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Andru [333]
3 years ago
14

Help me with this!!!

Mathematics
1 answer:
geniusboy [140]3 years ago
4 0

Answer:

-7

Step-by-step explanation:

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How do i solve this equation
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Your starting equation is: x/6 + 4 = 15 The first step would be to subtract 4 from both sides, to move all of the constants to the right side of the equation. x/6 = 11 The final step is to multiply both sides by 6, to get rid of the fraction on the left side of the equation. x = 66 Hope this helps!
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A halyard 16ft long is cut into two pieces. If the L is the length of one piece, express the length of the second piece in terms
jeka94
If you cut length L from the 16 ft halyard, you are subtracting the value of L from 16. This means that the second piece should be 16-L feet long.
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4 0
3 years ago
Write the linear system of differential equations in matrix form then solve the system.
zvonat [6]

In matrix form, the system is

\dfrac{\mathrm d}{\mathrm dt}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&1\\4&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}

First find the eigenvalues of the coefficient matrix (call it \mathbf A).

\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}1-\lambda&1\\4&1-\lambda\end{vmatrix}=(1-\lambda)^2-4=0\implies\lambda^2-2\lambda-3=0

\implies\lambda_1=-1,\lambda_=3

Find the corresponding eigenvector for each eigenvalue:

\lambda_1=-1\implies(\mathbf A+\mathbf I)\vec\eta_1=\vec0\implies\begin{bmatrix}2&1\\4&2\end{bmatrix}\begin{bmatrix}\eta_{1,1}\\\eta_{1,2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}

\lambda_2=3\implies(\mathbf A-3\mathbf I)\vec\eta_2=\vec0\implies\begin{bmatrix}-2&1\\4&-2\end{bmatrix}\begin{bmatrix}\eta_{2,1}\\\eta_{2,2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}

\implies\vec\eta_1=\begin{bmatrix}1\\-2\end{bmatrix},\vec\eta_2=\begin{bmatrix}1\\2\end{bmatrix}

Then the system has general solution

\begin{bmatrix}x\\y\end{bmatrix}=C_1\vec\eta_1e^{\lambda_1t}+C_2\vec\eta_2e^{\lambda_2t}

or

\begin{cases}x(t)=C_1e^{-t}+C_2e^{3t}\\y(t)=-2C_1e^{-t}+2C_2e^{3t}\end{cases}

Given that x(0)=1 and y(0)=2, we have

\begin{cases}1=C_1+C_2\\2=-2C_1+2C_2\end{cases}\implies C_1=0,C_2=2

so that the system has particular solution

\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}e^{3t}\\2e^{3t}\end{bmatrix}

8 0
3 years ago
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