Question

10 Determinati numerele a, b și c direct proportionale cu 0,25; 0,1(6) şi 0,(09) știind că 8a+4b-3c= 52. ​

Answer 1

Răspuns:

Explicație pas cu pas:

{a ; b ; c} d.p {0,26 ; 0,1(6) ; 0,(09)}

$\frac{a}{0,25} = \frac{b}{0,1(6)} = \frac{c}{0,(09)} = k,$ unde k = coeficient de propoționalitate

$0.25 = \frac{25 {}^{(25} }{100} = \frac{1}{4}$

$0,1(6) = \frac{16 - 1}{90} = \frac{15 {}^{(15} }{90} = \frac{1}{6}$

$0,(09) = \frac{9 {}^{(9} }{99} = \frac{1}{11}$

$\frac{a}{ \frac{1}{4} } = \frac{b}{ \frac{1}{6} } = \frac{c}{ \frac{1}{11} } = k$

$a = \frac{1}{4} k$

$b = \frac{1}{6} k$

$c = \frac{1}{11} k$

_________________

8a + 4b - 3c = 52

$8 \times \frac{1}{4} k + 4 \times \frac{1}{6} k - 3 \times \frac{1}{11} k = 52$

$\frac{8}{ 4} k + \frac{4 {}^{(2} }{6} k - \frac{3}{11} k = 5 2$

$2k + \frac{2}{3} k - \frac{3}{11} k = 52 \: \: \: \Big| \times 33$

66k + 22k - 9k = 1716

79k = 1716

$k = \frac{1716}{79}$

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$a = \frac{1}{4} \times \frac{1716}{79} = \frac{1716 {}^{(4} }{316} = \frac{429}{79}$

$b = \frac{1}{6} \times \frac{1719}{79} = \frac{1716 {}^{(6} }{474} = \frac{286}{79}$

$c = \frac{1}{11} \times \frac{1716}{79} = \frac{1716 {}^{(11} }{869} = \frac{156}{79}$