First - 8
second - 4
third - 12
fourth - 9
Answer:
- B. BC = 18 so ∆ABC ~ ∆DEF by SAS
Step-by-step explanation:

So, ∆ABC ~ ∆DEF by SAS.
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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
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