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:
![r=\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+(-3)^{2}} = 3\sqrt{2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%3D%5Csqrt%7B3%5E%7B2%7D%2B%28-3%29%5E%7B2%7D%7D%20%3D%203%5Csqrt%7B2%7D)
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:
Answer:
1 1/+6
Step-by-step explanation:
9/16 + 8/16=17/16=1 1/16
I tried my best at it and my guess is that in standard form is 0.00086
Answer:
x = 12
Step-by-step explanation:
The segment from the vertex to the base is a perpendicular bisector.
Using Pythagoras' identity on the right triangle on the right.
(
x )² + 3² = (
)² , that is
x² + 9 = 45 ( subtract 9 from both sides )
x² = 36 ( multiply both sides by 4 )
x² = 144 ( take the square root of both sides )
x =
= 12