(10-2)-1 = 7 whole numbers
There are 10 whole numbers in 10 to 19.
There are an infinite amount of numbers that that have squares between 10 and 65
Assuming the question is meant to be about the whole numbers between which root 98 lies, the answers are: -10 and -9, and 9 and 10.
10 and a half, being between the whole numbers 10 and 11, can't be a whole number itself.
(10-2)-1 = 7 whole numbers
There are 10 whole numbers in 10 to 19.
Only 5 whole numbers:6, 7, 8, 9, and 10.
Two of them.
Both 10 and 1000 are individual whole numbers.
You know that sum of the first n whole numbers is n(n+1)/2. ( it is the same as the first n natural numbers since the zero does not add anything) So lets say you want the sum of all the whole numbers between 3 and 10. ( I made it easy to illustrate the idea.) The sum of the whole numbers between 0 and 3 is 3(4)/2=6 The sum of the whole numbers between 0 and 10 is 10(11)/2=55 So the sum of the whole numbers between 3 and 10 is the (sum of the whole number between 0 and 10) -(sum of whole numbers between 0 and 3) which is 55-6=49 So in general, for whole numbers m and n with m
There are an infinite amount of numbers that that have squares between 10 and 65
56789
Assuming the question is meant to be about the whole numbers between which root 98 lies, the answers are: -10 and -9, and 9 and 10.
10 and a half, being between the whole numbers 10 and 11, can't be a whole number itself.
If you allow fractions and decimals, then there are an infinite number of them. If you only want to talk about whole numbers, then there are only nine of them.
There are no prime numbers in between these numbers. The only whole number between them is 6