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tino4ka555 [31]
3 years ago
10

Which of the equations below represents a line perpendicular to the y-axis?

Mathematics
2 answers:
timurjin [86]3 years ago
7 0

Answer:

B. y = -6

Step-by-step explanation:

As you can see the graph I done is perpendicular

Olenka [21]3 years ago
4 0

Answer:

B

Step-by-step explanation:

A line perpendicular to the y- axis is a horizontal line parallel to the x- axis, with equation

y = c ( where c is the value of the y- coordinates the line passes through )

The only equation in this form is

y = - 6 → B

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x = 27

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2/3x -2= 16

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2/3x = 18

multiply by 3 on both sides

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divide by two on both sides

x= 27

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3y''-6y'+6y=e*x sexcx
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From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

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y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
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