Since consecutive integers differ by one, let's represent them as x, x + 1, x + 2, and x + 3. So we have,
x + (x + 1) + (x + 2) + (x + 3) = 94
4x + 6 = 94
4x = 88
x = 22
Thus, the numbers are 22, 23, 24, and 25. (Check:)
9 and 11
There is no such thing as "solving integers". You can solve an equation, which means finding all the unknowns in that equation, but you can't solve an integer.
by finding the way to solve your question..
Your question is not well formed, but i assume you mean 3 consecutive integers that sum to -363. If that is the case solve the following equation: (n-1) + (n) + (n+1) = -363 to give you the middle integer.
By finding a pattern the first time you solve a problem, then applying this pattern (algorithm) to solve similar problems.
Start with 1, go from there.... you'll find them all eventually.
9 and 11
It isn't possible to solve that. The sum of three consecutive integers is always a multiple of 3 (try it out with a few small numbers); 125 is not a multiple of 3, ergo, the problem has no solution.
The term "consecutive" only makes sense for integers; you can't solve this with integers.
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.
what is the formula to use to solve the product of two consecutive odd integers
31.how do you solve?
You can solve this in two ways.1) Trial and error. That is, try multiplying two consecutive integers; if the product is too large, try smaller integers; if the product is too small, try larger consecutive integers. 2) Call the two consecutive integers "n" and "n+1", and solve the equation: n(n+1)=210
There is no such thing as "solving integers". You can solve an equation, which means finding all the unknowns in that equation, but you can't solve an integer.
If three consecutive integers have the sum of 96, then the problem can be expressed with the equation... N + (N+1) + (N+2) = 96 ...Simplify that and solve and you get... 3N + 3 = 96 3N = 93 N = 31 ... so the three integers are 31, 32, and 33.
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.
You can do this by trial-and-error. Or, give the lowest of the four consecutive integers a name, like "x". The three other integers will then be "x+2", "x+4", and "x+6". So, you have to solve the equation:x + (x + 2) + (x + 4) + (x + 6) = 4The answer is the lowest of the four consecutive even integers.You can do this by trial-and-error. Or, give the lowest of the four consecutive integers a name, like "x". The three other integers will then be "x+2", "x+4", and "x+6". So, you have to solve the equation:x + (x + 2) + (x + 4) + (x + 6) = 4The answer is the lowest of the four consecutive even integers.You can do this by trial-and-error. Or, give the lowest of the four consecutive integers a name, like "x". The three other integers will then be "x+2", "x+4", and "x+6". So, you have to solve the equation:x + (x + 2) + (x + 4) + (x + 6) = 4The answer is the lowest of the four consecutive even integers.You can do this by trial-and-error. Or, give the lowest of the four consecutive integers a name, like "x". The three other integers will then be "x+2", "x+4", and "x+6". So, you have to solve the equation:x + (x + 2) + (x + 4) + (x + 6) = 4The answer is the lowest of the four consecutive even integers.