2 nickels in a dime
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
The set of odd numbers is an arithmetic sequence. Let say that the sequence has n odd numbers where the first term is a1 and the last one is n. The formula to find the sum on nth terms for an arithmetic sequence is: Sn = (n/2)(a1 + an) or Sn = (n/2)[2a1 + (n - 1)d] where d is the common difference that for odd numbers is 2. Sn = (n/2)(2a1 + 2n - 2)
You can use the formula for the sum of an arithmetic series. Sn = (n/2)[2a + (n - 1)d].....n is the number of terms, a is the first term and d the common difference between terms. In this question, n = 45, a = 1 (as 1 is the first odd number) and d = 2 (the difference between consecutive odd numbers). Then, S(45) = (45/2)[2 + 44x2] = (45/2) x 90 = 2025.
That would mean d times 2.
Use the equation S/2 = a + ( n - 1)d Were a = 1 d = 2 (Consecutive odd numbers havie a difference of '2' n = 90 Substitute S = 2[(1) + (90 - 1) 2[ S = 2 + 2(89)2 S = 2 + 356 S = 358
2 n in a d is 2 = nickels in a dime.
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
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
group 3- (n-1)d1 ns2. Group 12 (n-1)d10 ns2, groups 4-11 do not necessarily have identical outer electron configurations. Where n represents your period that you are in.
int maxprod (int n) { return n/2; } int main (void) { int n, a, b; n= 7; /* for example */ a= maxprod (n); b= n-a; printf ("%d+%d=%d, %d*%d=%d\n", a, b, a+b, a, b, a*b); }
The fraction is 35/56.The way to find it is as follows:1) n/d = 5/82) (n-5) / (d+4) = 1/23) n = (5/8)*d [rearrange (1)]4) 2*(n-5) = (d+4) [rearrange (2)]5) 2*n - 10 = d + 4 [rearrange (4)]6) 2*n = d + 14 [rearrange (5)]7) 2*((5/8)*d) = d + 14 [substitue (3) into (6)]8) (10/8)*d = d+14 [rearrange (7)]9) (10/8)*d - d = 14 [rearrange (8)]10) (10/8)*d - (8/8)*d = 14 [rearrange (9)]11) (2/8)*d = 14 [rearrange (10)]12) d = 14 / (2/8) = 14 * 8/2 [rearrange (11)]13) d = 112/2 = 56 [simplify (12)]14) n = (5/8)*56 [substitute (13) into (3)]15) n = 280/8 = 35 [simplify (14)]16) THE ANSWER IS n/d = 35/56 [substitute (13) and (15) into the definition of the fraction]
#include#includevoid main(){int i,n;clrscr();ptintf("\n enter the no.\n");scanf("%d",&n);for(i=2;i
H.A.N.D. means have a nice day!
To calculate for spiral length, the formula is L = pi*N* (D + d) / 2. This is wherein N = (D - d) / (2*t) is the number of wraps of tape of thickness t on a roll of diameter D (when full) around a core of diameter d.
The simple, trite answer is unsigned long nfact (int n) { if (n<=1) return 1; else return n * nfact(n-1); } This will quickly overflow, however, because N Factorial quickly gows out of bound of any native data type. A more robust implementation would use a bin technique using linked lists of digits to do the multiplications. Here it is, including comparison between various methods. Note that for large values of N, you may need to increase your linker stack size beyond its default, typically, 1MB... #include <stdlib.h> #include <stdio.h> /* Microsoft 32-bit iterative */ unsigned long NFactLongIterative (unsigned long N) { unsigned long result = N; if (N < 2) return 1; if (N == 2) return 2; while (--N >= 2) result *= N; return result; } /* Microsoft 64-bit iterative */ unsigned long long NFactLongLongIterative (unsigned long long N) { unsigned long long result = N; if (N < 2) return 1; if (N == 2) return 2; while (--N >= 2) result *= N; return result; } /* Microsoft 64-bit recursive */ unsigned long long NFactLongLongRecursive (unsigned long long N) { if (N < 2) return 1; if (N == 2) return 2; return N * NFactLongLongRecursive (N - 1); } /* Portable double recursive */ double NFactDouble (double N) { if (N < 2) return 1; if (N == 2) return 2; return N * NFactDouble (N - 1); } /* Portable arbitrary length decimal iterative */ /* one node of a linked list of digits, the first node being low-order */ struct _decimal { int digit; struct _decimal *next; }; typedef struct _decimal decimal; /* Portable arbitrary length decimal iterative */ /* Initialize the list - necessary on second pass, if main recoded */ void decimal_initialize (decimal *d, int n) { decimal *next, *nextsave; d->digit = n; nextsave = d->next; d->next = NULL; next = nextsave; while (next != NULL) { nextsave = next->next; free (next); next = nextsave; } return; } /* Portable arbitrary length decimal iterative */ /* Append a digit at the high order position */ void decimal_add_digit (decimal *d, int n) { decimal *new_digit = (decimal*) malloc (sizeof (decimal)); while (d->next != NULL) d = d->next; new_digit->digit = n; new_digit->next = NULL; d->next = new_digit; return; } /* Portable arbitrary length decimal iterative */ /* Print the digits in reverse order - recursive */ void decimal_print_digits (decimal *d, int last_digit) { if (d->next != NULL) decimal_print_digits (d->next, false); printf ("%d", d->digit); if (last_digit) printf("\n"); return; } /* Portable arbitrary length decimal iterative */ /* multiply the list by N */ void decimal_multiply (decimal *d, int N) { int carry = 0; while (d != NULL) { d->digit = d->digit * N + carry; carry = d->digit / 10; d->digit %= 10; if (carry != 0 && d->next == NULL) decimal_add_digit (d, 0); d = d->next; } return; } /* Portable arbitrary length decimal iterative */ /* Primary interative algorithm */ void decimal_NFactIterative (decimal *d, int N) { if (N < 2) { decimal_initialize (d, 1); return; } if (N == 2) { decimal_initialize (d, 2); return; } while (N > 2) { decimal_multiply (d, N); N--; } return; } /* Example main line */ /* Generates all variations to show differences in results */ int main (int argc, char *argv[]) { int N; decimal Decimal = {2, NULL}; if (argc < 2) { printf ("Enter N (or use command line) : "); scanf_s ("%d", &N); } else { N = atoi (argv[1]); } printf ("Long: %u! = %u\n", N, NFactLongIterative (N)); printf ("LongLong: %u! = %I64u\n", N, NFactLongLongIterative (N)); printf ("Recursive: %u! = %I64u\n", N, NFactLongLongRecursive (N)); printf ("Double: %u! = %.0f\n", N, NFactDouble (N)); /* note: arbitrary is exact - if the others don't match, arithmetic overflow occurred */ printf ("Arbitrary: %u! = ", N); decimal_NFactIterative (&Decimal, N); decimal_print_digits (&Decimal, true); return 0; }
Park, Reverse, Neutral, Drive, 2nd, and 1st. These are the different gear positions in an automatic transmission
RNRF means Rock N' Roll Forever!!!!!!:D