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3 years ago

I need help on this math problem!

Mathematics

1 answer:

swat323 years ago

8 0

(-7, 2) is the location of W.

because W on the X axis is -7

and W on the Y axis is 2.

making it (-7, 2)

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X* square root of 3 =34

disa [49]

Answer:

17.49286

Step-by-step explanation:

Photomath :)

4 0

3 years ago

Read 2 more answers

A line passes through (-2,5) and has slope. What is an equation of the line in point-slope form?

shepuryov [24]

Answer:

Step-by-step explanation:

since the question does not have a given slope you can just put the point into the point slope formula y-y1 = m(x-x1)

y-5 = m(x-(-2))

y-5 = m(x+2)

since the slope is not given, leave it like that

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3 years ago

Evaluate each of the following values for the f(×)=[×]:

defon

It’s um um.......y......e......a.......h. Yeah

6 0

3 years ago

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

1 1\4x2 1\5 ≤ ≥ = 5\6 x 2 1\5

ValentinkaMS [17]

Answer:

≥

Step-by-step explanation:

1 1\4x2 1\5 = 2 3/4

5/6 x 2 1/5 = 1 5/6

7 0

3 years ago