ProffesorNobody

ohhh sorry... :(

Ok, finally figured it out-

so we have x^2+y^2+2y+6x-6=0

then we put the x wuith the x and the y with the ys. 

x^2+6x+y^2+2y=6

then we complete the square for both x and y

x^2+6x+9+y^2+2y+1=6+9+1

simplify

(x+3)^2+(y+1)^2=16

x=-3

y=-1

radius=4

ok wow i can read it now..... but im confused now

right? can somebody check my work?

Got it: We take the absolute value of the coefficient which subsequently leads to 5. And that's the answer.

Ok so Let us start with the smallest numbers that some can be...

which is naturally 1,2,3,4,5.

then, we place a ten down, and make the 4 next integers: 11,12,13,14,

1,2,3,4,5,10,11,12,13,14

calculating the average: 1+2+3+4+5=15

                                       10+11+12+13+14=60

15+60=65

65/10

6.5

actually it doesn't help...

maybe this helps: 

\(Let \omega be a nonreal root of z^3 = 1. Let a_1, a_2, \dots, a_n be real numbers such that \[\frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \dots + \frac{1}{a_n + \omega} = 2 + 5i.\]Compute \[\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \dots + \frac{2a_n - 1}{a_n^2 - a_n + 1}.\]\)

nope

I want to know how thou... I kinda guessed there..... :P