Answer:
Step-by-step explanation:
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:
Step-by-step explanation:
Amount paid by Joanne before promotion = 74 + 59 + 69 = $ 202
a) Amount saved if she switches to "3 for 129 plan" = 202 - 129 = $ 73
Percentage of saving = 
= 36.13 = 36%
b) Amount saved after the first year = 202 - 139 = $ 63
Percentage of saving = 
= 31.18 = 31%
Answer:
1/6
Step-by-step explanation:
We want to find how long it takes for the bacteria to lose half its size. We can do this by taking one point of the bacteria and finding how long it takes to go to half its size. When t=0, 9300 * (1/64)^t = 9300 * 1 = 9300 as anything to the power of 0 is 1. Therefore, we can solve for t when the end result of the bacteria is 9300/2= 4650, making our equation
4650 = 9300 * (1/64)^t
divide both sides by 9300
1/2 = (1/64)^t
First, we can tell that 2^6 = 64*. Because of this, we can say that (1/2)^6 = 1^6/2^6 = 1/64, so (1/64)^(1/6) = 1/2. We know this because
(1/2)^6 = 64
take the 6th root of both sides
(1/2) = (64)^(1/6)
. This means that t=1/6, so the bacterial culture loses 1/2 of its size every 1/6 seconds
* if this is harder to figure out, e.g. 3 and 729, we can plug (log₃729) into a calculator
Answer:
In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line (or alternatively, a quantity that can be represented as an infinite decimal expansion).
