Tre and Hector want to calculate the maximum possible throw at this field. They calculate that the farthest point on the field would be center fielder standing back at the dead center wall at point (322,322) Suppose the centerfielder threw the ball from here to home base. How far is this throw?

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

55/i dont know i do it for the points

step-by-step explanation:

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation:

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