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aflati a b c stiind ca sunt proportionale co 0,5,0,2,0 , (3) si a + b + c = 930


Răspuns :

{a, b, c} d. p. {0.5 ; 0.2 ; 0.(3)}
a + b + c = 930
[tex] \frac{a}{0.5} [/tex] = [tex] \frac{b}{0.2} [/tex] = [tex] \frac{c}{0.(3)} [/tex] = k
=> a = 0.5k
b = 0.2k
c = 0.(3)k
0.5k + 0.2k + 0.(3)k = 930
0.5 = [tex] \frac{1}{2} [/tex]
0.2 = [tex] \frac{1}{5} [/tex]
0. (3) = [tex] \frac{1}{3} [/tex]
[tex] (\frac{1}{2}) ^{15} [/tex]k + [tex] (\frac{1}{5}) ^{6} [/tex]k + [tex]( \frac{1}{3}) ^{10} [/tex]k = 930[tex] ^{30} [/tex]
15k + 6k + 10k = 27900
31k = 27900
k = 27900 : 31
k = 900 => a = [tex] \frac{1}{2} [/tex] x 900 = 450
                 b = [tex] \frac{1}{5} [/tex] x 900 = 180
                 c = [tex] \frac{1}{3} [/tex] x 900 = 300
:)