Write the equation of the sphere in standard form. x2 + y2 + z2 + 6x − 2y − 6z = 17

We are given equation of sphere x^2 + y^2 + z^2 + 6x − 2y − 6z = 17.

We know , standard form of a sphere (x−a)^2+(y−b)^2+(x−c)^2=r^2.

Where centered at (a,b,c) ( a , b , c ) with radius r.

Let us convert given equation of sphere in standard form by completing the square.

Writing x, y and z terms seprately.

(x^2+6x) + (y^2-2y) + (z^2-6z) = 17.

We have coefficents of x, y and z are 6, -2 and -6.

First we will find half of those coefficents and then square them.

We get 6/2 =3, -2/2 =-1 and -6/2=-3.

Now, squaring we get

(3)^2 = 9,  (-1)^2 = 1 and (-3)^2 = 9.

Adding those numbers inside first, second and third parentheses and adding same numbers on right side.

(x^2+6x+9) + (y^2-2y+1) + (z^2-6z+9) = 17+9+1+9.

Now, finding perfect square of each, we get

(x+3)^2 + (y-1)^2 + (z-3)^2 = 36.

We could write 36 as 6^2.

So, final answer would be

(x+3)^2 + (y-1)^2 + (z-3)^2 = 6^2 is the standard form of a sphere equation.

with btw u said 100pts but it clearly says

i answerd and just got 10pts

no hate

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6