6 Nights in a Biblical Story of Creation
6 Nights in a Biblical Story of Creation
6= Atomic Number of Carbon
Abduce
void main() { int a,b,c; clrscr(); printf("Enter the value of a:"); scanf("%d",&a); printf("\nEnter the value of b:"); scanf("%d",&b); printf("\nEnter the value of c:"); scanf("%d",&c); if(a>b) { if(a>c) { if(b>c) { printf("c is smallest\n"); printf("b is middle\n"); printf("a is largest\n"); } else { printf("b is smallest\n"); printf("c is middle\n"); printf("a is largest\n"); } } else { printf("b is smallest\n"); printf("a is middle\n"); printf("c is largest\n"); } } else if(b>c) { if(a>c) { printf("c is smallest\n"); printf("a is middle\n"); printf("b is largest\n"); } else { printf("a is smallest\n"); printf("c is middle\n"); printf("b is largest\n"); } } else { printf("a is smallest\n"); printf("b is middle\n"); printf("c is largest\n"); } getch(); }
Let the number be X, then B% = B/100 → B% of X = C → B/100 x X = C → X = C ÷ (B/100) = C x 100/B = 100C ÷ B So to find the number, divide C by B percent.
The recursive formula for a sequence typically defines each term based on previous terms. For a sequence denoted as ( A(n) ), ( B(n) ), and ( C(n) ), a common recursive approach might be: ( A(n) = A(n-1) + B(n-1) ) ( B(n) = B(n-1) + C(n-1) ) ( C(n) = C(n-1) + A(n-1) ) These formulas assume initial values are provided for ( A(0) ), ( B(0) ), and ( C(0) ). Adjustments can be made based on the specific context or properties of the sequence.
If A and B are multiples of C, then A + B is also a multiple of C: If A is a multiple of C then A = mC for some integer m If B is a multiple of C, then B = nC for some integer n → A + B = mC + nC = (m + n)C = kC where k = m + n and is an integer → A + B is a multiple of C
A + B is also a multiple of C. ------------------------------------------- let k, m and n be integers. Then: A = nC as A is a multiple of C B = mC as B is a multiple of C → A + B = nC + mC = (n + m)C = kC where k = n + m kC is a multiple of C. Thus A + B is a multiple of C.
Well, NFFBC ("N double F B C!") stands for "No Friends For Butt Crack!" its just a chant.
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Start dimension A[N], B[N] For c=1 to N Input A[N] Next For c=1 to N A[N] = A[N] *10 next For c+1 to N B[N] = A[N] Next for c=1 to N print B[N] Next End
You can use if-else statements to check for equality.Eg:char a, b, c;scanf("%c%c%c", &a, &b, &c);if( a==b && b==c)printf("They are equal\n");elseprintf("They are not equal\n");