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.
Its 0.909 simple and easy ;)
Given :
A restaurant had 3 full boxes of spoons and 4/6 of a box.
Each full box weighed 2 kilograms .
To Find :
The combined weight of the boxes the restaurant has.
Solution :
Mass of 1 box is 2 kg .
So , mass of 4/6 of a box is 4/3 kg .
Total mass :
Therefore , the combined weight of the boxes the restaurant has is 7.33 kg.
Hence , this is the required solution .
5:29 could be a possible equivalent ratio
