First do 135 divided by 3, which is 45 then you take two off the 45 which is 43. Also add 2 to the 45 to make 47. So 43+45+47= 135
2007-01-21 20:10:26
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answer #1
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answered by james 2
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i'm going to help you with the 1st one. which will, in turn, help you with the subsequent 3. So: permit x be the 1st integer. Then x+a million is the subsequent consecutive integer, outstanding? And their sum is x + (x+a million). so which you're able to sparkling up the equation: x + (x+a million) = 7^2 2x + a million = 40 9 2x = 40 8 x = 24 so x+a million = 25; to that end, 24 and 25 is the answer to the 1st subject. For #5, you're able to translate this word subject into algebra. back, x is the 1st integer, x+a million the 2nd. Ten below two times the smaller is 2x - 10. So the equation you desire is: x+a million = 2x-10 11 = x and 12 = x+a million
2016-12-16 10:27:02
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answer #2
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answered by ? 4
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I would divide 135 by 3 to get an average and it came out to 45 then since you are looking for consecutive odd integers I would take the one below and the one above the average which would make it 43,45,47
2007-01-22 00:55:49
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answer #3
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answered by Dave aka Spider Monkey 7
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The first is odd so we can write it as x+1
if x+1 is the smallest than the others are x+3 and x+5
Their sum is x+1+x+3+x+5 = 3x+9
so x=42
the integers are 43 , 45 , 47
2007-01-21 21:08:55
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answer #4
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answered by maussy 7
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x=the first odd integer
x+2=the second odd integer
x+4=the third odd integer
x+x+2+x+4=135
3x+6=135
3x+6-6=135-6
3x/3=129/3.x=43-the first integer. x+2=43+2=45 - the second integer..x+4=43+4=47-the third integer.
43+45+47=135.
2007-01-21 20:43:34
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answer #5
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answered by Max 6
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let the 1st integer be = x
let the 2nd integer be = x + 2
let the 3rd integer be = x + 4
now,solve:
x + x + 2 + x + 4 = 135
3x + 6 = 135
transpose, 6 will be negative:
3x = 135 - 6
3x = 129
divide both sides by 3:
x = 43
substitute:
1st integer : x = 43
2nd integer: 43 + 2 = 45
3rd integer: 43 + 4 = 47
2007-01-21 20:16:09
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answer #6
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answered by genius_06 3
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135/3 = 45
43 + 45 + 47 = 135
2007-01-21 21:11:19
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answer #7
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answered by Helmut 7
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x + (x+2) + (x+4) = 135
3x + 6 = 135
x = 43
Verify : 43 + 45 + 47 = 135
2007-01-21 20:18:07
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answer #8
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answered by shyamKrishna v 1
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X=1st odd integer
X+2=2nd odd integer
X+4=3rd odd integer
135=sum of the three
equation:
X+X+2+X+4=135
3X+6=135
3X=135-6
3X=129
Divide both sides by 3
X=43
X+1=45
X+2=47
2007-01-21 20:12:54
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answer #9
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answered by Princess Shai 3
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Using Model Method
[XXXXXXX] - 2
[XXXXXXX] {This is the middle Bar}
[XXXXXXX] + 2
Add up the above 3 Bars
3 x [XXXXXXX] = 135
Therefore the middle Bar is,
[XXXXXXX] = 135/3
[XXXXXXX] = 45
Therefore the three consecutive odd integers are:
43, 45 & 47
2007-01-21 21:41:36
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answer #10
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answered by ideaquest 7
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