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
8.008
8.08
8.081
8.09
since it says increasing order it means least to greatest
Answer:
78%
Step-by-step explanation:
70 = 0.50 (62) + 0.50 (x)
70 = 31 + 0.5x
39 = 0.5x
x = 78
You must get at least a 78% on the test to have at least 70% in the class.
x = number of 1-cent stamps
y = number of 8-cent stamps
z = number of 12-cent stamps
We have 31 stamps all together, so x+y+z = 31.
"I have 4 more 1-cent stamps than 8-cent stamps" means we have the equation x = y+8. Whatever y is, add 8 to it to get x. Solve for y to get y = x-8.
You also have "twice as many one cent stamps as 12 cent stamps", so x = 2z. Solving for z gets you z = 0.5x
-------------
x+y+z = 31
x+x-8+z = 31 ... y replaced with x-8
x+x-8+0.5x = 31 ... plug in z = 0.5x
2.5x-8 = 31
2.5x = 31+8
2.5x = 39
x = 39/2.5
x = 15.6
Your teacher made a typo somewhere because we should get a positive whole number result for x (since x is a count of how many 1-cent stamps we have).