44,45,46
44 + 45 + 46 = 135
44 + 45 + 46 = 135
44,45,46
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
The numbers are 135 and 136.
137
134 and 135
138
Those three odd integers are 43, 45, and 47.
Let's call the three consecutive integers x, x+1, and x+2. The sum of these integers is x + (x+1) + (x+2) = 3x + 3. We know that this sum is equal to 135, so we can set up the equation 3x + 3 = 135. Solving for x gives us x = 44. Therefore, the three consecutive integers are 44, 45, and 46.
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