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What is three consecutive integers whose sum is 117?
Three consecutive integers whose sum is 117 are 38, 39, and 40. N + (N+1) + (N+2) = 117 3N + 3 = 117 3N = 114 N = 38
What are the three consecutive integers of 354?
117, 118 and 119.
What are the three consecutive odd integers with the sum of 117?
The numbers are 37, 39 and 41.
What are two consecutive integers of the square root of 117?
The square roots of 117 are irrational numbers and so are not two integers - consecutive or otherwise.
Ind three consecutive odd integers that have a sum of 123?
x is lowest of the integers so x + (x + 2) + (x + 4) = 123 ie 3x + 6 = 123 so 3x = 117 and x = 39, so integers are 39, 41 and 43
Find three Consecutive odd integers whose sum is 117?
117/3 = 39, so the three consective odd integers whose sum is 117 are 37, 39, and 41.
What is three consecutive integers whose sum is 117?
Three consecutive integers whose sum is 117 are 38, 39, and 40. N + (N+1) + (N+2) = 117 3N + 3 = 117 3N = 114 N = 38
What are the three consecutive integers of 354?
117, 118 and 119.
What are the three consecutive odd integers with the sum of 117?
The numbers are 37, 39 and 41.
What are three consecutive integers whose sum equals -354?
(-117)+(-118)+(-119)=(-354)
What are two consecutive integers of the square root of 117?
The square roots of 117 are irrational numbers and so are not two integers - consecutive or otherwise.
Between which two integers does the square root of 117 lie?
10 and 11 or -11 and -10
What are 3 consecutive integers that have a sum of 117?
38, 39, 40
Ind three consecutive odd integers that have a sum of 123?
x is lowest of the integers so x + (x + 2) + (x + 4) = 123 ie 3x + 6 = 123 so 3x = 117 and x = 39, so integers are 39, 41 and 43
How is 117 composite?
117 = 3*39, a product of two integers which are larger than 1.
Why are 117 composite?
117 = 3*39, a product of two integers greater than 1.
What are three consecutive odd integers that have a sum of 123?
39, 41, 43 Let x represent the smallest of these numbers. From the problem, we know x+(x+2)+(x+4)=123 Solving for x: 3x+6=123 3x=117 x=117/3=39 So our integers are: 39, 41, 43
