Step-by-step explanation:
4(2x-1)+3(2x+5)
8x-4+6x+15
8x+6x-4+15
14x+11
Answer:
k=8
Step-by-step explanation:
yes
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
Step-by-step explanation:
x² + 3x + 2
x² + x + 2x + 2 because 1+2=3 and 2×1=2
By taking out common terms, we get
x(x+1)+2(x+1)
(x+1)(x+2)
Answer:
y(-4) = 5
y'(-4) = -7
Step-by-step explanation:
Hi!
Since the tangent line T and the curve y must coincide at x=-4
y(-4) = T(-4) = 5
On the other hand, the derivative of the curve evaluated at -4 y'(x=-4) must be the slope of the tangent line. Which inspecting the tangent line T(x) is -7
That is:
y'(-4) = -7
