Odd integers differ by 2.
Let x = 1st integer =23
x+2 = 2nd integer = 25
x+4 = 3rd integer = 27
x + 6 = 4th integer = 29
x + x + 2 + x + 4 + x + 6 = 104
4x + 12 = 104
4x = 92
x = 23
2007-10-27 10:37:51
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answer #1
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answered by dpirsq2 5
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If they are consecutive odd ones then they are 2 apart from each other. If the first one is N then the second one is N+2, the third is N+4, and the fourth is N+6. Add them up and they total 104.
so N + (N+2) + (N+4) + (N+6) = 104
Combining the N part and the number part you get 4N+12 = 104
So 4N = 92 and N = 92/4 = 23
That makes the numbers 23, 25, 27, 29.
2007-10-27 10:33:25
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answer #2
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answered by Rich Z 7
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23+25+27+29=104
2007-10-27 10:48:38
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answer #3
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answered by sciencegeek 2
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I love using guess and check. Since the perimeter is close to 100 each side should be about 25. So start with 23 and add 25, 27, and 29 --> you get 104! Hope this helps =)
2007-10-27 10:32:57
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answer #4
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answered by Sleeping Beauty 2
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since the sides are consecutive odd integers the sides should be x , x+2 , x+4 , x+6
perimeter = 104
so x + (x+2) + (x+4) + (x+6) = 104
4x + 12 = 104
4x = 92
x = 23
from this u can find the individual sides
23, 25 , 27, 29
2007-10-27 10:30:34
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answer #5
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answered by abs 2
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you have to make an equation-
It would be
x+(x+2)+(x+4)+(x+6)=104 feet.
Solve for the answer: 4x+12=104 feet
4x=92
x=23
This means the shortest side in 23 feet and the rest are 25,27, and 29. Have a good day.
2007-10-27 10:33:28
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answer #6
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answered by Maglio Ordonez 2
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I won't solve it for you, but I will help you get started. You have to look at the ones place first. You need to find the combination of numbers that when added, equatl 4, then you can move onto the 10's place. Obviously, you need to pick 4 of the following and then play with the numbers a bit: 1 3 5 7 9 . Have fun solving your work :)
2007-10-27 10:33:29
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answer #7
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answered by Kimberly V 3
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hhhhhhhhhhmmmmmmmmmm im sure someone wil help you good luck thnx for the 2 points
2007-10-27 10:29:31
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answer #8
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answered by Anonymous
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dont know
2007-10-27 10:29:54
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answer #9
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answered by Anonymous
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