Question

Impartiti numarul 118 in parti invers proportional cu numerele 2;5;7 ajutatima va rog urgent dau coroana?????????????????????

Answer 1

(a;b;c) i.p. (2;5;7)
a+b+c=118
a/1/2=b/1/5=c/1/7=K }⇒a=K/2
                                      b=K/5
                                      c=K/7
K/2+K/5+K/7=118 (trebuie sa aduci fractiile la acelasi numitor,prima fractie o amplifici cu 35,a doua cu 14,a treia cu 10)⇔
35K+14K+10K =118
           70
59K=118
70
59K=118x70
59K=8260
K=8260 :59
K=140
a=140:2⇒a=70
b=140:5⇒b=28
c=140:7⇒c=20

Answer 2

{ a, b, c } i.p { 2, 5, 7 }
a x 2 = b x 5 = c x 7 = k
a x 2 = k ⇒ a = k/2
b x 5 = k ⇒ b = k/5
c x 7 = k ⇒ c = k/7
a + b + c = 118
35)    14)    10)
k/2 + k/5 + k/7 = 118
35k/70 + 14k/70 + 10k/70 = 118
59k/70 = 118
59k = 118 x 70
59k = 8260
k = 8260 : 59
k = 140
a = 140/2 = 70
b = 140/5 = 28
c = 140/7 = 20