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  1. i did a little reaserch into this and i find b as the most plausable awnser i might be wrong tho so tell me if i am


  2. The integers are [tex]\boxed{52}[/tex], [tex]\boxed{53}[/tex] and [tex]\boxed{54}[/tex].

    Further explanation:


    The sum of the first two is 51 more than the third.


    Consider the first integers as [tex]x[/tex].

    So the second integers is [tex]x+1.[/tex]

    Therefore, the third integers is [tex]x+2.[/tex]

    Given that the sum of first two numbers is 51 more than the third.

    [tex]\begin{aligned}\left( x \right) + \left( {x + 1} \right) &= 51 + \left( {x + 2} \right)\\x + x + 1 &= 51 + x + 2\\2x + 1 &= 53 + x\\2x - x &= 53 - 1 \\x&= 52\\\end{aligned}[/tex]

    The first integer is 52.

    The second integer can be calculated as,

    [tex]\begin{aligned}{\text{Second integer}} &= x + 1\\&= 52 + 1\\&= 53\\\end{aligned}[/tex]

    The second integer can be calculated as,

    [tex]\begin{aligned}{\text{Third integer}}&= x + 2\\&= 52 + 2\\&= 54\\\end{aligned}[/tex]

    Hence, the integers are [tex]\boxed{52}, \boxed{53}[/tex] and [tex]\boxed{54}.[/tex]

    Learn more:

    Learn more about inverse of the Learn more about equation of circle Learn more about range and domain of the function

    Answer details:

    Grade: High school

    Subject: Mathematics

    Chapter: Linear equation

    Keywords: Three consecutive, consecutive, sum, addition, integers, more, 51 more, consecutive integers, first two.

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