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advanced solutions to contemporary problems
Thursday, November 1, 2012
(MP) π π‘ = (1 β π3) ( 1 + π1 Ξππ‘ +π2 1 + π1 Ξππ‘ + β π¦ π‘ + π3π π‘β1 + ππ‘
1 +
π
1
Ξ
π
π‘
+
π
2
4
π¦
π‘
β π¦
π‘
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