An N3 certificate requires a minimum of four passed technical subjects. N3 subjects are at a level of Matric. N3 certificate must have an addition of two languages to be fully equivalent to a Matric certification.
6+N3 is the sum of N*3 then you add six. 6*N3 is you multiply 6 times three times N times three.
Input N1, N2, N3 Max = N1 If N2 > Max then Max = N2 If N3 > Max then Max = N3 Display Max
The sequence is too short for a definitive answer. One possibility is Un = n2 - 2n + 2 for n = 1, 2, 3, ... another is Un = (n3 - n + 6)/6 or Un = (n3 - 3n2 + 5)/3 There are many more.
n3 + 3n2 + 4n + 12 = (n3 + 3n2) + (4n + 12) = n2(n + 3) + 4(n + 3) = (n2 + 4)(n + 3).
n3 + 1 = n3 + 13 = (n + 1)(n2 - n + 12) = (n + 1)(n2 - n + 1)
Let any number be n:- n3/n3 = n*n*n/n*n*n = 1 And in index form: n3/n3 = n3-3 = n0 = 1
class Program { static void Main(string[] args) { int n1, n2, n3,i; n1 = 0; n2 = 1; for (i = 1; i <= 20; i++) { n3 = n1 + n2; if (n3 <= 200) { Console.WriteLine(n3); n1 = n2; n2 = n3; } } Console.ReadKey(); } }
tn = n3
Let the number be n.Then (n3)3 = n3 x 3 = n9.........or n to the power nine.However, if the question is what the the one-third power of a number cubed, then:(n3)1/3 = n3 x 1/3 = n1 = n
N3-13
The simplest formula for nitride is N. Nitride refers to compounds containing nitrogen ions in a -3 oxidation state, such as lithium nitride (Li3N) or magnesium nitride (Mg3N2).
This can be solved by applying the laws of exponents - such that nx/ny = n(x-y), if the two exponents are the same, for example, n3/n3 = n3-3 = n0 = 1. Plugging an example number into the equation: 43/43 = 64/64 = 1 Therefore, 40 = 1. It follows therefore, that x0 = 1.
The element that completes n3 is lithium (Li).
Azide, N3- ( a linear ion isoelectronic with CO2), or conceivably nitride N3- a monoatomic ion
Formula: N3-
n3