![\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)
Step-by-step explanation:
Let,
- Money received by Jan = 4x
- Money received by Jane = 9x
- Money received by Jello = 6x
According to the question,
→ Money received by Jan = $200
→ 4x = $200
→ x = $200 ÷ 4
→ x = 50 ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀… ( 1 )
Now,
→ Money received by Jane = 9x
→ Money received by Jane = 9($50)
→ <u>Money received by Jane = $450</u> [Ans]
And,
→ Money received by Jello = 6x
→ Money received by Jello = 6($50)
→ <u>Money received by Jello = $300</u> [Ans]
Answer:
c = bx/(x+1)
Step-by-step explanation:
Add cx to get c-terms together, and divide by the coefficient of c.
bx = cx + c
bx = c(x +1) . . . . factor out c
bx/(x +1) = c
The x is the 0392-9383-8383 and then imma slap ur mom
