Answer:
A .cos(x)<1
Step-by-step explanation:
According to the first inequality
cos(x)<1
x < arccos 1
x<0
This therefore does not have a solution within the range 0 ≤ x ≤ 2pi
x cannot be leas than 0. According to the range not value, 0≤x which is equivalent to x≥0. Thus means otvis either x = 0 or x> 0.
For the second option
.cos(x/2)<1
x/2< arccos1
x/2<0
x<0
This inequality also has solution within the range 0 ≤ x ≤ 2pi since 0 falls within the range of values.
For the inequality csc(x)<1
1/sin(x) < 1
1< sin(x)
sinx>1
x>arcsin1
x>90°
x>π/2
This inequality also has solution within the range 0 ≤ x ≤ 2pi since π/2 falls within the range of values
For the inequality csc(x/2)<1
1/sin(x/2) < 1
1< sin(x/2)
sin(x/2)> 1
x/2 > arcsin1
X/2 > 90°
x>180°
x>π
This value of x also has a solution within the range.
Therefore option A is the only inequality that does not have a solution with the range.
So the problem ask to what could be the right choice among the three that could support the said statement in you problem that tells that an angle that has a supplement also has a complement and the best answer would be that it is ALWAYS be TRUE. I hope you are satisfied with my answer
Answer:
m = 44
Step-by-step explanation:
USe the pythagoream thyroem a^2 + b^2 = m^2. so 10^2 + m = 12^2
or 100 + m = 144. subtract 100 from both sides, you get m = 44
Answer:
The awnser is C
Step-by-step explanation:
Answer:
97
Step-by-step explanation:
