Answer:
The polygon will be triangle as shown in attached figure.
Step-by-step explanation:
Considering the points
As there are three points, when we locate the points in the coordinate plane and draw the line segments between them, it is clear that the polygon will be triangle as shown in attached figure.
Therefore, the polygon will be triangle as shown in attached figure.
Answer:
$315,840.00
Step-by-step explanation:
$564,000 x 0.56 = $315,840.00
![\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.
X=7 is the answer when solving for x
Answer:
7 dollars
Step-by-step explanation:
To solve, use this equation:
34.20 - (16 x 1.70) = x
First do 16 x 1.70
This equals 27.2
34.20 - 27.2 = x
Now subtract
34.20 - 27.20 = $7
The student card $7