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4 years ago

Geometry question slope intercept form

Mathematics

2 answers:

Anuta_ua [19.1K]4 years ago

8 0

The answer would be y = 2x-16

kirill [66]4 years ago

5 0

Okay so a perpendicular line has a slope of the opposite reciprocal so for example in this equation the slope is 2 so the slope of a perpendicular equation would be -1/2 now since you know that you can put that in the slope intercept equation
Now you have y=-1/2x+b and since you have a point (point A) you can plug it in and solve for b
-8=-1/2(4)+b
-8=-2+b
b=-6
Now you have both parts simply put it in and you have the equation
y=-1/2x-6
I also attached the image with both to prove they are perpendicular

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4 years ago

Consider the equation ay'' + by' + cy = d, where a, b, c, and d are constants. (a) find all equilibrium, or constant, solutions

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Rewrite equation so that it is homogeneous .
$ay'' +by' + cy = 0$

Solve characteristic equation:
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The solutions to the homogeneous equation are:
$y = k_1 e^{r_1 t} + k_2 e^{r_2 t}$

Finally you need the constant term in "cy" to equal 'd' to satisfy the particular solution.
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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:

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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, $c_1=2.5,c_2=1.5$

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

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

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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vova2212 [387]

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Divide the expression for the fifth term by the expression for the second term:

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Hence, the explicit rule for the sequence is:

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3 years ago

Read 2 more answers