English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

14 answers

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 · answer #1 · answered by james 2 · 3 0

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 · answer #2 · answered by ? 4 · 0 0

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 · answer #3 · answered by Dave aka Spider Monkey 7 · 0 0

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 · answer #4 · answered by maussy 7 · 0 0

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 · answer #5 · answered by Max 6 · 0 0

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 · answer #6 · answered by genius_06 3 · 1 0

135/3 = 45
43 + 45 + 47 = 135

2007-01-21 21:11:19 · answer #7 · answered by Helmut 7 · 0 0

x + (x+2) + (x+4) = 135
3x + 6 = 135
x = 43

Verify : 43 + 45 + 47 = 135

2007-01-21 20:18:07 · answer #8 · answered by shyamKrishna v 1 · 0 0

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 · answer #9 · answered by Princess Shai 3 · 0 1

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 · answer #10 · answered by ideaquest 7 · 0 0

fedest.com, questions and answers