Answer:
(4,1)
Step-by-step explanation:
let J(6,6) be x1, y1
let K(2,-4) be x2, y2
midpoint = (x1+x2)/2 , (y1+y2)/2
=(6+2)/2 ,(6-4)/2
= 8/2 ,2/2
=4, 1
Hi,
Work:
Equation;

Write all numerators above the least common denominator 32.

Calculate the sum of positive numbers.

Simplify (FRACTION RESULT)

Or (DECIMAL RESULT)

Hope this helps.
r3t40
Find the zeros<span> of Function (</span>x)= (x+3)^2(x-5)^6 and state the multiplicity. ... (x--5)^6<span> = 0=> </span>x--5<span>= 0 => </span>x<span> = </span>5<span>. Therefore real </span>zero's of the function<span> are --</span>3<span>,5.</span><span />