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How do you make a calculator in code block by using c?
#include<stdio.h> #include<conio.h> int main(void) { float a,b,c=0, d=0, e=0,f=0; printf("Please enter two numbers:\n"); scanf("%f %f", &a, &b); c=a+b; d=a-b; e=a*b; f=a/b; printf("The sum of %f and %f is :%f\n", a,b,c); printf("The subtraction of %f and %f is :%f\n", a,b,d); printf("The multiplication of %f and %f is :%f\n", a,b,e); printf("The division of %f by %f is :%f\n",a,b,f); getch(); }
What is the additvie identity property of 0?
any number (n) added to zero equals that same number (n). n + 0 = n
How do you expand binomials A plus b to the degree of n?
It isnC0*A^n*b^0 + nC1*A^(n-1)*b^1 + ... + nCr*A^(n-r)*b^r + ... + nCn*A^0*b^n where nCr = n!/[r!*(n-r)!]
Explain how additive inverses are used in subtraction?
Given a number n, the addititive inverse is -n since n+(-n)=0. Howevern+(-n) is the same as n-n which is 0 also.SO adding the additive inverse it the same as subtracting.
Examples of identity property of addition?
For any number n, n + 0 = 0 + n = n All you need to do is to substitute any number of your choice, for n.
What is a number added to its additive inverse will always have a sum of zero?
It is a tautological description of one of the basic properties of numbers used in the branch of mathematics called Analysis: Property 2: there exists an additive identity, called 0; for every number n: n + 0 = 0 + n = n. Property 3: there exists an additive inverse, of every number n denoted by (-n) such that n + (-n) = (-n) + n = 0 (the additive identity).
What property is shown by the equation 12 plus 0 equals 12?
It is the additive identity property of zero. (n+0=n)
What does a number to the power of zero equal?
Any number to the power '0' equals '1'. Proof ; Let a^(n) = b Then dividing a^(n) / a^(n) = b/b a^(n-n) = b/b a^(0) = 1
What property is n plus 72 equals 72?
It is the property of n as the additive identity.
Write a c program that interchanges the odd and even elements of an array?
#include<stdio.h> int main() { int num[]={1,2,3,4,5,6}; int i,temp; for(i=0;i<=5;i=i+2) { temp=num[i]; num[i]=num[i+1]; num[i+1]=temp; } for(i=0;i<=5;i=i+2) printf(''%d",num[i]); return 0; }
Program to check whether the given number is an Armstrong number?
<html> <head> <Script Language="JavaScript"> var a,n,b=0,t; n=parseInt(window.prompt("enter n","0")); t=n; while(n>0) { a=n%10; b=b+a*a*a; n=n/10; } if(b==t) { document.writeln("Armstrong no"); } else { document.writeln("Not an Armstrong no"); } </script> </head> </html>
Write a program to print Armstrong in c?
void main() { int n,b=0,t; clrscr(); printf("Enter the no"); scanf("%d",&n); t=n; while(n>0) { a=n%10; b=b+a*a*a; n=n/10; } if(b==t) { printf("Armstrong no"); } else { printf("Not an Armstrong no"); } getch(); }
