|
|
 | | Re: Geek Hierarchy
|
| (...) Hey, where are the toy-collecting geeks on this thing? I feel so underrepresented! :) Hilarious. -jeremiah- (22 years ago, 27-Mar-03, to lugnet.off-topic.geek)
| | | | | Geek Hierarchy
|
| Posted without further comment... (URL) (22 years ago, 27-Mar-03, to lugnet.off-topic.geek)
| | | | | Retina scanner?
|
| You do realise that that 'retina' scanner on CLSOTW is actually just photographing the cornea, not the retina, don't you? Jason Railton (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
| | | | | Re: On a scale of 0 to 10?
|
| (...) aHA! Much oblige! I now get (approx): s = 5.39 * e^(2.625q) - 0.39 Whew. Of course, now here's a totally different question. In order to get that point, I cheated. I couldn't solve: e^(B/4) + e^(-B) = 2 using algebra, but using other means, I (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
| | | | | Re: On a scale of 0 to 10?
|
| Hi all. The reason the equation is unsolvable is because B is an unneeded constant. Why? A*e^(q+B)=A*(e^q)*(e^B) It is impossible to distinquish between A & e^B. What you need to solve is an equation of the form s=A*exp(B*q) + C This is the form of (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
|
