OL ⊥ ON start overline, O, L, end overline, \perp, start overline, O, N, end overline \qquad m \angle LOM = 3x + 38^\circm∠LOM=3x+38 ∘ m, angle, L, O, M, equals, 3, x, plus, 38, degrees \qquad m \angle MON = 9x + 28^\circm∠MON=9x+28 ∘ m, angle, M, O, N, equals, 9, x, plus, 28, degrees Find m\angle LOMm∠LOMm, angle, L, O, M:

Step-by-step explanation:

If side OL is perpendicular to ON i.e OL ⊥ ON, then angle  ∠LON = 90°. If the is another line OM projecting from O with ∠LOM= (3x+38)° and ∠MON= (9x+28)°, then ∠MON +  ∠LOM =  ∠LON

Substituting the given angles int the expressions above to calculate the value of x;

(9x+28)° + (3x+38)° = 90°

12x+66 = 90°

12x = 90-66

12x = 24

x = 2°

Since ∠LOM= (3x+38)°, to get the value of the angle, we will substitute x = 2° into the expression as shown;

∠LOM= (3(2)+38)°

∠LOM= 6+38

∠LOM= 44°

Hence the measure of angle LOM is 44°

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

step-by-step explanation:

3024 rhe base is what up in worry about