First we need to find the gradient of K
which is y1-y2/x1-x2
(-1,3) and (5,-2)
so it becomes 3-(-2)/-1-5
m=-5/6
when two lines are perpendicular their gradients multiply to make -1
that means the gradient of L has to be 6/5
we can substitute the point on L (5,-2) and the gradient of 6/5 into y=mx+c
-2 = (6/5) x 5 + c
c = -8
the equation of line L is y= 6x/5 -8
Answer:
A. 2612
cause 2612 is a even number
Answer: D
Step-by-step explanation:
Answer:
#1. A. C. D.
#2. D. F.
3^4 equals 81 and so do answer D. & F.
We can use 3 facts:



So, to find our x and our y we can do:
So, now that we know our x and our y we can find our r.

You'll notice, though, that there were two solutions for
, which means we have a positive and a negative for our x. So:
