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Descrição: Basic of Foundation Engineering with solved problems provides an easy way of studying to students preparing for their final exams. It provides you with basic formulas and how the are applicable. ...
Solutions question no 1 2d - 3 = m............. eq 1 m = 63 - d .............eq 2....... > m + d = 63 given put in eq 1 2d - 3 = 63 - d 3d = 66 d = 22 Question no 2 f - 10 = 7(s-10) f - 10 = 7s - 70 f = 7s - 70 + 10 f = 7s - 60 2(f + 2) = !(s+2) 2f + " = !s + 10 2f = !s + 10 - " 2f = !s + 6 #epl$%e f &it' (7s-60) 2(7s - 60) = !s + 6 1"s - 120 = !s + 6 1"s = !s + 6 + 120 1"s - !s = 126 s = 126 s= s = 1" *s son,s p*esent $ge question no 3 f = 3s 2s + 1! = f 2s + 1! = 3s s = 1!
Question no " / S / 45/S S44 48 /# S 70 9 s$di$ = s mot'*t = m s + m = 70 ............eq 1 10 9 /4#:/# /# 45/ ;4S 1<" /# /#,S 45/ s - 10 = 1<" ( m - 10 ) ......eq 2 eq 2 multipl " ot' sides &e get "s - "0 = m - 10 ......eq 3 f*om eq 1 s = 70 - m ...........eq " put t'e v$lue of s in eq 3 &e get " ( 70 - m ) - "0 = m - 10 20 - 30 = !m 2!0 = !m m = !0
question no 6 ? "" - ( @ - ) A<(@ - ) = 6
question no 7 f = 1!s f + 1 = 3 (s +1 ) f + 1 = 3f<1! + 1
!f + 0 = f + 270 "f = 10 f = "!
question no f = !" f - " = !( s - " ) !" - " = !s - 20 !0 + 20 = !s s = 1" $fte* ! e$*s s = 1 question @<3 + 20 = @ @ + 60 = 3@ 2@ = 60 @= 30
