Answer:
Step-by-step explanation:
The solution of the system of equations contains one point
<h3>How to determine the number of solutions?</h3>
The system is given as:
x + y = 6
x - y = 0
Add both equations
2x = 6
Divide by 2
x = 3
Substitute x = 3 in x - y = 0
3 - y = 0
Solve for y
y = 3
So, we have x =3 and y = 3
Hence, the solution of the system of equations contains one point
Read more about system of equations at:
brainly.com/question/14323743
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You have to use this equation the a=5, h=4 and b=9 and you should get the answer of 28
FYI, that L in the denominator is factorial.
This is some pretty serious stuff for high school.
My inclination is that the answer is A, which is the definition of <em>e. </em> Let's see if we can show this.
Let's write
![\displaystyle f(n) = \left(1 + \dfrac 1 n \right)^n](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28n%29%20%3D%20%5Cleft%281%20%2B%20%5Cdfrac%201%20n%20%5Cright%29%5En)
Let's expand this with the binomial expansion
![\displaystyle f(n) =\sum_{k=0}^n {n \choose k} \dfrac{1}{n^k}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28n%29%20%3D%5Csum_%7Bk%3D0%7D%5En%20%7Bn%20%5Cchoose%20k%7D%20%5Cdfrac%7B1%7D%7Bn%5Ek%7D)
![\displaystyle f(n) =\sum_{k=0}^n \dfrac{n!}{k!(n-k)!} \cdot \dfrac{1}{n^k}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28n%29%20%3D%5Csum_%7Bk%3D0%7D%5En%20%5Cdfrac%7Bn%21%7D%7Bk%21%28n-k%29%21%7D%20%5Ccdot%20%5Cdfrac%7B1%7D%7Bn%5Ek%7D)
![\displaystyle f(n) =\sum_{k=0}^n \dfrac{n(n-1)\cdots(n-k+1)}{k! \, n^k}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28n%29%20%3D%5Csum_%7Bk%3D0%7D%5En%20%5Cdfrac%7Bn%28n-1%29%5Ccdots%28n-k%2B1%29%7D%7Bk%21%20%5C%2C%20n%5Ek%7D)
Let's focus on when n is really big and on the ks that are relatively small, which make up the bulk of the sum as the terms get small rapidly.
Then that numerator n(n-1)···(n-k+1) ≈ n^k as all the factors are about n.
![\displaystyle f(n)\approx \sum_{k=0}^n \dfrac{n^k}{k!\, n^k}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28n%29%5Capprox%20%5Csum_%7Bk%3D0%7D%5En%20%5Cdfrac%7Bn%5Ek%7D%7Bk%21%5C%2C%20n%5Ek%7D)
![\displaystyle f(n)\approx \sum_{k=0}^n \dfrac{1}{k!}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28n%29%5Capprox%20%5Csum_%7Bk%3D0%7D%5En%20%5Cdfrac%7B1%7D%7Bk%21%7D)
OK, we showed for large n this is approximately true, and it will be exactly true in the limit, so we choose
Answer: A
Answer:
$404.17
Step-by-step explanation:
(19700 - 19700/2 -5000) / 12 =