SOLUTION: MO bisects < LMN, m < LMO = 6x - 27 and m < NMO = 2x+33 solve for X and find m < LMN

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Question 1092388: MO bisects < LMN, m < LMO = 6x - 27 and m < NMO = 2x+33 solve for X and find m < LMN
Found 2 solutions by richwmiller, josgarithmetic:
Answer by richwmiller(17219) About Me  (Show Source):
Answer by josgarithmetic(39631) About Me  (Show Source):
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Angles LMO and NMO are congruent, of equal measures, so 6x-27=2x%2B33.

4x=33%2B27
4x=60
x=15
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