Dividindo o número 51 em partes diretamente proporcionais a 2, 3, 5, e 7 obtemos :
Soluções para a tarefa
Respondido por
2
vamos lá...
![\frac{a}{2} = \frac{b}{3} = \frac{c}{5} = \frac{d}{7} \\ a+b+c+d=51 \\ \\ \frac{a+b+c+d}{2+3+5+7} = \frac{51}{17} =3 \\ \\ \frac{a}{2} =3~~~~~~ \frac{b}{3} =3~~~~~~ \frac{c}{5} =3~~~~~~ \frac{~d}{7} =3 \\ \\ a=6~~~~~~b=9~~~~~~c=15~~~~~~d=21 \frac{a}{2} = \frac{b}{3} = \frac{c}{5} = \frac{d}{7} \\ a+b+c+d=51 \\ \\ \frac{a+b+c+d}{2+3+5+7} = \frac{51}{17} =3 \\ \\ \frac{a}{2} =3~~~~~~ \frac{b}{3} =3~~~~~~ \frac{c}{5} =3~~~~~~ \frac{~d}{7} =3 \\ \\ a=6~~~~~~b=9~~~~~~c=15~~~~~~d=21](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%7D%7B2%7D+%3D+%5Cfrac%7Bb%7D%7B3%7D+%3D+%5Cfrac%7Bc%7D%7B5%7D+%3D+%5Cfrac%7Bd%7D%7B7%7D++%5C%5C+a%2Bb%2Bc%2Bd%3D51+%5C%5C++%5C%5C++%5Cfrac%7Ba%2Bb%2Bc%2Bd%7D%7B2%2B3%2B5%2B7%7D+%3D+%5Cfrac%7B51%7D%7B17%7D+%3D3+%5C%5C++%5C%5C++%5Cfrac%7Ba%7D%7B2%7D+%3D3%7E%7E%7E%7E%7E%7E+%5Cfrac%7Bb%7D%7B3%7D+%3D3%7E%7E%7E%7E%7E%7E+%5Cfrac%7Bc%7D%7B5%7D+%3D3%7E%7E%7E%7E%7E%7E+%5Cfrac%7B%7Ed%7D%7B7%7D+%3D3+%5C%5C++%5C%5C+a%3D6%7E%7E%7E%7E%7E%7Eb%3D9%7E%7E%7E%7E%7E%7Ec%3D15%7E%7E%7E%7E%7E%7Ed%3D21)
Perguntas interessantes
Lógica,
11 meses atrás
Biologia,
11 meses atrás
Inglês,
11 meses atrás
Biologia,
1 ano atrás
Matemática,
1 ano atrás