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:
Segments LO and ON are perpendicular, providing the required
information for the value of the sum of ∠LOM and ∠MON.
Reasons:
The given parameter are;
is perpendicular to ;
m∠LOM = (3·x - 15°)
m∠MON = (5·x - 23°)
Required:
Find m∠MOM
Solution:
Given that is perpendicular to , we have;
m∠LON = 90° by definition of perpendicular lines
m∠LON = m∠LOM + m∠MON by angle addition postulate
Therefore;
m∠LOM + m∠MON = 90° by substitution property of equality
Which gives;
(3·x - 15°) + (5·x - 23°) = 90° by substitution property
8·x - 38° = 90°
x = 16°
m∠MON = 5·x - 23°
m∠MON = 5 × 16° - 23° = 57°
Learn more here:
Answer:57
Step-by-step explanation:
Answer:
15
Step-by-step explanation:
Answer:
The answer should be 15.
Step-by-step explanation:
Answer:
Step-by-step explanation:
Surface area is 2(πr²) + 2πrh. The first includes the top and bottom; the second is the area of lateral sides of the cylinder (radius r and height h).
The volume of this cylinder is V = πr²h.
Answer:
2 pi r square + pi dh
Step-by-step explanation:
2 pi r square + pi dh