Answer:

Step-by-step explanation:
<u>Eigenvalues of a Matrix</u>
Given a matrix A, the eigenvalues of A, called
are scalars who comply with the relation:

Where I is the identity matrix
![I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
The matrix is given as
![A=\left[\begin{array}{cc}3&5\\8&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D)
Set up the equation to solve
![det\left(\left[\begin{array}{cc}3&5\\8&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda \end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)
Expanding the determinant
![det\left(\left[\begin{array}{cc}3-\lambda&5\\8&-\lambda\end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3-%5Clambda%265%5C%5C8%26-%5Clambda%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)

Operating Rearranging

Factoring

Solving, we have the eigenvalues

Hello There!
2² - 6 = -2
3² - 6 = 3
4² - 6 = 10
Etc....the list goes on
Hope This Helps You!
Good Luck :)
- Hannah ❤
Answer:
1.649 approximately 2
Step-by-step explanation:
S.d = standard deviation = 0.5
Time taken = lead time = 2 weeks
Mean = demand for week = 5 boxes
We are required to find the safety stock to maintain at 99% service level.
At 99% level, the Z value is equal to 2.326.
Therefore,
Safety stock = z × s.d × √Lt
= 2.326 × 0.5 x √2
= 1.649
Which is approximately 2.
Answer:
<h2>125 tickets for an adult</h2><h2>and 175 tickets for a student</h2>
Step-by-step explanation:


Answer:

Step-by-step explanation:
The sum you are trying to understand is this.

Remember that in general when you have a geometric series
you have that
and that equality is true as long as
.
Therefore here we have
and 
Therefore we can use the formula and
