Ask a New Question

Question

convert (-sqrt2, -sqrt6) to polar coordinates with
r > 0 and 0 ≤ θ < 2π.
5 years ago

Answers

Reiny
tanθ = -√6/-√2 = √3 , where θ is in quad III
θ = 180°+60° = 240° or 4π/3 radians
r = √(-√2)^2 + (-√6)^2) = √8

(-sqrt2, -sqrt6) <-----> (√8, 4π/3)
5 years ago

Related Questions

1. (3-sqrt6)/(5-2sqrt6) 2. (2sqrt3-sqrt6)/(5sqrt3+2sqrt6) Don't you do something with the co... (sqrt3+5 sqrt6)(sqrt3-5 sqrt6) DRWLS-Just take the difference of the squares of the two terms, sq... if f(x)=x^2 find f(2-sqrt6) if f(x)=x^2 find f(2-sqrt6) I have the answer as 10-4sqrt6 is that right? 1. Simplify: sqrt6(sqrt2/2+sqrt3)-sqrt8 2. Simplify: sqrt(72x^3)-5x sqrt(2x) 3. Simplify: (2sqrt6-... Simplify: sqrt6(sqrt2/2+sqrt3)-sqrt8 I need all of the steps. I need this ASAP because this is d... evaluate sqrt(5+2*(sqrt6)) + sqrt(8-2*(sqrt15)) Subtract (2)/(4+sqrt6) - (2)/(4-sqrt6). Can someone explain how to do this to me? (sqrt6 + sqrt3)(sqrt2 - 2) sqrt6 * sqrt8
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use