N2 / N3 qualification is a trade theory (Electrician, Motor mechanic, Plumber, Mechanotechnologist, Geologist and related career) qualification before you do your Apprentice. I South Africa, before you are admitted for and apprentice service Training) you need to have a N3 qualification.
When an electron falls from n4 to n1, it releases more energy because it is transitioning between high energy states. This higher energy transition corresponds to a shorter wavelength of light being emitted, according to the energy of the photon being inversely proportional to its wavelength. In contrast, when an electron falls from n2 to n1, the energy released is less, resulting in a longer wavelength of light emitted.
You're expressing those two reactants as ions. The chemical equation for the reaction between barium and nitrogen is: 3Ba + N2 --> Ba3N2. Or, if you like, the net ionic equation for the formation of the insoluble precipitate barium nitride is: 3Ba+2 + 2N-3 --> Ba3N2.
Ne (neon) completes n3.
The formula for the nitride ion is N3-.
The ion for nitride is N3-. It consists of a nitrogen atom with a charge of -3.
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(); } }
declare n1 number; n2 number; n3 number; begin n1:=3 n2 :=5 n3:= sum(n1,n2); dbms_output.put_line( n3); end
Input N1, N2, N3 Max = N1 If N2 > Max then Max = N2 If N3 > Max then Max = N3 Display Max
n1=val(text1.text) n2=val(text2.text) n3=val(text3.text) text4.text=(n1*n2*n3)/100
Nrem (n1, n2, n3)
The following is a schematic program. Details will depend on the language.n1 = 1output n1n2 = 1output n2.start loopn3 = n1 + n2output n3got enough numbers?if yes go to end elsen1
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
As there is only one number 288818. Average = (n1 + n2 +n3 ....... nx)/x = (28818)/1 = 28818
It is a three-way case statement. ( Arithmetic IF statement) IF(exp) n1,n2,n3 meaning go to n1 if expr < 0 ( value is negative) go to n2 if expr = 0 ( value is zero) go to n3 if expr > 0 ( value is positive) IF(expr1-expr2) n1,n2,n3 meaning go to n1 if expr < expr2 go to n2 if expr = expr2 go to n3 if expr > expr2 reference: Fortran 77 with 90 Applications for scientific and engineers second edition RAMA N. REDDY CAROLA A. ZEIGLER
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2-1) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
void main() { int i; float n1,n2; abc: printf("Enter two nos "); scanf("%f%f",&n1,&n2); printf("\n %f + %f = %f " ,n1,n2,n1+n2); printf("\n %f - %f = %f " ,n1,n2,n1-n2); printf("\n %f x %f = %f " ,n1,n2,n1*n2); printf("\n %f / %f = %f " ,n1,n2,n1/n2); printf("\npress 5 to make another calculation"); scanf("%d",&i); if (i==5) goto abc; }