Answer:
∠BAD=20°20'
∠ADB=34°90'
Step-by-step explanation:
AB is tangent to the circle k(O), then AB⊥BO. If the measure of arc BD is 110°20', then central angle ∠BOD=110°20'.
Consider isosceles triangle BOD (BO=OD=radius of the circle). Angles adjacent to the base BD are equal, so ∠DBO=∠BDO. The sum of all triangle's angles is 180°, thus
∠BOD+∠BDO+∠DBO=180°
∠BDO+∠DBO=180°-110°20'=69°80'
∠BDO=∠DBO=34°90'
So ∠ADB=34°90'
Angles BOD and BOA are supplementary (add up to 180°), so
∠BOA=180°-110°20'=69°80'
In right triangle ABO,
∠ABO+∠BOA+∠OAB=180°
90°+69°80'+∠OAB=180°
∠OAB=180°-90°-69°80'
∠OAB=20°20'
So, ∠BAD=20°20'
Answer: option B. it has the highest y-intercept.
Explanation:
1) point -slope equation of the line
y - y₁ = m (x - x₁)
2) Replace (x₁, y₁) with the point (5,3):
y - 5 = m (x - 3)
3) Expand using distributive property and simplify:
y - 5 = mx - 3m ⇒ y = mx + 5 - 3m
4) Compare with the slope-intercept equation of the line: y = mx + b, where m is the slope and b is the y-intercept
⇒ slope = m
⇒ b = 5 - 3m = y - intercept.
Therefore, for the same point (5,3), the greater m (the slope of the line) the less b (the y-intercept); and the smaller m (the slope) the greater the y - intercept.
Then, the conclusion is: the linear function with the smallest slope has the highest y-intercept (option B).
Okay if there are 8 teachers and 6 reams of paper you would have to divide 6 by 8 and you would get 0.75
6/8= 0.75
each teacher would get 0.75 (3/4) of a ream of paper
