Suppose tan(b) = –2, and the terminal side of b is located in quadrant II. What is cot(b)?

–2
Negative one-half
One-half
2

What is the missing step in this proof ?

What are the solutions of the system? y = x^2 + 2x +3y = 4x - 2a. (-1, -6) and (-3, -14)b. (-1, 10) and (3, -6)c. no solutiond. (-1, -6) and (3, 10)

Cd is the perpendicular bisector of both xy and st, and cy=20. find xy.