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Harlamova29_29 [7]
3 years ago
15

Point A is located at (-2, 2), and point M is located at (1,0). If point M is the midpoint of AB, find the location of point B.

Mathematics
1 answer:
bezimeni [28]3 years ago
5 0

Answer:

B. B = (4,-2)

Step-by-step explanation:

GIven that A = (-2, 2) and M = (1, 0), and that point M is the midpoint of AB, the midpoint can be determined as a vectorial sum of A and B. That is:

M = \frac{1}{2}\cdot A + \frac{1}{2}\cdot B

The location of B is now determined after algebraic handling:

\frac{1}{2}\cdot B = M - \frac{1}{2}\cdot A

B = 2\cdot M -A

Then:

B = 2\cdot (1,0)-(-2,2)

B = (2\cdot 1, 2\cdot 0)-(-2,2)

B = (2,0) -(-2,2)

B = (4,-2)

Which corresponds to option B.

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Answer:

2.4 hours

Step-by-step explanation:

Get common denominators -

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4 years ago
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Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

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Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
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So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
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but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
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<h2>I'm always happy to help :)</h2>
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Step-by-step explanation:

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