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anonymous

  • one year ago

Solve this problem on paper using all four steps. Find four consecutive integers whose sum is 114. The integers are , , , and .

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  1. anonymous
    • one year ago
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    do 114/4=28.5 then pick 2 numbers greater than and 2 less than in a row 27+28+29+30=114

  2. Haseeb96
    • one year ago
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    let suppose n is the interger four consecutive integers are n, n+1,n+2,n+3 solution :- n+ n+1+n+2+n+3=114 4n +6=114 4n =114-6 4n = 108 n = 108 /4 n= 27 so four consecutive integers will be 27, 28, 29, 30 Answer

  3. anonymous
    • one year ago
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    @jjmack2 dont forget to medal and fan those who help u

  4. Michele_Laino
    • one year ago
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    hint: your problem can be modeled by the subsequent equation: \[\Large n + \left( {n + 1} \right) + \left( {n + 2} \right) + \left( {n + 3} \right) = 114\]

  5. Michele_Laino
    • one year ago
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    where n is an integer number

  6. Michele_Laino
    • one year ago
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    please, solve that equation for n

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