Question

devida o número 380 em partes inversamente proporcionais a 2,5 e 4​

Answer 1

$\large\boxed{\begin{array}{l}\begin{array}{l|l}\rm2,4,5&\rm2\\\rm1,2,5&\rm2\\\rm1,1,5&\rm5\\\rm1,1,1\end{array}\\\rm m\cdot m\cdot c\,(2,4,5)=2\cdot2\cdot5=20\\\rm\dfrac{20}{2}=10~~\dfrac{20}{4}=5~~\dfrac{20}{5}=4\\\rm constante\,de\,proporcionalidade:\\\rm k=\dfrac{total}{\sum~das~raz\tilde oes}\\\\\rm k=\dfrac{380}{10+5+4}=\dfrac{380}{19}=20\\\\\rm 1^a\,parte=20\cdot10=200\\\rm 2^a\,parte=20\cdot5=100\\\rm 3^a\,parte=20\cdot4=80\end{array}}$

Answer 2

$> resolucao \\ \\ \\ > \frac{x}{2} + \frac{x}{5} + \frac{x}{4} = 380 - - mmc = 20 \\ \\ > \frac{10x + 4x + 5x}{20} = \frac{7600}{20} \\ \\ > 19 \times = 7600 \\ \\ x = \frac{7600}{19} \\ \\ x = 400 \\ \\ \\ > \frac{x}{2} = \frac{400}{2} = 200 \\ \\ > \frac{x}{5} = \frac{400}{5} = 80 \\ \\ \frac{x}{4} = \frac{400}{4} = 100 \\ \\ \\ > resposta = \: as \: partes \: sao \: 200 \: . \: 80 \: e \: 100 \\ \\ \\ > > > > >$