Given
\qquad \overline{OL}\perp\overline{ON}
OL
⊥
ON
start overline, O, L, end overline, \perp, start overline, O, N, end overline
\qquad m \angle LOM = 3x - 15^\circm∠LOM=3x−15
∘
m, angle, L, O, M, equals, 3, x, minus, 15, degrees
\qquad m \angle MON = 5x - 23^\circm∠MON=5x−23
∘
m, angle, M, O, N, equals, 5, x, minus, 23, degrees
Find m\angle MONm∠MONm, angle, M, O, N:

{}^{\circ}
∘
degrees

m∠MON=27

Step-by-step explanation:

From the diagram, we see that together m∠LOM and m∠MON form m∠LON, so

m∠LOM+ m∠MON=m∠LON

Since we are given that OL  ⊥  ON , we know m∠LON=90  

Substitute in the expressions that were given for each measure:

3x−15  + 5x−23  = 90  

Combine like terms:

8x - 38 = 90

Add 38 degrees to both sides:

8x = 128

Divide both sides by 8 to find x:

x = 16

Substitute 16 for x in the expression that was given for m∠MON:

m∠MON= 5(16) - 23

Simplify:

m∠MON  = 80 - 23

So m∠MON=57  

(took a while to decipher all the weird code from Khan Academy. Hope I was able to help)

2 is the final ! : )

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?