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

How far is the throw?

go and ask your mom to you

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)

step-by-step explanation: