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What else can I help you with?
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
What does N squared mean?
Oh, dude, N squared is just a fancy way of saying you multiply a number by itself. So, like, if N is 3, then N squared is 3 times 3, which equals 9. It's like math's way of saying, "Hey, let's make this more confusing than it needs to be."
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 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 least positive integer n such as that 210 divides n!?
To determine the least positive integer ( n ) such that ( 210 ) divides ( n! ), we first factor ( 210 ) into its prime components: ( 210 = 2 \times 3 \times 5 \times 7 ). For ( n! ) to be divisible by ( 210 ), ( n ) must be at least as large as the largest prime factor, which is ( 7 ). Thus, the least positive integer ( n ) such that ( 210 ) divides ( n! ) is ( n = 7 ).
What is a number that divides a different number?
A number that divides n is called a "divisor" of n.
What is the answer for (two n minus three)(n squared plus 5 n minus 1)?
It is 2n^3 + 7n^2 - 13n + 3
How do you solve by factoring if you only have something like n squared plus 5n equals 0?
n^2 + 5n = 0Where there is no constant term, then there is one monomial and one binomial factors. In this case the monomial is n and the binomial is (n + 5) So n*(n + 5) = 0 which leads to the solutions n = 0 or n = 5.
