As of October 2023, the Governor of the Central Bank of Nigeria (CBN) is Godwin Emefiele. He has been in office since June 2014, overseeing the country's monetary policy and financial stability. Emefiele has played a significant role in implementing various economic policies, particularly during periods of economic challenges. However, please verify for any updates or changes in leadership.
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.
B n joshi
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");
#include<stdio.h> main() { int a=0,b=1,n,c,i; printf("enter number"); scanf("%d",&n); printf("%d\n",a); for(i=1;i<=n;i++) { c=a+b; printf("%d\n",c); b=a; a=c; } getch(); }
if (a > b && a > c) printf("%d\n", a); else if (b > c) printf("%d\n", b); else printf("%d\n", c);
/*WAP to display Fibonacci series*/ #include<stdio.h> #include<conio.h> void main() { int i,a=0,b=1,c; scanf("%d",&n); printf("%d\n%d",a,b); for(i=0;i<n;i++) { c=a+b; a=b; b=c; printf("\n%d",c); } getch(); }
A over B+ C over B=A+C over B=N