Answer:
A. <ECF and <BCF
Step-by-step explanation:
Complementary angles are angles that add up to give 90°
m<BCE = m<BCA = 90° (right angles)
m<ECF + m<BCF = m<BCA
m<ECF + m<BCF = 90° (Substitution)
Therefore, <ECF and <BCF are complementary angles.
Answer:
EH = 9
FH = 7
EG = 6
EF = EH - FH
EF = 9 -7
EF = 2
FG = EG -EF
FG = 6 -2
FG = 4
Step-by-step explanation:
It looks like the given equation is
sin(2x) - sin(2x) cos(2x) = sin(4x)
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
which lets us rewrite the equation as
sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)
Move everything over to one side and factorize:
sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0
sin(2x) - 3 sin(2x) cos(2x) = 0
sin(2x) (1 - 3 cos(2x)) = 0
Then we have two families of solutions,
sin(2x) = 0 or 1 - 3 cos(2x) = 0
sin(2x) = 0 or cos(2x) = 1/3
[2x = arcsin(0) + 2nπ or 2x = π - arcsin(0) + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
(where n is any integer)
[2x = 2nπ or 2x = π + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
[x = nπ or x = π/2 + nπ]
… … … or [x = 1/2 arccos(1/3) + nπ or x = -1/2 arccos(1/3) + nπ]
Answer:
no ni**a
Step-by-step explanation:
do your own work you dumb fa**ot b*t*h
