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awesomeness123
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I WILL FAN/GIVE YOU A MEDAL/WRITE A TESTIMONIAL FOR YOU IF YOU HELP ME!! ANSWERS BELOW~!
Select the equations that are parallel and perpendicular to y = x + 5 and that pass through the point (2, 1).
 one year ago
 one year ago
awesomeness123 Group Title
I WILL FAN/GIVE YOU A MEDAL/WRITE A TESTIMONIAL FOR YOU IF YOU HELP ME!! ANSWERS BELOW~! Select the equations that are parallel and perpendicular to y = x + 5 and that pass through the point (2, 1).
 one year ago
 one year ago

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awesomeness123 Group TitleBest ResponseYou've already chosen the best response.0
parallel: y = x  1 perpendicular: y = x + 2 parallel: y = x  1 perpendicular: y = x + 1 parallel: y = x + 1 perpendicular: y = x  3 parallel: y = 2x  2 perpendicular: y = 2x  1
 one year ago

awesomeness123 Group TitleBest ResponseYou've already chosen the best response.0
@jim_thompson5910
 one year ago

awesomeness123 Group TitleBest ResponseYou've already chosen the best response.0
@phi
 one year ago

awesomeness123 Group TitleBest ResponseYou've already chosen the best response.0
@saifoo.khan
 one year ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.2
A line parallel to: y = x + 5 will have same slope.
 one year ago

awesomeness123 Group TitleBest ResponseYou've already chosen the best response.0
so do i have to like use yy1=m(xx1) for all of them to find which one has the same slope
 one year ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.2
Yes. One equation can be found out by: y  (1) = 1(x(2))
 one year ago

awesomeness123 Group TitleBest ResponseYou've already chosen the best response.0
then in case there is two with the same slope, the reciprocal will be the slope of the perpendicular line right?
 one year ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.2
The first one will be with same slope. The second one will be the reciprocal one. Which will be 1.
 one year ago

awesomeness123 Group TitleBest ResponseYou've already chosen the best response.0
thanks!
 one year ago
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