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Q: Three consecutive integers whose sum is 135?

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Those three odd integers are 43, 45, and 47.

Given the sum of three consecutive integers 135, we write out the expression as:x + (x + 1) + (x + 2) = 135 where:x = smallest integerx + 1 = middle integerx + 2 = largest integerSolve for x by rearranging the terms:3x + 3 = 1353x = 132x = 44Therefore, the largest integer is x + 2 = 46.

43+45+47=135. i had this question for my online math class. but i don't know how to make that into an equation. something like 3x+2...maybe.. here is how you do it: We need three consecutive odd integers so we will let the first one be x, the next is x+2 and then third is x+4 3x+6=135 3x=129 x-43 x+2=45 x+4=47 So the three numbers are 43, 45 and 47

44,45,46

44,45,46

44 + 45 + 46 = 135

44 + 45 + 46 = 135

let the three consecutive no.s be k-1,k and k+1,where k is a positive whole no. so (k-1)+k+(k+1)=135 according to the question or, 3k=135 or, k=45 which gives the no.s as 44,45,46 44+45+46=135

45, 90 and 135

135 and 136.

Let x=1st integer x+1=2nd integer x+2=3rd integer 5(x+x+1+x+2)=150 5(3x+3)=150 15x+15=150 15x=135 x=9 So you put what x is 1st integer (x)= 9 Since it's three consecutive integers, you add one to the 2nd integer 2nd integer (x+1)= 10 Then you do the same thing for the third integer 3rd integer (x+2)= 11

First of all, let's set up the equation.x=the first number=44x+1=the second number=45x+2=the third number=46x+x+1+x+2=1353x+3=1353(x+1)=135x+1=45x=44

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