Substitution.
Here is an example.
Let x be equal to 3 and y equal to 3.
From this we can conclude that the values of both x and y are equal to three therefore x and y have the same value and are equal.
Here in your case we have:
Hope this helps.
r3t40
![\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)
All real numbers for the domain
In minutes it is 1500 min because if you just multiply 25 and 60 it equals 1500. I got the 60 from 60 sec. in one minute.
Answer:
Reflection
Step-by-step explanation:
The shape A'B'C' is just reflected from the shape ABC.
Hope this helps :)
