Question

aflati nr i.p cu 0,0(3),1_2 si si 0,(3) stiind ca media lor aritmetica este175

Answer 1

0,0(3)=3/90=1/30 si 0,(3)=3/9=1/3
⇒x/30=y/2=z/3=k⇒x=30·k;y=2·k;z=3·k
(x+y+z)/3=175⇒x+y+z=175·3=525
30·k+2·k+3·k=525⇒35·k=525⇒k=15
x=30·k=30·15=450
y=2·k=2·15=30
z=3·15=45
proba:450+30+45=525

Answer 2

{a,b,c}i.p.{0,0(3);$\frac{1}{2}$;0,(3)}
0,0(3)=$\frac{3}{90}= \frac{1}{30}$
0,(3)=$\frac{3}{9}= \frac{1}{3}$
=>{a,b,c}i.p.{$\frac{1}{30}; \frac{1}{2}; \frac{1}{3}$}=>
$\frac{a}{30}= \frac{b}{2}= \frac{c}{3}$=K⇒
a=30K
b=2K
c=3K
(a+b+c):3=175⇒a+b+c=525
30K+2K+3K=525
35K=525
K=15
a=30·15=450
b=2·15=30
c=3·15=45
sper ca te-am ajutat :)