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From: KY on 14 Jan 2010 01:15 Eliminate a, b, c; a + b + c = 0, ((b - c)/a + (c - a)/b + (a - b)/c)*(a/(b - c) + b/(c - a) + c/(a - b)) = k
From: ksoileau on 14 Jan 2010 11:32 On Jan 14, 10:15 am, KY <wkfkh...(a)yahoo.co.jp> wrote: > Eliminate a, b, c; > a + b + c = 0, > ((b - c)/a + (c - a)/b + (a - b)/c)*(a/(b - c) + b/(c - a) + c/(a - b)) = k a = -b-c, k = 9
From: preedmont on 14 Jan 2010 11:27 "KY" <wkfkh056(a)yahoo.co.jp> wrote in message news:1082409701.70238.1263485757008.JavaMail.root(a)gallium.mathforum.org... > Eliminate a, b, c; > a + b + c = 0, > ((b - c)/a + (c - a)/b + (a - b)/c)*(a/(b - c) + b/(c - a) + c/(a - b)) = > k trivial, k = sqrt( (a*b-c)*(b*c-a)*(a*c-b))
From: Richard P. Cranium on 14 Jan 2010 13:13 Buh. Low. Me.
From: Amy on 14 Jan 2010 16:55
"KY" <wkfkh056(a)yahoo.co.jp> wrote in message news:1082409701.70238.1263485757008.JavaMail.root(a)gallium.mathforum.org... > Eliminate a, b, c; > a + b + c = 0, > ((b - c)/a + (c - a)/b + (a - b)/c)*(a/(b - c) + b/(c - a) + c/(a - b)) = > k 2 equations, 4 unknowns.................... Eliminate KY |