If the amount of negative integers being multiplied is even, the result will be positive. If it is odd (like 101), the product will be negative.
a negative minus a positive is a negative ex: -12-13=-25 -9-2=-11 -100-1=-101
The only pair, with positive integers, is 1 and 101. But otherwise, there are infinitely many sets For example: 0.1*10*1*101 is one possibility
It is rational. Any integer is a rational number, since it can be written in the form (integer)/1.Negative 101 can be written as (-101)/(1) , (101)/(-1) , (-303)/(3) , (78,163)/(-763) ,or an infinite number of other ratios of integers.
There are 48.
101+102=203
The sum of the first positive odd integers less than 101 is 10,000.
there are 999 - 100 + 1 = 900 positive triple digit positive integers, between 100 and 999.(e.g. there are 102 - 100 + 1 = 3 triple digit integers between 100 and 102,namely 100, 101 and 102.)multiply that by 2 to take in consideration of the negative integers,you have 1800 triple digit integers.
(7*100*101)/2 = 35,350 jpacs * * * * * What? How can there be 35,350 integers in the first 100 integers? There are 14 of them.
Yes
a negative minus a positive is a negative ex: -12-13=-25 -9-2=-11 -100-1=-101
The only pair, with positive integers, is 1 and 101. But otherwise, there are infinitely many sets For example: 0.1*10*1*101 is one possibility
It is rational. Any integer is a rational number, since it can be written in the form (integer)/1.Negative 101 can be written as (-101)/(1) , (101)/(-1) , (-303)/(3) , (78,163)/(-763) ,or an infinite number of other ratios of integers.
To find the sum of the first 100 integers, you first add 1 plus 100 (the first and last numbers of the set) and get 101. Do the same with the next two integers, 2 and 99 and you'll get 101. Since you are adding two integers at a time and there are 100 integers between 1 and 100, you'll get 101, fifty times. Therefore, a shortcut would be to simply multiply 101 times 50 and get 5,050.
Yes 101 is an integer
101 and 102
101+102=203
101 and 102