Hello :
<span> sin 2x = sin x and 0 ≤ x ≤ 2π.
all solutions :
2x= x +2k</span>π or x= π -x +2kπ ..... k in : Z
x = 2kπ or : x = π/2 + kπ
but :
<span>0 ≤ x ≤ 2π
</span><span>all values of x such that sin 2x = sin x and 0 ≤ x ≤ 2π are :
</span>k = 0 : x=0 , x= π/2
k=1 : x=2 π , x=3π/2
Answer:
Option A
Step-by-step explanation:
the y value has a constant slope of 5 while the other tables have changing slopes
4. SOLVE FOR X:
Using the Alternate Interior Angles Theorem, we know that the 67 degree angle is congruent with the (12x - 5) degree angle. With this information, all I have to do is set the two equal to each other and solve for x.
67 = 12x - 5
67 + 5 = 12x - 5 + 5
72/12 = 12x/12
6 = x
x = 6
SOLVE FOR Y:
Using the Vertical Angles theorem, we know that angle y must be congruent to the 67 degree angle.
y = 67 degrees.
5. SOLVE FOR Y:
Alternate exterior angles: 6(x - 12) = 120
6x - 72 + 72 = 120 + 72
6x/6 = 192/6
x = 32
SOLVE FOR Y:
6((32) - 12) + y = 180
192 - 72 + y = 180
120 + y - 120 = 180 - 120
y = 60
Answer:
Step-by-step explanation:
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