To write the sum of a number k and 7, you would use the mathematical expression k + 7. This expression represents adding the value of k to 7. In algebraic terms, this is known as an addition operation where k is the addend and 7 is the other addend. The result of this addition operation would give you the sum of k and 7.
It is: 5k-7
1
(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)
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
A number, k, increased by two and a half.
k = f(n) = 7n