Matemática, perguntado por tecapeteca10, 3 meses atrás

99 dividido por números inversamente proporcionais a 2,4e6

Soluções para a tarefa

Respondido por Helvio
2

Dividindo 99  por números inversamente proporcionais a 2, 4 e 6

\large \text {$2 = 16,5     $}\\\\\large \text {$4 = 33    $}\\\\\large \text {$6 =  49,5$}\\\\

                       \large \text {$Grandezas~ inversamente~ proporcionais$}

a = 2

b = 4

c = 6

\large \text {$    $}\large \text {$\dfrac{a}{\dfrac{1}{2} }  $} = \large \text {$\dfrac{b}{\dfrac{1}{4} }  $} = \large \text {$\dfrac{c}{\dfrac{1}{6} }  $}\\\\\\\large \text {$\dfrac{a}{\dfrac{2}{1} }  $} = \large \text {$\dfrac{b}{\dfrac{4}{1} }  $} = \large \text {$\dfrac{c}{\dfrac{6}{1} }  $}\\\\\\\large \text {$ \dfrac{a}{2}    $} = \large \text {$ \dfrac{b}{4}    $} = \large \text {$ \dfrac{c}{6}    $}

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\large \text {$ \dfrac{a}{2}  = x   $}\\\\\\\large \text {$ a = 2x  $}

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\large \text {$ \dfrac{b}{4}  = x   $}\\\\\\\large \text {$b = 4x  $}

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\large \text {$ \dfrac{c}{6}  = x   $}\\\\\\\large \text {$ c = 6x  $}

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Substituindo os valores de a, b e c na expressão:

\large \text {$a + b  + c = 99    $}\\\\\\\large \text {$2x + 4x  + 6x = 99    $}\\\\\\\large \text {$12x = 99    $}\\\\\\\large \text {$x = \dfrac{99}{12}    $}\\\\\\\large \text {$x = 8,5   $}\\\\\\

Substituindo o valor de x em a, b e c

\large \text {$a = 2x    $}\\\\\large \text {$a = 2 ~. ~8,5    $}\\\\\large \text {$a =16,5    $}

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\large \text {$b = 4x    $}\\\\\large \text {$b = 4 ~. ~8,5    $}\\\\\large \text {$b =33    $}

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\large \text {$c = 6x    $}\\\\\large \text {$c = 6 ~. ~8,5    $}\\\\\large \text {$c =49,5    $}

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Prova:

16,5 + 33 + 49,5  = 99

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Anexos:

SocratesA: Ótima resposta Mestre.
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