Positive 8 and negative 8 both are.
Radius of circle would be R=a×✓2/2=√2 and Circle area is then A=R^2π=✓2^2π=2π or 2×3.14=6.28...
Answer:
The one opposite of 100°, as the bigger the angle, the bigger the opposing side, is the biggest.
By the same logic, the one opposite 45° is the next biggest.
Continuing with that, the one opposite the black one (which would equal 35°, I think) is the smallest.
I think that
the triangle is 6
the square is -5
and the star is -20
but i would get a second answer bc i cant certainly say that<span />
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.
