This problem asks us to find 2 numbers, n1 and n2, with the following relations between them: * n2 = 4 n1 * n1 + n2 = 45 Substituting the first equation into the second one gives us: * n1 + 4n1 = 45 which gives us * n1 = 9. We can now use this solution to find n2 with the first equation * n2 = 4 n1 = 36 So the first number is 4, the second number is 36.
it is the formula of refraction. this is based of snell's law and its about the incident ray and the refracted ray. page 3 of this should be of help to you
There are infinitely many possible answers. 1 + 121 is one possible answer.
It is: 1*17 = 17 which is a prime number
by just doing the problem
n1 has 1 n2 has 4
#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
This problem asks us to find 2 numbers, n1 and n2, with the following relations between them: * n2 = 4 n1 * n1 + n2 = 45 Substituting the first equation into the second one gives us: * n1 + 4n1 = 45 which gives us * n1 = 9. We can now use this solution to find n2 with the first equation * n2 = 4 n1 = 36 So the first number is 4, the second number is 36.
#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
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
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To swap two numbers N1 and N2, using a third variable T... T = N1; N1 = N2; N2 = T;
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