Question #b415d

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Question #b415d

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Answer 1 · 2017-03-09T13:29:42

ops!....I read three....disregard my answer!

Explanation:

We can call our consecutive integers:
$n, n + 1, n + 2$ $n, n+1,n+2$
so we get:
$n + (n + 1) + (n + 2) = - 66$ $n+(n+1)+(n+2)=-66$
solve for $n$ $n$
$3 n + 3 = - 66$ $3n+3=-66$
$3 n = - 66 - 3$ $3n=-66-3$
$3 n = - 69$ $3n=-69$
$n = - \frac{69}{3} = - 23$ $n=-69/3=-23$
so our integers will be:
$n = - 23$ $n=-23$
$n + 1 = - 23 + 1 = - 22$ $n+1=-23+1=-22$
$n + 2 = - 23 + 2 = - 21$ $n+2=-23+2=-21$

Answer 2 · 2017-03-09T13:30:27

$- 18, - 17, - 16, - 15$ $-18,-17,-16,-15$

Explanation:

$Let the 1st integer be n$ $"Let the 1st integer be "n$

Since they are consecutive then the next 3 integers will be.

$n + 1, n + 2, n + 3$ $n+1,n+2,n+3$

$\Rightarrow n + (n + 1) + (n + 2) + (n + 3) = - 66$ $rArrn+(n+1)+(n+2)+(n+3)=-66$

simplifying left side gives.

$4 n + 6 = - 66$ $4n+6=-66$

subtract 6 from both sides.

$4ncancel(+6)cancel(-6)=-66-6$

$rArr4n=-72$

divide both sides by 4

$(cancel(4) n)/cancel(4)=(-72)/4$

$rArrn=-18$

$rArrn+1=-18+1=-17$

$rArrn+2=-18+2=-16$

$rArrn+3=-18+3=-15$

$"The 4 consecutive integers are "-15,-16,-17,-18$

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Question #b415d

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