Answer:
(3 square root of 2 , 135°), (-3 square root of 2 , 315°)
Step-by-step explanation:
Hello!
We need to determine two pairs of polar coordinates for the point (3, -3) with 0°≤ θ < 360°.
We know that the polar coordinate system is a two-dimensional coordinate. The two dimensions are:
- The radial coordinate which is often denoted by r.
- The angular coordinate by θ.
So we need to find r and θ. So we know that:
(1)
x = rcos(θ) (2)
x = rsin(θ) (3)
From the statement we know that (x, y) = (3, -3).
Using the equation (1) we find that:
Using the equations (2) and (3) we find that:
3 = rcos(θ)
-3 = rsin(θ)
Solving the system of equations:
θ= -45
Then:
r = 3\sqrt{2}[/tex]
θ= -45 or 315
Notice that there are two feasible angles, they both have a tangent of -1. The X will take the positive value, and Y the negative one.
So, the solution is:
(3 square root of 2 , 135°), (-3 square root of 2 , 315°)
Answer:
Step-by-step explanation:
let x be necromancers and y be wizards
if you know the amount of wizards and want to find the amount of necromancers. you just use y + 6
but if you know the amount of necromancers and want to find out the amount of wizards you just do x - 6
Hello from MrBillDoesMath
Answer:
[email protected] = - sqrt(7)/ 4
which is choice B
Discussion:
This problem can be solved by drawing triangles and looking at ratios of sides or by using the trig identity:
([email protected])^2 + (sin2)^2 = 1
If [email protected] = 3/4
, the
([email protected])^2 + (3/4)^2 = 1 => (subtract (3/4)^2 from both sides)
([email protected])^2 = 1 - (3/4)^2 = 1 - 9/16 = 7/16
So...... taking the square root of both sides gives
[email protected] = +\- sqrt(7)/ sqrt(16) = +\- sqrt(7)/4
But is [email protected] positive or negative? We are told that @ is in the second quadrant and cos(@) is negative in this quadrant, so our answer must be negative
[email protected] = - sqrt(7)/ 4
which is choice B
Thank you,
Mr. B
Answer:
Tangent segment ? I think. I don't know for sure because of the way you wrote the question- I got a bit confused.
