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Find three consecutive numbers whose sum is 36.

solution please?​


Sagot :

✒MATHEMATICS

SOLUTION:

Write the place value of the following digits in the numeral 437.218

Let the three consecutive number are n , n + 1 , n + 2.

So,

  • n + (n + 1) + (n + 2) = 36
  • 3n + 3 = 36
  • 3n = 36 - 3
  • n = 33/3
  • n = [tex]{\boxed{\green{\sf{11 }}}}[/tex]

then solve for three consecutive,

  • (i) n = [tex]{\boxed{\green{\sf{11 }}}}[/tex]
  • (ii) n + 1 = 11 + 1 = [tex]{\boxed{\green{\sf{12 }}}}[/tex]
  • (iii) n + 2 = 11 + 2 = [tex]{\boxed{\green{\sf{13 }}}}[/tex]

ANSWER :

  • Therefore, the three consecutive numbers whose sum is 36 are 11, 12 and 13.

hope this helps

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Answer:

Three consecutive integers whose sum is 36 are 11, 12, and 13.

Step-by-step explanation:

Let the three consecutive numbers be :

( x ),( x+1 ), ( x+2 )

As per the condition given:

(x)+(x+1)+(x+2)=36

3x+3=36

3x=36

x=11

So the numbers are as follows: 11,12,13