It is: 5k-7
(k x 4) + m = 4k + m Just put the numbers together with the unknown for products, then put the other parts you need to add behind (think of grouping)
1
If you want to say "Nine is less than a number k," you would write 9<K. If you're asking for an equation, e.g. "Nine less than a number k is equal to that number divided by three," you would write k - 9.
7
k + 7
I 277273
11 + k
17
it would have a part in it like this: for (i=0; i<n; ++i) { . for (j=0; j<l; ++j) { . . sum= 0; . . for (k=0; k<m; ++k) { . . . sum += a[i][k] * b[k][j]; . . } . . c[i][j] = sum; . } }
enter the number whose digits are to be added num is the given value num=0! k=num%10 sum=sum=k k=num/10 num=k print the sum of the digits
(100)/(k+m)
int main() { int i,j,sum,k; for(i=1;i<=5;i++) { k=1; sum =0; for(j=1;j<=i;j++) { sum = sum+(i*k); k=k*10; } cout<< sum; cout<< "\n"; } }
#include<stdio.h> int main() { int count,i,j,k,n,*a,sum=0; printf("Enter the value of 'n':"); scanf("%d",&n); a=malloc(n*sizeof(int)); for(i=1;i<=n;i++) { count=0; j=1; while(j<=i) { if(i%j==0 && i!=2) count++; j++; } if(count==2 count==1) { for(k=0;k<=n;k++) a[k]=i; } } for(k=0;k<=n;k=k+1) printf("%d\t",a[k]); for(k=0;k<=n;k=k+2) sum+=a[k] * a[k]; printf("The sum of squares of alternative prime numbers is=%d",sum); getchar(); return 0; }
It is suppose to be k/3 minus 7
k = f(n) = 7n
A number, k, increased by two and a half.