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3 years ago

Please help will mark brainliest

Mathematics

1 answer:

tamaranim1 [39]3 years ago

4 0

Answer:

8 miles I think

Step-by-step explanation:

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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

Wat is 1+1 ummmmmmmmmmmmmmmm i dont know

Lisa [10]

Answer:

Step-by-step explanation:

7 0

3 years ago

Can anyone explain how to solve this?

eduard

I’m not 100 percent sure someone can comment on this and confirm but when it’s written like a fraction it means division. If they gave you the meaning of each variable (the letters) multiply each variable by each coefficient (the number next to the variable). Multiply all of those. Then divide it by the product of 4 and the variable

Hope this helped!

8 0

3 years ago

Day Care Tuition A random sample of 50 four-year-olds attending day care centers provided a yearly tuition average of $3987 and

olasank [31]

Answer:

We have to find the 90% confidence interval for the mean

Given sample size

Sample mean

Population standard deviation

Here the confidence level is 90% then

And

The 90% confidence interval is

Here is the critical value at 0.05

From the tables

Now the 90% confidence interval is =

=(3840.44, 4133.56)

Hence the 90% confidence interval for the mean is (3840.44, 4133.56)

5 0

3 years ago

Help this is due in 1 hour .

olga55 [171]

Answer:

Step-by-step explanation:

just out b i dont have time to explain

4 0

3 years ago

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