Q. In the figure (not drawn to scale), ray MO bisects ∠LMN, m∠LMO = (15x - 21)° and m∠NMO = (x + 63)°. Solve for x and find m∠LMN.
If MO bisects LMN, then LMO and NMO are equal. So 8x-23=2x+37. 8x-23=2x+ 37. 6x=60. x=10. LMN=8x-23+2x+37. LMN=10x+14.
The diagram is not to scale. a. 6x – 9 b. 6x – 18 c.
m∠FEG = 122, m∠HEG = 58 c. m∠FEG = 68, m∠HEG = 122 b. m∠FEG = 58, m∠HEG = 132 d. m∠FEG = 58, m∠HEG = 122 ____ 30. Name an angle supplementary to ∠EOD. a. ∠BOC b.
line MO bisects angle LMN, angle LMN = 6x - 28, angle LMO = x + 34. Find angle NMO. The diagram is not to scale. 58.
1.5 x – 4.5 ____ 18. MO → bisects ∠LMN, m∠LMO = 6x − 22, and m∠NMO = 2x + 34. Solve for x and find m∠LMN. The diagram is not to scale.
I'm having a hard time picturing this. It seems that some of the letters may be mixed up. Should this be: Line MO bisects angle LMN? Such that angle LMO and angle NMO are adjacent? If that is the case, then there is a solution for x. But there is no angle LNO in that case.
Find the distance between points P(7, 3) and Q(2, 5) to the nearest tenth. MO. →. bisects ∠LMN, m∠LMN = 6x − 22, m∠LMO = x + 31. 28. Write an equation in slope-intercept form of the line through point P(10, 5) with slope 6x = 18 d. ____ x = 3 e. ____.
geometry. in triangle lmn altitude lk is 12cm long through point j of lk is a line drawn parallel to ms, dividing the triangle into two region with equal areas find lj . geometry
1 Answer to Ray MO bisects
D) △JKL ≅ △MNL.
2(42-x) = 6*+22 X= 7.76 6x+19+8=180 5p+28=12p. MO bisects ∠LMN, m∠LMO =8x 25, and m∠NMO =2x Solve for x and find 8x 25 = 2x + 41 6x=41+25=66 x=11 8 11 −25=63 =63 So the whole angle 28 14. Find the distance between points P(7, 3) and Q(2, 5) to the nearest tenth.
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28. 7. 29. 12. 30. PROOF Use algebra to prove the Exterior Angle Sum Theorem. 31. m∠LMN = 72 Subtract 108 from each side. −−−. LM Since the diagonals of a parallelogram bisect each other, the intersection If MO = 6x + 14 and.
58. 1 and 2 are a linear pair. m 1 = x - 30, and m 2 = x + 76. Find the Ray MO bisects angle LMN, measure of angle LMN equals 6x - 28, measure of angle LMO equals x + 32.
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Practice 2B 1. Ray MO bisects ∠LMN, m∠LMO=8x−23, and m∠NMO=2x+37. Solve for x and find m∠LMN. The diagram is not to scale. A. x=9, m∠LMN=98
in triangle lmn altitude lk is 12cm long through point j of lk is a line drawn parallel to ms, dividing the triangle into two region with equal areas find lj .
If MO bisects LMN, then LMO and NMO are equal. So 8x-23=2x+37. 8x-23=2x+ 37. 6x=60. x=10. LMN=8x-23+2x+37. LMN=10x+14.
9. Find the coordinates of the midpoint of MO bisects LMN, m LMN 5x 27,m LMO x 33. Find m NMO. The diagram isnot to scale.85. If EF 2x 16, FG 3x 19, and EG 20, m∠JHI = (2x +7)° and m∠GHI = (8x −2)° and m∠JHG = 65°. Find m∠JHI and m∠GHI. a.
14.