Answer: 90 Earth years.
Step-by-step explanation:
Analizing the information provided in the exercise, you need to find the Least Common Multiple (LCM) of the given numbers.
You can follow these steps:
1. You must descompose 6, 9, 15 and 18 into their prime factors:
2. Finally, you need to choose the commons and non commons with their greatest exponents and multiply them. Then you get:
Therefore, it will take 90 Earth years for them to all line up.
I think the scale of the factor of area is 2
hmmm first off let's convert the √3 +i to trigonometric form, and then use De Moivre's root theorem, bearing in mind that √3 and i or 1i are both positive, meaning we're on the I Quadrant.
![\bf (\stackrel{a}{\sqrt{3}}~,~\stackrel{b}{1i})\qquad \begin{cases} r=&\sqrt{(\sqrt{3})^2+1^2}\\ &\sqrt{3+1}\\ &2\\ \theta =&tan^{-1}\left( \frac{1}{\sqrt{3}}\right)\\\\ &tan^{-1}\left( \frac{\sqrt{3}}{3} \right)\\ &\frac{\pi }{6} \end{cases}~\hfill \implies ~\hfill 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right]](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Ba%7D%7B%5Csqrt%7B3%7D%7D~%2C~%5Cstackrel%7Bb%7D%7B1i%7D%29%5Cqquad%20%5Cbegin%7Bcases%7D%20r%3D%26%5Csqrt%7B%28%5Csqrt%7B3%7D%29%5E2%2B1%5E2%7D%5C%5C%20%26%5Csqrt%7B3%2B1%7D%5C%5C%20%262%5C%5C%20%5Ctheta%20%3D%26tan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5Cright%29%5C%5C%5C%5C%20%26tan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D%20%5Cright%29%5C%5C%20%26%5Cfrac%7B%5Cpi%20%7D%7B6%7D%20%5Cend%7Bcases%7D~%5Chfill%20%5Cimplies%20~%5Chfill%202%5Cleft%5B%20cos%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%2Bi~sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%5Cright%5D)
![\bf ~\dotfill\\\\ \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~%5Cdotfill%5C%5C%5C%5C%20%5Cqquad%20%5Ctextit%7Bpower%20of%20two%20complex%20numbers%7D%20%5C%5C%5C%5C%5C%20%5B%5Cquad%20r%5Bcos%28%5Ctheta%29%2Bisin%28%5Ctheta%29%5D%5Cquad%20%5D%5En%5Cimplies%20r%5En%5Bcos%28n%5Ccdot%20%5Ctheta%29%2Bisin%28n%5Ccdot%20%5Ctheta%29%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \left[ 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right] \right]^3\implies 2^3\left[ cos\left( 3\cdot \frac{\pi }{6}\right) +i~sin\left( 3\cdot \frac{\pi }{6}\right) \right] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 8\left[cos\left( \frac{\pi }{2} \right) +i~sin\left( \frac{\pi }{2} \right) \right]~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%202%5Cleft%5B%20cos%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%2Bi~sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%5Cright%5D%20%5Cright%5D%5E3%5Cimplies%202%5E3%5Cleft%5B%20cos%5Cleft%28%203%5Ccdot%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%2Bi~sin%5Cleft%28%203%5Ccdot%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%208%5Cleft%5Bcos%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%5Cright%29%20%2Bi~sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%5Cright%29%20%5Cright%5D~%5Chfill)
Answer:
(7, 8)
Step-by-step explanation:
Midpoint Formula: 
Simply set our midpoint and given point equal into a formula to find the other endpoint:
4 = (1 + x)/2
8 = 1 + x
x = 7
6 = (4 + y)/2
12 = 4 + y
y = 8
32 divided by 2.39 = about 13.3
