Find m∠BOC, if m∠MOP = 110°.
Answer:
m∠BOC= 40 degrees
Step-by-step explanation:
A diagram has been drawn and attached below.
- OM bisects AOB into angles x and x respectively
- ON bisects ∠BOC into angles y and y respectively
- OP bisects ∠COD into angles z and z respectively.
Since ∠AOD is a straight line
x+x+y+y+z+z=180 degrees

We are given that:
m∠MOP = 110°.
From the diagram
∠MOP=x+2y+z
Therefore:
x+2y+z=110°.
Solving simultaneously by subtraction

x+2y+z=110°.
We obtain:
x+z=70°
Since we are required to find ∠BOC
∠BOC=2y
Therefore from x+2y+z=110° (since x+z=70°)
70+2y=110
2y=110-70
2y=40
Therefore:
m∠BOC= 40 degrees
Answer:
2.8 miles higher the plane go without leaving the troposphere?
Step-by-step explanation:
Troposphere height on different locations= 6-12 miles
plane altitude at present flight = 5.8 miles.
Troposphere deepness from the present flight= 8.6 miles
So from present altitude till troposphere= 8.6-5.8= 2.8 miles
Answer:
4/75
Step-by-step explanation:
you will get 5 and 1/3 percent equals (5 + 1/3) / 100 which is equal to 5/100 + (1/3) / 100 which is equal to 5/100 + 1/300. place both fractions under a common denominator. you will get 5/100 + 1/300 = 15/300 + 1/300 = 16/300. simplify the fraction to get 16/300 = 8/150 = 4/75
Answer:
6=6
True for all a
Step-by-step explanation:
