![\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:
u have to use derieved form of formula of sin and cos
Answer and Explanation:
Since this is an isosceles right triangle, you can use the ratio of side lengths specific to this type of triangle where both legs are the hypotenuse over root 2:
u = v = 164√2 / √2 = 164
so:
u = 164
v = 164
Hello,
Vertex=(-2,5)
x=0==>y=-3(0+2)²+5=-7
Answer:
c. 130
Step-by-step explanation:
Let call B the quantity of voters who voted yes for both propositions.
From the question we know that twice as many voters voted "yes" for R as for S, that can be written as the following equation:
R+B=2(S+B)
Where R is the number who voted "yes" for R but "no2 for S and S is the number who voted "yes" for S but "no" for R.
Replacing R by 750 and S by 310 and solving for B, we get:
750+B=2(310+B)
750+B=620+2B
2B-B=750-620
B=130
So, 130 voters voted yes for both propositions