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
First find the length of the diagonal of a square of side 4 inches:
d^2 = 4^2 + 4^2 = 2*4^2 = 32. Then the diagonal of the cube has length
sqrt( 32 + 4^2) = sqrt(32+16) = sqrt(48) = 4sqrt(3).
Answer:
I think the answer is B. f(x) = -1/3x - 4
Step-by-step explanation:
Use the given functions to set up and simplify
4−16.
XF(x)=X
Fx
1 − 7 = −6
2 − 10 = −8
3 − 13 = −10
4 − 16 = −12
