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Elena L [17]
3 years ago
9

Factor each expression and explain how you solved it.

Mathematics
1 answer:
rosijanka [135]3 years ago
7 0
42. g^2+2g-7g-14= g(g+2)-7(g+2) = (g-2)(g+7)
44. 6k^2-6kL-4kL+4L^2= 6k(k-L)-4L(k-L)= (6k-4L)(k-L)
46. r^2-4r+10r-40 = r(r-4)+10(r-4)= (r+10)(r-4)
48. t^2-3t-10t-30 = t(t-3)-10(t-3)= (t-10)(t-3)
50. 3x^2-3x+6x-6 = 3x(x-1)+6(x-1) = (3x+6)(x-1)


REALLY hoping this helps bc this took me some time
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A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2
andrey2020 [161]

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}

Step-by-step explanation:

Consider the provided matrix.

v_1=\begin{bmatrix}-3\\1 \end{bmatrix}

v_2=\begin{bmatrix}-1\\1 \end{bmatrix}

\lambda_1=4, \lambda_2=2

The general solution of the equation x'=Ax

x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}

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}

x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}

Substitute initial condition x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}

\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}

Reduce matrix to reduced row echelon form.

\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}

Therefore, c_1=2.5,c_2=1.5

Thus, the general solution of the equation x'=Ax

x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}

x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}

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}

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