The question does not make sense.
#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(); }
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)!]
any number (n) added to zero equals that same number (n). n + 0 = n
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.
Certainly! Here are some key formulas and properties related to exponents and identity elements: Exponents Formulas: *Product of Powers:* [ a^m \cdot a^n = a^{m+n} ] When multiplying two exponents with the same base, you add the exponents. *Quotient of Powers:* [ \frac{a^m}{a^n} = a^{m-n} \quad (\text{for } a \neq 0) ] When dividing two exponents with the same base, you subtract the exponents. *Power of a Power:* [ (a^m)^n = a^{m \cdot n} ] When raising an exponent to another power, you multiply the exponents. *Power of a Product:* [ (ab)^n = a^n \cdot b^n ] When raising a product to a power, you raise each factor to the power. *Power of a Quotient:* [ \left( \frac{a}{b} \right)^n = \frac{a^n}{b^n} \quad (\text{for } b \neq 0) ] When raising a quotient to a power, you raise both the numerator and the denominator to the power. *Zero Exponent:* [ a^0 = 1 \quad (\text{for } a \neq 0) ] Any non-zero number raised to the power of zero is 1. *Negative Exponent:* [ a^{-n} = \frac{1}{a^n} \quad (\text{for } a \neq 0) ] A negative exponent indicates the reciprocal of the base raised to the opposite positive exponent. Identity Elements: *Additive Identity:* [ a + 0 = a \quad \text{and} \quad 0 + a = a ] The number 0 is the additive identity because adding 0 to any number ( a ) leaves ( a ) unchanged. *Multiplicative Identity:* [ a \times 1 = a \quad \text{and} \quad 1 \times a = a ] The number 1 is the multiplicative identity because multiplying 1 by any number ( a ) leaves ( a ) unchanged. These formulas and properties are fundamental in algebra and are used frequently in solving equations and simplifying expressions. If you need further details or examples, please let me know!
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
For any number n, n + 0 = 0 + n = n All you need to do is to substitute any number of your choice, for n.
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).
#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; }
<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>
The inverse operation of addition would be subtraction. Example : 2 + 3 = 5 Performing the inverse operation yields the original number. Example : 5 - 3 = 2
It is the additive identity property of zero. (n+0=n)