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
(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
k-7
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; . } }
To find the number of positive integers less than 1,000,000 whose digits sum to 7, we can use the combinatorial method known as "stars and bars." We want to distribute 7 identical "stars" (the sum) into 6 "bars" (the digits), allowing for leading zeros. The number of non-negative integer solutions to the equation ( x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 7 ) is given by the formula ( \binom{n+k-1}{k-1} ), where ( n ) is the sum and ( k ) is the number of digits. Thus, the number of solutions is ( \binom{7+6-1}{6-1} = \binom{12}{5} = 792 ). Therefore, there are 792 positive integers less than 1,000,000 with digits that sum to 7.
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; }
Seven fewer than a number ( k ) can be expressed mathematically as ( k - 7 ). This means you take the value of ( k ) and subtract 7 from it. For example, if ( k ) is 10, then seven fewer would be ( 10 - 7 = 3 ).
It is suppose to be k/3 minus 7