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What are the different combinations you can use to make 0.35 only using quarters dimes and nickels?
1 q 1 d 1 q 2 n 2 d 3 n 3 d 1 n 1 d 5 n 7 n
How to write a program to print first n odd numbers in descending ordre?
Pick one 1: for (i=2*n-1; i>=1; i-=2) printf ("%d\n", i); 2: for (i=n; i>=1; i--) printf ("%d\n", 2*i-1); 3: for (i=0; i<n; i++) printf ("%d\n", 2*(n-i)-1);
What is the sum of the first 22 odd numbers?
The sum of a sequence is given by sum = n/2(2a + (n-1)d) where: n = how many a = first number of sequence d = difference between terms of sequence. For the first 22 odd numbers these are: n = 22 a = 1 d = 2 → sum = 22/2(2×1 + (22 - 1)×2)) = 22² = 484 The sum of the first n odd numbers is always n²: sum = n/2(2×1 + (n-1)2) = n/2(1 + (n-1))×2 = n(n) = n²
What is first difference?
In a sequence of numbers, a(1), a(2), a(3), ... , a(n), a(n+1), ... he first differences are a(2) - a(1), a(3) - a(2), ... , a(n+1) - a(n) , ... Alternatively, d the sequence of first differences is given by d(n) = a(n+1) - a(n), n = 1, 2, 3, ...
What is the sum of the arithmetic sequence?
The sum of an arithmetical sequence whose nth term is U(n) = a + (n-1)*d is S(n) = 1/2*n*[2a + (n-1)d] or 1/2*n(a + l) where l is the last term in the sequence.
How many ways can you make change for a dollar using nickels dimes andor quarters?
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
How many different ways can you make change for a half dollar without using pennies?
Assuming nickles, dimes, and quarters, there are ten different ways to make change for a half dollar. Just enumerate the combinations... 10 n 8 n 1 d 6 n 2 d 4 n 3 d 2 n 4 d 5 d 5 n 1 q 3 n 1 d 1 q 1 n 2 d 1 q 2 q
How do you evaluate an arithmetic series if the sigma notation is not given My hw gives me a1 an and n but I don't know how to solve?
This site no longer allows me to enter subscripts so I will use brackets: a(n) to indicate the nth term.a(n) = a(1) + (n-1)*d where d is the common difference between the terms of the arithmetic sequence.Therefore, d = [a(n) - a(1)]/(n-1)Then, the appropriate arithmetic series isS(n) = 1/2*n[2*a(1) + (n-1)*d] where all the terms on the right hand side are known.
Function to find the transpose of a sparse matrix?
// transpose for the sparse matrix void main() { clrscr(); int a[10][10],b[10][10]; int m,n,p,q,t,col; int i,j; printf("enter the no of row and columns :\n"); scanf("%d %d",&m,&n); // assigning the value of matrix for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("a[%d][%d]= ",i,j); scanf("%d",&a[i][j]); } } printf("\n\n"); //displaying the matrix printf("\n\nThe matrix is :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("%d\t",a[i][j]); } printf("\n"); } t=0; printf("\n\nthe non zero value matrix are :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { // accepting only non zero value if(a[i][j]!=0) { t=t+1; b[t][1]=i; b[t][2]=j; b[t][3]=a[i][j]; } } } printf("a[0 %d %d %d\n",m,n,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,b[i][1],b[i][2],b[i][3]); } a[0][1]=n; a[0][2]=m; a[0][3]=t; int s[10],u[10]; if(t>0) { for(i=1;i<=n;i++) { s[i]=0; } for(i=1;i<=t;i++) { s[b[i][2]]=s[b[i][2]]+1; } u[1]=1; for(i=2;i<=n;i++) { u[i]=u[i-1]+s[i-1]; } for(i=1;i<=t;i++) { j=u[b[i][2]]; a[j][1]=b[i][2]; a[j][2]=b[i][1]; a[j][3]=b[i][3]; u[b[i][2]]=j+1; } } printf("\n\n the fast transpose matrix \n\n"); printf("a[0 %d %d %d\n",n,m,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,a[i][1],a[i][2],a[i][3]); } getch(); }
What is the formula to find sum of n even numbers?
Sum = n/2[2Xa1+(n-1)d] where n is last number, a1 is the first number & d is the common difference between the numbers, here d=2 for the even /odd numbers. Sum = n/2 [2Xa1+(n-1)2]
What is the sum of the numbers 31 through 59?
Which is n/2(2a + (n - 1)d) where a is the first term, d is the difference between the terms and n is how many terms there are. a = 31 d = 1 n = 29 29/2(62 + 28) = 29 x 45 = 1305
What is the sum of 1 2 3 4 5 to 100?
There is a formula, called the formula for an arithmetic series, which you can use to calculate this: n/2 * [2a + (n - 1)d] Where: n is the number of terms (100) a is the first term (1) d is the difference between each term (1) Substituting these values we get: 100/2 * [2 + (100 - 1) * 1] = 50 * (2 + 99) = 5,050
