Answer:
Step-by-step explanation:
take two points on the line ,i take (-4,0) and (0,-2)
slope=(y2-y1)/(x2-x1)=(-2-0)/(0+4)=-2/4=-1/2
Answer:
2520°
Step-by-step explanation:
The sum of the angles in any triangle is 180°
shape. sides. triangles. sum int angles.
triangle. 180×1=180°
quad. 180×2=360°
pentagon. 180×3=540°
n-sides. n( n−2) 180(n−2)
16-sides. 180×14=2520°
Answer:
x=2.75
Step-by-step explanation:
The answer will be correct.
You find the eigenvalues of a matrix A by following these steps:
- Compute the matrix
, where I is the identity matrix (1s on the diagonal, 0s elsewhere) - Compute the determinant of A'
- Set the determinant of A' equal to zero and solve for lambda.
So, in this case, we have
![A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C-2%260%5Cend%7Barray%7D%5Cright%5D%20%5Cimplies%20A%27%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C-2%260%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1-%5Clambda%26-2%5C%5C-2%26-%5Clambda%5Cend%7Barray%7D%5Cright%5D)
The determinant of this matrix is
![\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4](https://tex.z-dn.net/?f=%5Cleft%7C%5Cbegin%7Barray%7D%7Bcc%7D1-%5Clambda%26-2%5C%5C-2%26-%5Clambda%5Cend%7Barray%7D%5Cright%7C%20%3D%20-%5Clambda%281-%5Clambda%29-%28-2%29%28-2%29%20%3D%20%5Clambda%5E2-%5Clambda-4)
Finally, we have
![\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}](https://tex.z-dn.net/?f=%5Clambda%5E2-%5Clambda-4%3D0%20%5Ciff%20%5Clambda%20%3D%20%5Cdfrac%7B1%5Cpm%5Csqrt%7B17%7D%7D%7B2%7D)
So, the two eigenvalues are
![\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}](https://tex.z-dn.net/?f=%5Clambda_1%20%3D%20%5Cdfrac%7B1%2B%5Csqrt%7B17%7D%7D%7B2%7D%2C%5Cquad%20%5Clambda_2%20%3D%20%5Cdfrac%7B1-%5Csqrt%7B17%7D%7D%7B2%7D)
Answer:
Hey... Ans is in the pic..
Hope it helped u if yes mark me mark me BRAINLIEST!
Tysm!
:)