Choose a nonzero integer for n to show -n can be evaluated as a positive number?
The product of two consecutive positive integers can be found by multiplying the smaller integer by the larger integer. If the smaller integer is represented as ( n ), then the larger integer would be ( n + 1 ). Therefore, the product of two consecutive positive integers is ( n \times (n + 1) ).
1,560 is.
10
Positive exponents: an = a*a*a*...*a where there are n (>0) lots of a. Negative exponents: a-n = 1/(a*a*a*...*a) where there are n (>0) lots of a.
Let's use N to represent any number.N x N = NN x -N = -N-N x -N = NSo the rules are:A positive integer times a positive integer will be a positive integerA positive integer times a negative integer will be a negative integerA negative integer times a negative integer will be a positive integer.
No, there is not. Given any positive integer n, n+1 is also a positive integer and it is larger.
Choose a nonzero integer for n to show -n can be evaluated as a positive number?
What a delightful little problemette ! It has to be the sum of the integers from 1 to 'n' . If 'n' is an even number, then that's n/2 times (n+1), (as in the young Gauss).
The product of two consecutive positive integers can be found by multiplying the smaller integer by the larger integer. If the smaller integer is represented as ( n ), then the larger integer would be ( n + 1 ). Therefore, the product of two consecutive positive integers is ( n \times (n + 1) ).
1,560 is.
Since n is positive, |n| = n, so you have 2n - n = n. The difference is n.
They are all the members of the set { n } where 1 ≤ n ≤ 180 . If we require that (n) = the greatest integer in (n), then there are 180 members in the set. If that condition is not required, then there are an infinite number of them.
605
n=27
10
No, it cannot.Consider a polygon which has n vertices, where n is an integer greater than or equal to 3.The sum of the interior angles of such a polygon is 180*(n - 2) degrees.Since n is an integer, (n - 2) must be an integer and so 180 must be a factor of the sum of the angles.180 does not divide 4300 and so it cannot be the sum of interior angles.It is, of course possible for a polygonal shape on a curved surface to have an angle sum of 4300 degrees, but such a shape would not be a polygon.