To swap two numbers N1 and N2, using a third variable T... T = N1; N1 = N2; N2 = T;
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(); } }
public int findSum(int n1, int n2) { return n1 + n2; }
Because a transformer does not generate power, it transformers it. Power is equivalent to the voltage times current. A transformer with a ratio of N1:N2 takes voltage (V1) and current (I1) at one winding and transforms it into (N2/N1)*V1 voltage and (N1/N2)*I1 current at the other winding. So the input power is V1 * I1, and the output power is (N2/N1)*V1*(N1/N2)*I1 = V1*I1 (ignoring the small amount of losses associated with the transformer).
n1=val(text1.text) n2=val(text2.text) n3=val(text3.text) text4.text=(n1*n2*n3)/100
The N6 certificate is on NQF level 5, and the Diploma is on NQF level 6.
NQF stands for National Quality Forum in regards to Meaningful Use.
Level 6
#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; }
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
level 7 as i know
#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; }
the value of the exponent n1
the value of the exponent n1
In South Africa, Grade 12 is equivalent to NQF level 4. This is the final year of the General Education and Training Phase and signifies the completion of secondary education.
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; }
The sum of the first 10 positive integers, using the formula N1 + (N1 + 1) + ... + N2 = N2 * (N2 + 1) / 2 - (N1 - 1) * N1 / 2 is: 55