Answer:
we know that
the relationship between the 2-dimensional polar and Cartesian coordinates is
r = √(x² + y²)
Θ = tan⁻¹ (y/x)
so
Part a) (2, −2)---------> this point belong to the IV quadrant
r = √(x² + y²)------ r = √(2² + (-2)²)-----> r=√8
Θ = tan⁻¹ (y/x)---- Θ = tan⁻¹ (2/2)----> 45°
remember that the point belong to the IV quadrant
so
Θ=360-45-----> Θ=315°
the answer part A) is
(r,Θ)=(√8,315°)
Part b) (-1, 3)---------> this point belong to the II quadrant
r = √(x² + y²)------ r = √(-1² + (3)²)-----> r=√10
Θ = tan⁻¹ (y/x)---- Θ = tan⁻¹ (3/1)----> 71.57°
remember that the point belong to the II quadrant
so
Θ=180-71.57-----> Θ=108.43°
the answer part B) is
(r,Θ)=(√10,108.43°)
mark me branlest plz hope it helps
Subtract tchem both and wilk get the answer
102.62-101.94=0.68
We know angle P is 90 degrees because of Thales Theorem, and angle N is 60 degrees. Because the angles in a triangle have to add up to 180 degrees, angle PMN has to be 30 degrees.
Angle QMO is half of angle PMN, so it's 15 degrees. Angle MQO is equal to angle QMO because the radii make Triangle OQM isosceles, so angle MQO is 15 degrees.
Answer:
$669
Step-by-step explanation:
