Answer:
θ = 5π/6 rad and 11π/6 rad
Step-by-step explanation:
Given the expression cotθ+√3=0
Subtract √3 from both sides
cotθ+√3-√3=0-√3
cotθ = -√3
Since cotθ = 1/tanθ
1/tanθ = -√3
Reciprocate both sides:
tanθ = -1/√3
θ = tan^-1(-1/√3)
θ = -30°
Since the angle is negative, and tanθ is negative in the second and fourth quadrant.
In the second quadrant;
θ = 180-30
θ = 150°
Since 180° = πrad
150° = 150π/180
150° = 5π/6 rad
In the fourth quadrant;
θ = 360-30
θ = 330°
Since 180° = πrad
330° = 330π/180
330° = 11π/6 rad
Hence the solutions are 5π/6 rad and 11π/6 rad.
Answer:
20%
Step-by-step explanation:
150 and 120 have a difference of 30
30 ÷ 150
.2 then multiply by 100
20 there is your answer
Hope this helped :)
Equation of line passing through given points :
Let's proceed with two point form ~
Assume :
So, the equation of required line is : y = 3 ~
Answer:
When squaring a fraction, square both numerator AND denominator
Answer:
C. Lines with arrows that point outward radiate from the charge in all
directions.
