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AfilCa [17]
3 years ago
9

Please help me ASAP!!! I will give 20 points..... bad answers will be reported

Mathematics
1 answer:
belka [17]3 years ago
3 0
150? I think thats the anwser
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Which equation matches the graph?
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Step-by-step explanation:

The slope is -2/1

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2x – 4y = 14<br> x = 2y – 2<br><br> What is the solution to the system?
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No Solution.

Step-by-step explanation:

2x-4y=14

x=2y-2

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2(2y-2)-4y=14

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How do u do this problem: W/6+5/8=-1 3/8 <br> I got this answer: -1 53/64
Kobotan [32]
We have question
\frac{W}{6}+ \frac{5}{8} =-1 \frac{3}{8}    
\frac{W}{6} + \frac{5}{8} =- \frac{11}{8}  <span> /*8      (multiply both sides by 8)</span>
\frac{W8}{6} + \frac{5}{1} =- \frac{11}{1}  /*6      (multiply both sides by 6)
8W+5*6=-11*6      /-30 both sides
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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}

6 0
3 years ago
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