i.p - invers proportional cu ...
0.75 = [tex] \frac{75}{100} [/tex]=[tex] \frac{3}{4} [/tex]
0.5 = [tex] \frac{5}{10} [/tex] = [tex] \frac{1}{2} [/tex]
{x,y} i.p {[tex] \frac{3}{4} [/tex] ; [tex] \frac{1}{2} [/tex]}
[tex] \frac{3x}{4} [/tex] = [tex] \frac{y}{2} [/tex] = k
x = [tex] \frac{4k}{3} [/tex]
y= 2k
[tex] \frac{4k}{3} [/tex] + 2k= 40 / ×3
4k + 6k = 120
10k = 120
k= 12
x= [tex] \frac{3 x 12}{4} [/tex]
x= 16
y = 2 x 12
y = 24
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