You can do this by trial-and-error.Or, call one integer "n", the next one "n+1", and solve the equation:
n(n+1) = 440
Note: Whichever of these two methods you use, it will be clear that this problem has no integer solution.
Consecutive negative integers that sum to 440 would be integers that are sequentially negative and add up to that positive number. For example, the integers -1, -2, -3, and so forth are negative integers, but their sum cannot reach 440 since they are all negative. If you meant the consecutive negative integers that multiply to give -440, those could be -20 and -22, as they are consecutive and their product is 440.
There are no such integers. 20*21 = 420 is too small while 21*22 = 462 is too large.
As the product of its prime factors: 2*2*2*5*11 = 440 or as 23*5*11 = 440
440 = 2 x 2 x 2 x 5 x 11 OR 23 x 5 x 11
Any number (positive or negative) minus zero is just the original number. e.g. 3 - 0 = 3 -440 - 0 = -440 -2 - 0 = -2 What you are really saying here is take away nothing from a number.
There are no two consecutive integers, negative or positive, whose product is 440.
I think you mean consecutive even integers: 20 & 22
There are no such integers. 20*21 = 420 is too small while 21*22 = 462 is too large.
Multiply 440 x .07 and then add the product to 440: (440 x .07) + 440 = 30.80 + 440 = 470.80
The product is 440
As the product of its prime factors: 2*2*2*5*11 = 440 or as 23*5*11 = 440
2 x 2 x 2 x 5 x 11 = 440
440 + 440 + 440 + 440 + 440 + 440 = 2,640
440 = 2 x 2 x 2 x 5 x 11 OR 23 x 5 x 11
Any number (positive or negative) minus zero is just the original number. e.g. 3 - 0 = 3 -440 - 0 = -440 -2 - 0 = -2 What you are really saying here is take away nothing from a number.
0.000 000 000 000 810 440 156
100% of 440= 100% * 440= 1 * 440= 440