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.
44 + 45 + 46 = 135
44 + 45 + 46 = 135
It is: 4.21875 times 32 = 135
135 = 13,500 percent
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.
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.
44,45,46
44,45,46
44 + 45 + 46 = 135
44 + 45 + 46 = 135
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
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
135
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
135 and 136.