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What must you multiply n squared to get n cubed?
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
How do you add n squared plus n 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.
What is the 5 Th term of 10-N squared?
10 - 52 = -15
What is the square of n squared plus 2n minus 1?
(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
If a divides b which one is the denomitor?
If a divides b then a is a factor of b and b is a multiple of a.Either of them could be the denominator. In a/b, b is the denominator while in b/a, a is the denominator.------------------------------------------------------------------------------------------------------------------If a divides b then b=n*a for some number n. Thus n=b/a
How is it that 3 is the only prime that is before a square number?
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).
What must you multiply n squared to get n cubed?
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
Prove the lemma for each integer n 3 divides n if and only if 3 divides n squared?
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.
What is the smallest counting number n such that the product 245n is a perfect square?
n = 5 245 * 5 = 1225 which is 35 squared
How do you add n squared plus n 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.
What are the multiples of n 2?
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.
What is a number that divides a different number?
A number that divides n is called a "divisor" of n.
What is the 5 Th term of 10-N squared?
10 - 52 = -15
How do you mathematically express x equals n squared where it repeats until n equals 0 eg for n equals 4 x equals n squared plus n-1 squared plus n-2 squared plus n-3 squared?
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
What is the answer for (two n minus three)(n squared plus 5 n minus 1)?
It is 2n^3 + 7n^2 - 13n + 3
If n is any prime no.and a2 is divisible by n then n will also divide a justify?
If n is any prime number and a² is divisible by n, then n will also divide a. Consider that 'n' exists as a prime number. If n divides a² then n also divides a. Let, n = 3 and a = 9 a² = 9² a² = 81 Here , 81/3 = 27 n divides a² Now, 9/3 = 3 Here, n also divides a.
How do you do n squared plus n?
(n*n)+n
