![\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
![\bf 12\left[cos\left( \frac{2\pi }{3} \right) +i\ sin\left( \frac{2\pi }{3} \right) \right]](https://tex.z-dn.net/?f=%5Cbf%2012%5Cleft%5Bcos%5Cleft%28%20%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%5Cright%29%20%2Bi%5C%20sin%5Cleft%28%20%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%5Cright%29%20%5Cright%5D)
Answer:
Step-by-step explanation:
3b⁵ + 15b⁴ - 18b⁷ = 3b⁵ + [3*5*b⁴] - [6*3*b⁷]
=3b⁴*( b + 5 - 6b³)
Answer:
1/12
Step-by-step explanation:
Answer: y= 12x +40
Step-by-step explanation:
Answer:
negative
Step-by-step explanation:
The graph falls to both the left and right, so the leading coefficient has to be negative, and it must be raised to an even power.
