Question

Folosind proprietatea fundamentala a proportiei,aflati:
a)a×b,stiind ca 3/a=b/5;
b)100·c×d,stiind ca c/0,3=0,2/d;
c)x/y,stiind ca 3·x=5·y;
d)m/n,stiind ca 6·n=2,5·m;
e)a/b,stiind ca 10a-3b=0;
f)2xy-5,daca x/2=6/y.

Answer 1

a)a×b,stiind ca 3/a=b/5;

axb=3x5
axb=15

b)100·c×d,stiind ca c/0,3=0,2/d;

100xcxd=100x0,3x0,2
100xcxd=100x0,06
100xcxd=6

c)x/y,stiind ca 3·x=5·y;
x/y=5/3

d)m/n,stiind ca 6·n=2,5·m;
m/n=6/2,5
m/n=2,4

e)a/b,stiind ca 10a-3b=0;
10a=3b
a/b=3/10

f)2xy-5,daca x/2=6/y.
x/2=6/y
xy=12
2xy-5=
2*12-5=
24-5=19