Answer:
I think the anser maybe 17 don't trust me bc im not quit sure
If the endpoints of a diameter are (6,3) and (2,1) the midpoint is the center of the circle so:
(x,y)=((6+2)/2, (3+1)/2)=(4,2)
Now we need to find the radius....the diameter is:
d^2=(6-2)^2+(3-1)^2
d^2=16+4
d^2=20 since d=2r, r=d/2, and r^2=d^2/4 so
r^2=5
The standard form of the circle is (x-h)^2+(y-k)^2=r^2 and we know:
(h,k)=(4,2) from earlier so:
(x-4)^2+(y-2)^2=5
Answer:
f(x) = (8/3)x(x - 1).
Step-by-step explanation:
The vertex form of the function is
y = a(x - 1/2)^2 - 2/3 where a is some constant.
Now the roots are 0 and 1 so
For x = 0
0 = a(-1/2)^2 - 2/3
1/4 a = 2/3
a = 2/3 / 1/4 = 8/3.
So the function is
f(x) = (8/3)(x - 1/2)^2 - 2/3.
3 f(x) = 8(x - 1/2)^2 - 2
= 8(x^2 - x + 1/4) - 2
= 8x^2 - 8x + 2 - 2
3 f(x) = 8x(x - 1)
f(x) = (8/3)x(x - 1).