2007-10-11 13:03:09 · 7 answers · asked by ichigo 1 in Science & Mathematics ➔ Mathematics
15,16,17
2007-10-11 13:05:59 · answer #1 · answered by Lynn A 4 · 1⤊ 0⤋
Easy. Let x be the first integer. Then:
(x) + (x + 1) + (x + 2) = 48
Simplifying:
3x + 3 = 48
3x = 45
x = 15
So your three consecutive integers are 15, 16, and 17.
2007-10-11 20:06:49 · answer #2 · answered by Lucas C 7 · 2⤊ 0⤋
Let n be the first, so that the next two are n+1, n+2
n + (n+1)+ (n+2) = 48
3n + 3 = 48
3n = 48 - 3 = 45
n = 45/3 = 15
n+1 = 16
n+2 = 17
2007-10-11 20:08:01 · answer #3 · answered by kellenraid 6 · 0⤊ 0⤋
let the first integer be x
second integer is (x+1)
third integer is (x+2)
sum them to get: x + (x+1) + (x+2) =48
3x = 45
x = 15
therefore: (x+1) = 16
and (x+2) = 17
2007-10-11 20:08:16 · answer #4 · answered by minorchord2000 6 · 0⤊ 0⤋
x + (x + 1) + (x + 2) = 48
3x + 3 = 48
3x = 45
x = 15
15 , 16 and 17 is answer.
2007-10-13 16:59:15 · answer #5 · answered by Como 7 · 0⤊ 0⤋
15
16
17
2007-10-11 20:06:12 · answer #6 · answered by ωĨŞΣ Ĝųγ 4 · 1⤊ 0⤋
15,16,17
2007-10-11 20:06:38 · answer #7 · answered by Renaissance Man 5 · 0⤊ 0⤋