Question

divida 459 em partes inversamente proporcionais aos numeros 3 , 4 , 10 , 6

Answer 1

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$\begin{array}{c|l}\sf3,4,10,6&\sf2\\\sf 3,2,5,3&\sf2\\\sf 3,1,5,3&\sf3\\\sf1,1,5,1&\sf5\\\sf1,1,1,1\end{array}\longrightarrow \bf mmc(3,4,10,6)=4\cdot3\cdot5=60$

$\sf\dfrac{1}{3}\cdot60=20\\\sf\dfrac{1}{4}\cdot60=15\\\sf\dfrac{1}{10}\cdot60=6\\\sf\dfrac{1}{6}\cdot60=10\\\underline{\rm c\acute alculo~da~constante~de~proporcionalidade:}\\\sf k=\dfrac{459}{20+15+6+10}=\dfrac{459}{51}=9$

$\boxed{\begin{array}{l}\sf os~n\acute umeros~procurados~s\tilde ao:\\\sf20\cdot9=180\\\sf15\cdot9=135\\\sf6\cdot9=54\\\sf10\cdot9=90\end{array}}\blue{\checkmark}$