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ycow [4]
3 years ago
7

Solve the absolute value equation. |X-2|+4=9

Mathematics
1 answer:
Luda [366]3 years ago
5 0

Answer:

x=7\text{ or } x=-3

Step-by-step explanation:

So we have the equation:

|x-2|+4=9

First, subtract 4 from both sides:

|x-2|=5

Definition of Absolute Value:

x-2=5\text{ or } x-2=-5

Add 2 to both sides for both equations:

x=7\text{ or } x=-3

And we're done!

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4 years ago
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

which has roots at r=1\pm i. This admits the two fundamental solutions

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
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u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
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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
7 0
3 years ago
Find two consecutive odd integers whose product is 99
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Second odd number will be x+2

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xsquare +2x - 99 = 0
xsquare +11x - 9x -99 = 0
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So the two odd numbers were 9 and 11
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