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December 19, 2014

December 19, 2014

Posted by **kayla** on Tuesday, May 24, 2011 at 3:50pm.

- 11th grade-math -
**Henry**, Wednesday, May 25, 2011 at 4:55pmn^2 = n.

n^2 - n = 0,

n = (-b+-sqrt(b^2-4ac)) / 2a.

n = (1+-sqrt(1-4*1*0)0 / 2,

n = (1+-sqrt(1-0)) / 2,

n = (1+-1) / 2,

n = (1+1) / 2 = 1.

n = (1-1) / 2 = 0/2 = 0.

Solution set: n = 1, and 0.

- 11th grade-math -
**Henry**, Wednesday, May 25, 2011 at 5:06pmCorrection:

n^2 = 6n.

n^2 - 6n = 0,

n = (6 +- sqrt(36-4*1*0)) / 2,

n = (6 +- sqrt(36-0)) / 2,

n = (6 +-sqrt(36)) / 2,

n = (6 +- 6) / 2,

n = (6+6) / 2 = 6.

n = (6-6) / 2 = 0/6 = 0.

Solution set: n = 6, and 0.

- 11th grade-math -
**ChingChongWong**, Thursday, June 2, 2011 at 9:39amYOU FAIL AT LIFE

- 11th grade-math -
**Sneha**, Sunday, June 5, 2011 at 1:03pmn^2= 6n

n^2-6n=0

n(n-6)=0

n=0,6

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