![\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)
<h3>
Answer: (x+1)(x+3)</h3>
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Explanation:
Let's assume it factors into (x+a)(x+b)
The goal is to find the two numbers a and b.
FOIL out (x+a)(x+b) to get x^2+(a+b)x+ab
Note how a+b is the middle term and ab is the last term.
In the original expression, 4 is the middle term and 3 is the last term.
So we need to find two numbers that
There are two ways to multiply to 3 and they are
- 1 times 3 = 3
- -1 times -3 = -3
But only the first way has the factors add to 4. So that means a = 1 and b = 3.
Therefore (x+a)(x+b) = (x+1)(x+3)
And x^2+4x+3 = (x+1)(x+3)
Answer:
Me too
Step-by-step explanation:
Is there a good
The answer is D Because in a tangent
20×1=20
20×2=40
20×3=60
20×4=80
20×5=100
20×6=120
20×7=140
20×8=160
20×9=180
20×10=200
(Only include this if the table is up to 12)
20×11=220
20×12=240
