Consider the line y=5x-3. Find the equation of the line that is parallel to this line and passes through the point (2, -5).
Find the equation of the line that is perpendicular to this line and passes through the point (2, -5).

y=5x-15, y=-1/5x-23/5

Step-by-step explanation:

Parallel lines have the same slope, so we know the line that is parallel to y=5x-3 must have slope 5.

We now have y=5x+b. We are given the point (2, -5) so substitute the x and y coordinates into the equation.

y=5x+b

-5=5(2)+b

-5=10+b

b=-15

So the equation of the parallel line is y=5x-15.

Perpendicular lines must have slopes that multiply to -1, so we know the line perpendicular to y=5x-3 must have slope -1/5.

We now have y=-1/5x+b. We are given the point (2, -5) so substitute the x and y coordinates into the equation.

y=-1/5x+b

-5=-1/5(2)+b

-5=-2/5+b

b=-23/5

So the equation of the perpendicular line is y=-1/5x-23/5.

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.

why can't you?

step-by-step explanation: