Answer: 52
Use the pythagorean theorum to calculate WX, since it's a rhombus, all sides are equal, multiply by four, finished.
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:
4.9h−2.7d−13
Step-by-step explanation:
Picture?? This question is incomplete just get a picture then I can answer it
Answer:
h(2) = 7/4
h(-3) = -2
h(-2) = 11/(-8)
h(-3) - h(-2) = -(5/8)
Step-by-step explanation:
h(x) = (2x^2-x+1) / (3x-2)
h(2) = (2*2^2 - 2 + 1) / (3*2 - 2)
= (8 - 2 + 1) / (6 - 2)
= 7/4
h(-3) = {2*(-3)^2 - (-3) + 1} / {3*(-3) - 2}
= (18 + 3 + 1) / (-9 - 2)
= 22/(-11)
= -2
h(-2) = {2*(-2)^2 - (-2) + 1} / {3*(-2) - 2}
= (8 + 2 + 1) / (-6 - 2)
= 11/(-8)
h(-3) - h(-2) = (-2) - {11/(-8)}
= -(5/8)
Hope this will help. Please give me brainliest.
