![\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:
9
Step-by-step explanation:
Simplifying
2y + -6 = 12
Reorder the terms:
-6 + 2y = 12
Solving
-6 + 2y = 12
Solving for variable 'y'.
Move all terms containing y to the left, all other terms to the right.
Add '6' to each side of the equation.
-6 + 6 + 2y = 12 + 6
Combine like terms: -6 + 6 = 0
0 + 2y = 12 + 6
2y = 12 + 6
Combine like terms: 12 + 6 = 18
2y = 18
Divide each side by '2'.
y = 9
Simplifying
y = 9
Answer:
There is an 18% chance that the top 3 cards are all kings
Step-by-step explanation:
Divide 54 by 100
0.54
Divide by 3
0.18
18%
It’s a variable because when you justo I can’t stay the same every time it changes
Answer:
(a) See attachment for tree diagram
(b) 24 possible outcomes
Step-by-step explanation:
Given
Solving (a): A possibility tree
If urn 1 is selected, the following selection exists:
![B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]](https://tex.z-dn.net/?f=B_1%20%5Cto%20%5BR_1%2C%20R_2%2C%20R_3%5D%3B%20R_1%20%5Cto%20%5BB_1%2C%20R_2%2C%20R_3%5D%3B%20R_2%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_3%5D%3B%20R_3%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_2%5D)
If urn 2 is selected, the following selection exists:
![B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]](https://tex.z-dn.net/?f=B_2%20%5Cto%20%5BB_3%2C%20R_4%2C%20R_5%5D%3B%20B_3%20%5Cto%20%5BB_2%2C%20R_4%2C%20R_5%5D%3B%20R_4%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_5%5D%3B%20R_5%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_4%5D)
<em>See attachment for possibility tree</em>
Solving (b): The total number of outcome
<u>For urn 1</u>
There are 4 balls in urn 1
Each of the balls has 3 subsets. i.e.
So, the selection is:
<u>For urn 2</u>
There are 4 balls in urn 2
Each of the balls has 3 subsets. i.e.
So, the selection is:
Total number of outcomes is:
