Option 3:
m∠ABC = 66°
Solution:
Given
and ABH is a transversal line.
m∠FAB = 48° and m∠ECB = 18°
m∠ECB = m∠HCB = 18°
<u>Property of parallel lines:
</u>
<em>If two parallel lines cut by a transversal, then the alternate interior angles are equal.</em>
m∠FAB = m∠BHC
48° = m∠BHC
m∠BHC = 48°
<u>Exterior angle of a triangle theorem:
</u>
<em>An exterior angle of a triangle is equal to the sum of the opposite interior angles.</em>
m∠ABC = m∠BHC + m∠HCB
m∠ABC = 48° + 18°
m∠ABC = 66°
Option 3 is the correct answer.
Answer: 10u3 + 4u2 - 2u + 10
Step-by-step explanation:
( 4u3 + 4u2 + 2) + ( 6u3 - 2u + 8)
opening the bracket
4u3 + 4u2 + 2 + 6u3 - 2u + 8
collecting like terms
4u3 + 6u3 + 4u2 - 2u + 2 + 8
10u3 + 4u2 - 2u + 10
(-1,8) and (5,6) are points on the line.
Slope of line = (8-6)/(-1-5) = 2/(-6) = -⅓
y-6 = -⅓(x-5)
y = -⅓x + 5/3 + 6 = -⅓x + 7⅔