Question

Se imparte nr 1268 in parti invers proportiinale cu nr 0,(6),0,75,0,8,0,8(3)

Answer 1

x+y+z+t=1268
0,(6)=6/9=2/3
0,75=75/100=3/4
0,8=8/10=4/5
0,8(3)=(83-8)/90=75/90=5/6
2x/3 = 3y/4 = 4z/5 = 5t/6 = k
x=3k/2
y=4k/3
z=5k/4
t=6k/5
______ +
Numitor comun 60
90k+80k+75k+72k=1268•60
317k=1268•60
k=(1268•60)/317
k=240
x=(240•3)/2=360
y=(240•4)/3=320
z=(240•5)/4=300
t=(240•6)/5=288

Answer 2

Succes in continuare!