Plz ! in desperate need of ! thx! remember to show all of your work! also, make it clear which problem is which when showing your work! thx so much!

First Picture:

AB and CD are straight lines. Their intersection has formed 4 angles: AOD, COB, AOC and DOB.
AOD and COB are vertically opposite angles, which means that AOD = COB = 152 degrees.

COB is formed by 3 other angles: COE, EOF and FOB.
Which means that COE + EOF + FOB = COB

We already proved that COB = 152 degrees, and on the figure, we have the algebraic measurement of the 3 angles forming COB::
COE = 3x degrees, EOF = x degrees and FOB = x + 12 degrees.

COE + EOF + FOB = COB
3x + x + (x+12) = 152
3x + x + x + 12 = 152
5x + 12 = 152

We need to solve for x.

Subtract 12 from each side:
5x + 12 = 152
5x + 12 - 12 = 152 - 12
5x = 140

Divide both sides by 5:
5x = 140
(5x)/5 = 140/5
x = 140/5

x = 28

Second picture:

Based on the picture, we know that the angle LOM = 90 degrees, MON = x degrees and NOL = 5x degrees.

The angles forms a circle:
LOM + MON + NOL = 360 degrees
So 90 + x + 5x = 360
90 + 6x = 360

We need to solve for x:

Subtract 90 from both sides:
90 + 6x = 360
90 - 90 + 6x = 360 - 90
6x = 270

Divide both sides by 6:
6x = 270
(6x)/6 = 270/6
x = 270/6

x = 45

Hope this helps! :)

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