%5Cpi%28n%2B1%29%7D%5Cleft%28%5Cdfrac%7Bn%2B1%7De%5Cright%29%5E%7Bn%2B1%7D%5Cleft%28%28n%2B1%29%5E%7Bk%2B1%7D-n%5E%7Bk%2B1%7D%5Cright%29%7D)
Let's examine this limit in parts. First,
![\dfrac{\sqrt{2\pi(k+n+1)}}{\sqrt{2\pi(n+1)}}=\sqrt{\dfrac{k+n+1}{n+1}}=\sqrt{1+\dfrac k{n+1}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%7B2%5Cpi%28k%2Bn%2B1%29%7D%7D%7B%5Csqrt%7B2%5Cpi%28n%2B1%29%7D%7D%3D%5Csqrt%7B%5Cdfrac%7Bk%2Bn%2B1%7D%7Bn%2B1%7D%7D%3D%5Csqrt%7B1%2B%5Cdfrac%20k%7Bn%2B1%7D%7D)
As
![n\to\infty](https://tex.z-dn.net/?f=n%5Cto%5Cinfty)
, this term approaches 1.
Next,
![\dfrac{\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\left(\dfrac{n+1}e\right)^{n+1}}=(k+n+1)^k\left(\dfrac{k+n+1}{n+1}\right)^{n+1}=e^{-k}(k+n+1)^k\left(1+\dfrac k{n+1}\right)^{n+1}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cleft%28%5Cdfrac%7Bk%2Bn%2B1%7De%5Cright%29%5E%7Bk%2Bn%2B1%7D%7D%7B%5Cleft%28%5Cdfrac%7Bn%2B1%7De%5Cright%29%5E%7Bn%2B1%7D%7D%3D%28k%2Bn%2B1%29%5Ek%5Cleft%28%5Cdfrac%7Bk%2Bn%2B1%7D%7Bn%2B1%7D%5Cright%29%5E%7Bn%2B1%7D%3De%5E%7B-k%7D%28k%2Bn%2B1%29%5Ek%5Cleft%281%2B%5Cdfrac%20k%7Bn%2B1%7D%5Cright%29%5E%7Bn%2B1%7D)
The term on the right approaches
![e^k](https://tex.z-dn.net/?f=e%5Ek)
, cancelling the
![e^{-k}](https://tex.z-dn.net/?f=e%5E%7B-k%7D)
. So we're left with
![\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%28k%2Bn%2B1%29%5Ek%7D%7B%28n%2B1%29%5E%7Bk%2B1%7D-n%5E%7Bk%2B1%7D%7D)
Expand the numerator and denominator, and just examine the first few leading terms and their coefficients.
![\displaystyle\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{n^{k+1}+(k+1)n^k+\cdots+1+n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B%28k%2Bn%2B1%29%5Ek%7D%7B%28n%2B1%29%5E%7Bk%2B1%7D-n%5E%7Bk%2B1%7D%7D%3D%5Cfrac%7Bn%5Ek%2B%5Ccdots%2B%28k%2B1%29%5Ek%7D%7Bn%5E%7Bk%2B1%7D%2B%28k%2B1%29n%5Ek%2B%5Ccdots%2B1%2Bn%5E%7Bk%2B1%7D%7D%3D%5Cfrac%7Bn%5Ek%2B%5Ccdots%2B%28k%2B1%29%5Ek%7D%7B%28k%2B1%29n%5Ek%2B%5Ccdots%2B1%7D)
Divide through the numerator and denominator by
![n^k](https://tex.z-dn.net/?f=n%5Ek)
:
![\dfrac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}=\dfrac{1+\cdots+\left(\frac{k+1}n\right)^k}{(k+1)+\cdots+\frac1{n^k}}](https://tex.z-dn.net/?f=%5Cdfrac%7Bn%5Ek%2B%5Ccdots%2B%28k%2B1%29%5Ek%7D%7B%28k%2B1%29n%5Ek%2B%5Ccdots%2B1%7D%3D%5Cdfrac%7B1%2B%5Ccdots%2B%5Cleft%28%5Cfrac%7Bk%2B1%7Dn%5Cright%29%5Ek%7D%7B%28k%2B1%29%2B%5Ccdots%2B%5Cfrac1%7Bn%5Ek%7D%7D)
So you can see that, by comparison, we have
![\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\lim_{n\to\infty}\frac1{k+1}=\frac1{k+1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%28k%2Bn%2B1%29%5Ek%7D%7B%28n%2B1%29%5E%7Bk%2B1%7D-n%5E%7Bk%2B1%7D%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac1%7Bk%2B1%7D%3D%5Cfrac1%7Bk%2B1%7D)
so this is the value of the limit.
Answer:
<h2><em>
no,set of side lengths does not satisfy triangle inequality theorem</em></h2>
Step-by-step explanation: