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Documents

p y b x b x p b 1 100 - mr.sc. Bojan Kovacic · 2 x 1 p% x , x 2 = x 1 + p 100 ·x 1 = 1+ p 100 ·x 1. (5) 3 y , yˆ 2= b·xa = b· p 1+ p 100 ·x 1 a = b· 1+ p 100 a ·xa 1 = 1+