Matematică
batranete
2015-11-03 16:27:45
efectuati: a) 11x(48x3,4)+5[19+4x(8,3-5,40]-20,3 b) { 3+0,5x[2,3-0,2x(3,1-5,12:3,20]} va roggg frumos
Răspunsuri la întrebare
duranico
2015-11-03 22:14:39

11x(48x3,4)+5[19+4x(8,3-5,40)]-20,3= 11x(48x3,4)+5[19+4x2,9]-20,3= 11x(48x3,4)+5x[19+11,6]-20,3= 11x(48x3,4)+5x30,6-20,3= 11x163,2+153-20,3= 1795,2+153-20,3= 1948,2-20,3= 1927,9 {3+0,5x[2,3-0,2x(3,1-5,12:3,20)]}= {3+0,5x[2,3-0,2x(3,1-1,6)]}= {3+0,5x[2,3-0,2x1,5]}= {3+0,5x[2,3-0,3]}= {3+0,5x2}= {3+1}= 4

Adăugați un răspuns