Monday

September 1, 2014

September 1, 2014

Posted by **lilboo** on Friday, September 24, 2010 at 8:15am.

m<AOB=4x-2, m<BOC=5x+10 m<COD=3x-8

- geometry -
**Aish T**, Wednesday, August 24, 2011 at 9:48pmm<AOE = 180 and M<BOC and m<DOE are congruent angles

So

m<AOB + m<BOC + m<COD + m<DOE = 180

(4x-2)+(5x+10)+(3x-8)+(5x+10) = 180

17x+10=180

17x = 170

x = 170

- geometry -
**Raven Wood**, Sunday, September 2, 2012 at 3:56pmm<AOE = 180 and M<BOC and m<DOE are congruent angles

So

m<AOB + m<BOC + m<COD + m<DOE = 180

(4x-2)+(5x+10)+(3x-8)+(5x+10) = 180

17x+10=180

17x=170

x=10

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