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PIUVICU JEU 12, DUC 24 JCPECHuel 201 1. Given a temperature field defined the function T = 280.0 +0.002x +0.004y2 +0.00982, c

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T= 280+ 0.0020² +0.004 g + o. 0092, 100 Te at T (100, 100, 50) I, at T (0, 0, 0) 160,000 & Ti = 280 T2= 280+ 20 + 40 +0,49 z3) а) = (х + ) у D) 1 - 0 - 25 с) - zf + ул. заy = 6lxgry? ) ( + (x²y3) g T. U = dui + 2 Uj by +(-2y) = My Ja dy i. J.5 = 4y . How is divergent JXU = | 7 =(0)-910) +R(2x -

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