Question 1118052
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In the first place, *[tex \Large i] is NOT equal to *[tex \Large 0.05].  It is very close to that number but close is not equal and therefore it is dead wrong to use the equals sign.


Let *[tex \Large u\ =\ i\ +\ 1] then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 109.3\ =\ \frac{10}{u}\ +\ \frac{110}{u^2}]


Multiply by *[tex \Large u^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 109.3u^2\ -\ 10u\ -\ 110\ =\ 0]


Multiply by 10 to eliminate the decimal and then use the quadratic formula to solve for *[tex \Large u]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ \frac{50\ \pm 40\sqrt{753}}{1093}]


But *[tex \Large i\ =\ u\ -\ 1] so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i\ =\ \frac{50\ \pm 40\sqrt{753}}{1093}\ -\ 1]


I'm going to go way out on a limb here and assume that you are looking for a non-negative value, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i\ =\ \frac{50\ + 40\sqrt{753}}{1093}\ -\ 1\ \approx\ 0.04999\ \approx\ 0.05]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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