The nth term of an AP with initial term a (= u{1}) and common difference d is given by: u{n} = a + (n - 1)d In this case: a = 6 d = (12 - 6) = 6 → u{n} = 6 + (n - 1)6 But this can be simplified: u{n} = 6 + (n - 1)6 = 6 + 6n - 6 = 6n
d = n + 7 and (n + 17)/(d - 6) = 2 so (n + 17)/ n + 7 - 6 = 2 (n + 17)/(n + 1) = 2 2n + 2 = n + 17 n = 15 and d = 22 Check n + 17 = 32, d - 6 = 16, 32/16 = 2 Original number is 15/22
There are 22 ways to make change from a dollar using nickels, dimes, and quarters. 1. 4 q 2. 10 d 3. 20 n 4. 2 q , 5 d 5. 3 q , 2 d , 1 n 6. 1 q , 7 d, 1 n 7. 9 d, 2 n 8. 8 d, 4 n 9. 7 d, 6 n 10. 6 d , 8 n 11. 5 d , 10 n 12. 4 d , 12 n 13. 2 d , 16 n 14. 1 d , 18 n 15. 5 n , 3 q 16. 3 n , 1 q , 6 d 17. 7 n , 1 q , 4 d 18. 9 n , 1 q , 3 d 19. 11 n , 1 q , 2 d 20. 13 n , 1 q , 1 d 21. 14n , 3 d 22. 15n , 1 q
Suppose the number is n/d. Then n/d >= d => n <= d2 if d < 0 or n >= d2 if d > 0.
#include <stdio.h> main() { int m, n, c, d, A[10][10],temp=0; printf("\nEnter the number of rows and columns for matrix A:\n"); scanf("%d%d", &m, &n); printf("\nEnter the elements of matrix A:\n"); for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < n ; d++ ) scanf("%d", &A[c][d]); printf("\nMatrix entered is:\n"); for ( c = 0 ; c < m ; c++ ) { for ( d = 0 ; d < n ; d++ ) printf("%d\t", A[c][d]); printf("\n"); } printf("\n\nMaximum element of each row is:\n"); for(c=0;c<m;c++) { for(d=0;d<n;d++) { if(A[c][d]>temp) temp=A[c][d]; } printf("\n\t\tRow %d: %d",c+1,temp); temp=0; } temp=A[0][0]; printf("\n\nMinimum element of each coloumn is:\n"); for(c=0;c<n;c++) { for(d=0;d<m;d++) { if(A[d][c]<temp) temp=A[d][c]; } printf("\n\t\tColoumn %d: %d",c+1,temp); temp=A[d][c] ; } return 0; }
The nth term of an AP with initial term a (= u{1}) and common difference d is given by: u{n} = a + (n - 1)d In this case: a = 6 d = (12 - 6) = 6 → u{n} = 6 + (n - 1)6 But this can be simplified: u{n} = 6 + (n - 1)6 = 6 + 6n - 6 = 6n
sha (as in banana)to (as in boat)d (short 'd')e (as a whatever [a cake, a car])bo (as in boat)frè (as in best)n (short 'n')
d = n + 7 and (n + 17)/(d - 6) = 2 so (n + 17)/ n + 7 - 6 = 2 (n + 17)/(n + 1) = 2 2n + 2 = n + 17 n = 15 and d = 22 Check n + 17 = 32, d - 6 = 16, 32/16 = 2 Original number is 15/22
There are 22 ways to make change from a dollar using nickels, dimes, and quarters. 1. 4 q 2. 10 d 3. 20 n 4. 2 q , 5 d 5. 3 q , 2 d , 1 n 6. 1 q , 7 d, 1 n 7. 9 d, 2 n 8. 8 d, 4 n 9. 7 d, 6 n 10. 6 d , 8 n 11. 5 d , 10 n 12. 4 d , 12 n 13. 2 d , 16 n 14. 1 d , 18 n 15. 5 n , 3 q 16. 3 n , 1 q , 6 d 17. 7 n , 1 q , 4 d 18. 9 n , 1 q , 3 d 19. 11 n , 1 q , 2 d 20. 13 n , 1 q , 1 d 21. 14n , 3 d 22. 15n , 1 q
6 numbers in a half dozen
It is the signal for abandon ship. Addition to answer. A B A N D O N ...................SHIP! 1 2 3 4 5 6 7 Short........... 1 Long
I N D E E D= 6 letters
Yes. If n is odd, then n + c where c is an even constant will be odd. n + d where d is an odd constant will be even.
The core if it would contain lines like these: printf ("sizeof (char)=%d\n" "sizeof (short)=%d\n" "sizeof (int)=%d\n" "sizeof (long)=%d\n" "sizeof (long long)=%d\n" "sizeof (size_t)=%d\n" "sizeof (void *)=%d\n" "sizeof (ptrdiff_t)=%d\n" "sizeof (va_list)=%d\n" "sizeof (intptr_t)=%d\n" , (int)sizeof (char) , (int)sizeof (short) , (int)sizeof (int) , (int)sizeof (long) , (int)sizeof (long long) , (int)sizeof (size_t) , (int)sizeof (void *) , (int)sizeof (ptrdiff_t) , (int)sizeof (va_list) , (int)sizeof (intptr_t) );
1. START 2. INPUT N 3. LET D=1 4. N%D? (if NO goto step 5 else goto step 6) 5. PRINT D 6. LET D=D+1 7. SQRT(N)<D? (if NO goto step 4 else goto step 8) 8. STOP
Group 3: ns^2 (n-1)d^1 Group 4: ns^2 (n-1)d^2 Group 5: ns^2 (n-1)d^3 Group 6: ns^1 (n-1)d^5 Group 7: ns^1 (n-1)d^6
Inside.