Write an inequality that represents the verbal expression. All real numbers greater than or equal to 38.

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

2 years ago

Which equation is a point slope form equation for line AB ?

zepelin [54]

(-2,3)(4,-1)
slope = (3 + 1)/(-2 - 4) = 4/-6 = -2/3

equation
y + 1 = -2/3(x - 4)

7 0

3 years ago

Hope researches the impact of one variable on another. She graphs the data from her research and sees that the data forms a shap

zaharov [31]

Answer:

She can determine that while there is an association between the variables, there is no correlation, and she cannot determine causation.

Step-by-step explanation:

Since there is an arch shape to her graph, we know that as one variable changes, the other changes in the same manner. This means there is an association between the variables.

However, since the graph is not linear, there is no correlation between the variables. Since there is no correlation, we cannot determine causation.

Mark me as brainiest

6 0

2 years ago

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A share of stock dropped 6 points on Tuesday. On Wensday, the same stock gained 3 points. What was the total loss for the two da

Bogdan [553]

The answer is three (3)

3 0

2 years ago

Please help me out with this so I can keep my kitten.

notka56 [123]

Answer:

$y = \frac{32}{3}$

Step-by-step explanation:

Firstly, move over the negative 3/4 fraction (don't forget to swap the operation i.e subtract to add):

$\frac{y}{8} = \frac{7}{12} + \frac{3}{4}$

Now, to add the two fractions, simply multiply the numerator and denominator by 3:

$\frac{3*3}{4*3} = \frac{9}{12}$

Now add this to the other fraction:

$\frac{9}{12} + \frac{7}{12} = \frac{16}{12}$

This can be simplified down by dividing both the numerator and denominator by 4:

$\frac{4}{3}$

Which now simplifies the original equation to:

$\frac{y}{8} = \frac{4}{3}$

Remove the y out of the fraction:

$\frac{1}{8}y = \frac{4}{3}$

Now multiply both sides by 8:

$(\frac{1}{8}y) * 8 = (\frac{4}{3}) * 8$

$y = \frac{4*8}{3}$

$y = \frac{32}{3}$

Hope this helps!

5 0

2 years ago

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