0.2
They are numbers of the form 210 + 2*n where n = 1, 2, 3, ... 15.
Assuming x is your number, x * 10^n = x moved n decimal places. When n is positive, move the decimal point n places to the right. When n is negative, move the decimal point n places to the left. When n is 0, do nothing.
n-n
One way to accomplish this is to allow the printf statement to do the work for you. Example: printf("Decimal %d = hex %02x\n", number, number); Or you could use the windows calculator. Select the Dec radian, type your number in, then select Hex radian.
this isn't the fact that the binary number 11 will have decimal equivalent 11... instead it have decimal number 1011 for decimal equivalant 11.here is the alogorithms for converting the decimal number to binary equivalent...!!!#include#includevoid main(){ int i,n,j,b[100];clrscr();printf("Enter a Number:");scanf("%d",&n);i=0;while(n>0){b[i]=n%2;n=n/2;i++;}printf("\n\nBinary Equivalent:");j=i-1;for(i=j;j>=0;j--)printf("%d",b[j]);getch();}using this algorithms... the binary equivalent of any number is taken out...!!! enjoy... have any query... email at :- "devilllcreature@yahoo.com" thank you....!!!
If you're looking for 220 percent of some number n n * 2.2 = 220 percent of n n * x.yz = xyz percent of n In other words, just shift the decimal 2 places to the left for any problem like this.
They are numbers of the form 210 + 2*n where n = 1, 2, 3, ... 15.
no. irrational numbers are always infininately long, otherwise the could be represented as a fraction by multiplying by 10^n and dividing by 10^n where n is a number large enough to make the number a number with no decimals.
The product of a decimal number ( n ) and 10 is always a whole number when ( n ) has only one decimal place. This is because multiplying a decimal with one decimal place (e.g., 2.5) by 10 shifts the decimal point one place to the right, resulting in a whole number (e.g., 25). If ( n ) has more than one decimal place, the product will not be a whole number.
If a number has an antilog whose integer part is n, then the number has n-1 digits before the decimal point.
Assuming x is your number, x * 10^n = x moved n decimal places. When n is positive, move the decimal point n places to the right. When n is negative, move the decimal point n places to the left. When n is 0, do nothing.
n-n
very simple way yr:int main(){int o;printf("\n Enter an octal number:");scanf("%o",&o);printf("The required decimal number is %d",o);getch();}
include <iostream> using namespace std; int main() { int n; // number to convert to binary while (cin >> n) { if (n > 0) { cout << n << " (decimal) = "; while (n > 0) { cout << n%2; n = n/2; } cout << " (binary) in reverse order" << endl; } else { cout << "Please enter a number greater than zero." << endl; } } return 0; }//end main
The 20th triangular number is 20*21/2 = 210
37/300:= 37 ÷ 300= 0.1233 n decimal
To determine the least positive integer ( n ) such that ( 210 ) divides ( n! ), we first factor ( 210 ) into its prime components: ( 210 = 2 \times 3 \times 5 \times 7 ). For ( n! ) to be divisible by ( 210 ), ( n ) must be at least as large as the largest prime factor, which is ( 7 ). Thus, the least positive integer ( n ) such that ( 210 ) divides ( n! ) is ( n = 7 ).