Answer:
The required position of the particle at time t is: ![x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}](https://tex.z-dn.net/?f=x%28t%29%3D%5Cbegin%7Bbmatrix%7D-7.5e%5E%7B4t%7D%2B1.5e%5E%7B2t%7D%5C%5C2.5e%5E%7B4t%7D-1.5e%5E%7B2t%7D%5Cend%7Bbmatrix%7D)
Step-by-step explanation:
Consider the provided matrix.
![v_1=\begin{bmatrix}-3\\1 \end{bmatrix}](https://tex.z-dn.net/?f=v_1%3D%5Cbegin%7Bbmatrix%7D-3%5C%5C1%20%5Cend%7Bbmatrix%7D)
![v_2=\begin{bmatrix}-1\\1 \end{bmatrix}](https://tex.z-dn.net/?f=v_2%3D%5Cbegin%7Bbmatrix%7D-1%5C%5C1%20%5Cend%7Bbmatrix%7D)
![\lambda_1=4, \lambda_2=2](https://tex.z-dn.net/?f=%5Clambda_1%3D4%2C%20%5Clambda_2%3D2)
The general solution of the equation ![x'=Ax](https://tex.z-dn.net/?f=x%27%3DAx)
![x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}](https://tex.z-dn.net/?f=x%28t%29%3Dc_1v_1e%5E%7B%5Clambda_1t%7D%2Bc_2v_2e%5E%7B%5Clambda_2t%7D)
Substitute the respective values we get:
![x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}](https://tex.z-dn.net/?f=x%28t%29%3Dc_1%5Cbegin%7Bbmatrix%7D-3%5C%5C1%20%5Cend%7Bbmatrix%7De%5E%7B4t%7D%2Bc_2%5Cbegin%7Bbmatrix%7D-1%5C%5C1%20%5Cend%7Bbmatrix%7De%5E%7B2t%7D)
![x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}](https://tex.z-dn.net/?f=x%28t%29%3D%5Cbegin%7Bbmatrix%7D-3c_1e%5E%7B4t%7D-c_2e%5E%7B2t%7D%5C%5Cc_1e%5E%7B4t%7D%2Bc_2e%5E%7B2t%7D%20%5Cend%7Bbmatrix%7D)
Substitute initial condition ![x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}](https://tex.z-dn.net/?f=x%280%29%3D%5Cbegin%7Bbmatrix%7D-6%5C%5C1%20%5Cend%7Bbmatrix%7D)
![\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D-3c_1-c_2%5C%5Cc_1%2Bc_2%20%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D-6%5C%5C1%20%5Cend%7Bbmatrix%7D)
Reduce matrix to reduced row echelon form.
![\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D%201%26%200%20%26%20%5Cfrac%7B5%7D%7B2%7D%5C%5C%200%26%201%20%26%20%5Cfrac%7B-3%7D%7B2%7D%5Cend%7Bbmatrix%7D)
Therefore, ![c_1=2.5,c_2=1.5](https://tex.z-dn.net/?f=c_1%3D2.5%2Cc_2%3D1.5)
Thus, the general solution of the equation ![x'=Ax](https://tex.z-dn.net/?f=x%27%3DAx)
![x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}](https://tex.z-dn.net/?f=x%28t%29%3D2.5%5Cbegin%7Bbmatrix%7D-3%5C%5C1%5Cend%7Bbmatrix%7De%5E%7B4t%7D-1.5%5Cbegin%7Bbmatrix%7D-1%5C%5C1%20%5Cend%7Bbmatrix%7De%5E%7B2t%7D)
![x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}](https://tex.z-dn.net/?f=x%28t%29%3D%5Cbegin%7Bbmatrix%7D-7.5e%5E%7B4t%7D%2B1.5e%5E%7B2t%7D%5C%5C2.5e%5E%7B4t%7D-1.5e%5E%7B2t%7D%5Cend%7Bbmatrix%7D)
The required position of the particle at time t is: ![x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}](https://tex.z-dn.net/?f=x%28t%29%3D%5Cbegin%7Bbmatrix%7D-7.5e%5E%7B4t%7D%2B1.5e%5E%7B2t%7D%5C%5C2.5e%5E%7B4t%7D-1.5e%5E%7B2t%7D%5Cend%7Bbmatrix%7D)
The correct solution: 4x^2 -27x +167 - 996/(x+6)
Step-by-step explanation:
Make bottom numbers same
convert to improper fraction
1 and 5/6=1+5/6=6/6+5/6=11/6
1/2 and 11/6
2 times 3=6
1/2 times 3/3=3/6
1/2+11/6=3/6+11/6=(4+11)/6=14/6
answer is A
Answer: 13.
Simply Count between Coordinates
Which Adds To 13.
Brainly Please.
Answer:
144
Step-by-step explanation:
6^2 = 6x6 = 36
36 x 4 = 144