Answer:
Problem B: x = 12; m<EFG = 48
Problem C: m<G = 60; m<J = 120
Step-by-step explanation:
Problem B.
Angles EFG and IFH are vertical angles, so they are congruent.
m<EFG = m<IFH
4x = 48
x = 12
m<EFG = m<IFH = 48
Problem C.
One angle is marked a right angle, so its measure is 90 deg.
The next angle counterclockwise is marked 30 deg.
Add these two measures together, and you get 120 deg.
<J is vertical with the angle whose measure is 120 deg, so m<J = 120 deg.
Angles G and J from a linear pair, so they are supplementary, and the sum of their measures is 180 deg.
m<G = 180 - 120 = 60
4 kilometres is 4000 meters
4000 meters - 3400 meters = 600 meters
and as John is the one that rode 4 kilometres he rode further and by 600 meters
First part its 4C2 = 4*3 / 2 = 6
Second part
12 * 100
---------- = 4.3 %
280
Answer:
A right circular cone
Step-by-step explanation:
Answer: 2x+12
Step-by-step explanation:
Multiply 2 by each number and letter in the parenthesis.
2*x+2*6=2x+12
