Outfield exploration: 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 centerfielder threw the
ball from here to home base.
9. How far is this throw? (2 points

go on the tube you might find some information there

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point