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3 years ago

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The figure shows a pair of parallel line segmants on a coordinant grid. the line segments are translated 2 units to the left to

form A'B' and C'D'. which statement describes A'B' and C'D'?
A) line segment A'B' and C'D' do not intersect and are closer together than AB and CD.
B) line segment A'B' and C'D' intersect at (-2,0) and are two times farther apart than AB and CD.
C) line segment A'B' and C'D' intersect at (0,-2) and are two times farther apart than AB and CD
D) line segment A'B' and C'D' do not intersect and are the same distance apart as AB and CD

2 answers:

Nitella [24]3 years ago

6 0

It is D because the segments don’t intersect and are just moving 2 units to the left. They are the same distance apart.

Hope it helps and is correct :)

quester [9]3 years ago

4 0

The answer is D.) I just took the test and got it correct.

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3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

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The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

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Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

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