1). Make your equation equal to 0
2). Then plug in your numbers to the quadratic equation
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

Answer:
The product of the y-coordinates of the solutions is equal to 3
Step-by-step explanation:
we have
-----> equation A
------> equation B
Solve by graphing
Remember that the solutions of the system of equations are the intersection point both graphs
using a graphing tool
The solutions are the points (2,3) and (6,1)
see the attached figure
The y-coordinates of the solutions are 3 and 1
therefore
The product of the y-coordinates of the solutions is equal to
(3)(1)=3
Answer:
Step-by-step explanation:
30).
degree ( 4 + 1 = <em>5</em> ; 1 + 1 = 2 and 2 + 2 = 4 )
31) D) Choice D
Answer:
c
Step-by-step explanation: