![\bf \begin{array}{clclll} -6&+&6\sqrt{3}\ i\\ \uparrow &&\uparrow \\ a&&b \end{array}\qquad \begin{cases} r=\sqrt{a^2+b^2}\\ \theta =tan^{-1}\left( \frac{b}{a} \right) \end{cases}\qquad r[cos(\theta )+i\ sin(\theta )]\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bclclll%7D%0A-6%26%2B%266%5Csqrt%7B3%7D%5C%20i%5C%5C%0A%5Cuparrow%20%26%26%5Cuparrow%20%5C%5C%0Aa%26%26b%0A%5Cend%7Barray%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ar%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5C%5C%0A%5Ctheta%20%3Dtan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7Bb%7D%7Ba%7D%20%5Cright%29%0A%5Cend%7Bcases%7D%5Cqquad%20r%5Bcos%28%5Ctheta%20%29%2Bi%5C%20sin%28%5Ctheta%20%29%5D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)

now, notice, there are two valid angles for such a tangent, however, if we look at the complex pair, the "a" is negative and the "b" is positive, that means, "x" is negative and "y" is positive, and that only occurs in the 2nd quadrant, so the angle is in the second quadrant, not on the fourth quadrant.
thus
Answer:
yes
Step-by-step explanation:
(there really isn't much else to say, it's quite simple)
Answer:
x=90 y=24
Step-by-step explanation:
I'm pretty sure if you're using slope this should be it.
Answer:
-206675344746
Step-by-step explanation:
1000*0+321*15678*9(-4563) = -206675344746