Question

Imparte nr 7 in parti invers proportionale cu nr: 2;3; 0,5 si 0,(3). Va rog mult

Answer 1

Răspuns:

Explicație pas cu pas:

(a,b,c,d) i.p. (2; 3; 0,5; 0,(3)).  a+b+c+d=7,  0,(3)=3/9=1/3.

⇒2a=3b=0,5c=(1/3)d=k, ⇒a=k/2,  b=k/3,  c=k:0,5=k:(1/2)=k·2=2k,  d=k:(1/3)=k·3=3k.

a+b+c+d=7, ⇒k/2 + k/3 + 2k + 3k =7 |·6, ⇒6k/2 +6k/3 + 12k + 18k = 42, ⇒

⇒3k+2k+30k=42, ⇒ k=42/35, ⇒ k=6/5.

Atunci, a=6/5:2=6/5· (1/2)=3/5=0,6.

b=6/5:3=6/5 ·(1/3)=2/5=0,4

c=2·(6/5)=12/5=2,4

d=3·(6/5)=18/5=3,6

Răspuns: 0,6; 0,4; 2,4; 3,6.