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Ira Lisetskai [31]
3 years ago
12

3x-2y=13 2x+3y=-16 are these lines parallell or perpendicular or neither

Mathematics
1 answer:
Lelechka [254]3 years ago
6 0
3x - 2y = 13
-2y = -3x + 13
y = 3/2x - 13/2...slope is 3/2, y int is -13/2

2x + 3y = -16
3y = -2x - 16
y = -2/3x - 16/3...slope = -2/3, y int is -16/3

perpendicular lines will have negative reciprocal slopes. To find the negative reciprocal of a number, flip the number and change the sign.
On the 1st equation, the slope was 3/2....find its negative reciprocal.....flip it, making it 2/3....change the sign...making it -2/3. So the negative reciprocal of 3/2 is -2/3. 

So if one equation has a slope of 3/2 and another equation has a slope of -2/3, then these lines are perpendicular <==
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By definition of covariance,


\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]


\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]


We have


\mathbb E[(aX-b)(cY-d)]=\mathbb E[acXY-adX-bcY+bd]

=ac\mathbb E[XY]-ad\mathbb E[X]-bc\mathbb E[Y]+bd


\mathbb E[aX-b]=a\mathbb E[X]-b


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\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd


Putting everything together, we find the covariance reduces to


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