n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
Oh, dude, adding n squared plus n squared is like adding apples to apples, you know? It's just like, you take two n squared terms and you add them together to get 2n squared. It's not rocket science, man. Just double up those n squares and you're good to go.
10 - 52 = -15
It means N times N. 3 squared is 9.
(n2 + 2n - 1) (n2 + 2n - 1) = n4 + 2n3 - n2 + 2n3 + 4n2 - 2n - n2 - 2n + 1 = n4 + 4n3 + 2n2 - 4n + 1 try with n = 5: (5 squared + 10 - 1) squared = 34 squared = 1156 with formula (5^4) + (4 *(5^3)) + (2 * (5^2)) - (4 * 5) + 1 = 625 + 500 + 50 - 20 + 1 = 1156
Take any positive integer n. If you square it, and subtract 1, you get (x squared - 1). If you take (n - 1) and (n + 1), and multiply them together, you get n squared - n + n - 1, which is the same as (n squared - 1).
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
This is an immediate consequence of the Fundemental Theorem of Arithmetic. The "only if" part of the assertion is trivial because if n = 3k then n squared = 3(3k^2). If we know that every number can be written as the product of primes in exactly one way, then it follows that p is in the prime factorization of AB only if p is in the prime factorization of A or p is in the prime factorization of B. Take p = 3 and A = B = n, and that proves the theorem. The Fundemental Theorem is really not necessary in proving this assertion; an alternate proof using "enumeration of cases" is possible. If 3 does not divide n then n is of the form 3k+1 or 3k+2 for some integer k. It is easy to check that (3k+1)^2 and (3k+2)^2 are both not divisible by 3.
n = 5 245 * 5 = 1225 which is 35 squared
Oh, dude, adding n squared plus n squared is like adding apples to apples, you know? It's just like, you take two n squared terms and you add them together to get 2n squared. It's not rocket science, man. Just double up those n squares and you're good to go.
If that's n squared, the multiples are n squared, 2 n squared, 3 n squared and so on. If that's n + 2, the multiples are n + 2, 2n + 4, 3n + 6 and so on.
A number that divides n is called a "divisor" of n.
10 - 52 = -15
It means N times N. 3 squared is 9.
Simple equation lad. In your example, you said n=4 and x= n squared + n - 1 + (n-2)squared + (n-3)squared. You simply write n²+n-1+(n-2)²+(n-3)² I hope that is what you mean by what you say. P.S. To get the to the power of sign, hold alt and press 0178 for ², and 0179 for ³
Oh honey, that's just n times n minus n. So, n squared minus n is just n times (n-1). Hope that clears things up for ya, darling.
It is 2n^3 + 7n^2 - 13n + 3