They are numbers of the form m, m+1 and m+2 where m is an integer. However, sometimes it can be easier - particularly with an odd number of consecutive integers - to write them as n-1, n and n+1 where n is an integer (= m+1).
Let x be the first integer. Then the sum of the three consecutive integers is x + (x+1) + (x+2), which equals 3x + 3. We are given that this sum is 43, so we can write the equation 3x + 3 = 43. Solving this equation, we find that x = 13. Therefore, the three consecutive integers are 13, 14, and 15.
-318 Negatives are integers.
Integers by definition do not have any decimals, so that is impossible.
120 is an integer. Integers aren't fractions or mixed numbers. You can write 120 as 120/1 or 119 and 4/4, but it wouldn't be in simplest form. The simplest form of any integer is the integer itself.
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.
They are numbers of the form m, m+1 and m+2 where m is an integer. However, sometimes it can be easier - particularly with an odd number of consecutive integers - to write them as n-1, n and n+1 where n is an integer (= m+1).
Let x be the first integer. Then the sum of the three consecutive integers is x + (x+1) + (x+2), which equals 3x + 3. We are given that this sum is 43, so we can write the equation 3x + 3 = 43. Solving this equation, we find that x = 13. Therefore, the three consecutive integers are 13, 14, and 15.
Let's denote the first even integer as x. The next consecutive even integer would be x + 2. Since they sum up to 60, we can write the equation x + (x + 2) = 60. Simplifying this equation gives us 2x + 2 = 60. Solving for x, we find x = 29. Therefore, the two consecutive even integers are 29 and 31, and their sum is 60.
-318 Negatives are integers.
Integers by definition do not have any decimals, so that is impossible.
(j)*(j+2)
Find two consecuitive integers whose sum is 89. To solve this problem, let x be the smaller of these integers. What is the larger of these two consecutive integers? In terms of x, write a formula that represents the sum of these two consecutive integers.
i dont now
That isn't possible; three consecutive integers, or three consecutive positive integers, always have a sum that is a multiple of 3. In general, you can solve this quickly by trial and error. In this case, you will quickly find that a certain set of three consecutive integers will give you a sum that is TOO LOW, while the next-higher even integers will give you a sum that is TOO HIGH. You can also write an equation and solve it: n + (n + 2) + (n + 4) = 32. If you solve it, you will find that the solution is fractional, not integral.
120 is an integer. Integers aren't fractions or mixed numbers. You can write 120 as 120/1 or 119 and 4/4, but it wouldn't be in simplest form. The simplest form of any integer is the integer itself.
465,000,000 is an integer. Mixed numbers are between integers greater than one.