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3 years ago

6

Solve the system of equations for -2x+2y=4

1 answer:

Greeley [361]3 years ago

8 0

You need another equation to be able to solve this

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Please help right now!!!!! Solve the system of equations below. x + y = 7 2x + 3y = 16

motikmotik

Answer:

$x=5$ and $y=2$ .

Step-by-step explanation:

We have been given a system of equations. We are asked to solve our given system.

$x+y=7...(1)$

$2x+3y=16...(2)$

From equation (1), we will get:

$x=7-y$

Upon substituting this value in equation (2), we will get:

$2(7-y)+3y=16$

$14-2y+3y=16$

$14+y=16$

$14-14+y=16-14$

$y=2$

Now, we will substitute $y=2$ in equation (1).

$x+2=7$

$x+2-2=7-2$

$x=5$

Therefore, the point $(5,2)$ is solution for our given equation.

7 0

3 years ago

Calculate the measure of segment KL<br><br><br> HELP PLS

iVinArrow [24]

10x+3 = 3x+24

10x-3x= 24-3

7x=21

X= 21/7

Final answer:

X=3

I’m not so sure of my answer but wish you the best and good luck!!

5 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

I posted a question similar to this but I entered it wrong too many times :(

slava [35]

Answer:

$\sqrt[4]{\frac{5}{7} }$

Step-by-step explanation:

^4√5/^4√7

~Apply radical rules

^4√5/7

Best of Luck!

5 0

3 years ago

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AlekseyPX

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3 years ago

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