/*Java program to find out square root of a given number * without using any Built-In Functions */ public class SquareRootDemo2 { public static void main(String[] args) { //Number for which square root is to be found double number = -12; //This method finds out the square root findSquareRoot(number); } /*This method finds out the square root without using any built-in functions and displays it */ public static void findSquareRoot(double number) { boolean isPositiveNumber = true; double g1; //if the number given is a 0 if(number==0) { System.out.println("Square root of "+number+" = "+0); } //If the number given is a -ve number else if(number<0) { number=-number; isPositiveNumber = false; } //Proceeding to find out square root of the number double squareRoot = number/2; do { g1=squareRoot; squareRoot = (g1 + (number/g1))/2; } while((g1-squareRoot)!=0); //Displays square root in the case of a positive number if(isPositiveNumber) { System.out.println("Square roots of "+number+" are "); System.out.println("+"+squareRoot); System.out.println("-"+squareRoot); } //Displays square root in the case of a -ve number else { System.out.println("Square roots of -"+number+" are "); System.out.println("+"+squareRoot+" i"); System.out.println("-"+squareRoot+" i"); } } }
the normal amount would be 17
By using Pythagoras' theorem.
17The prime numbers between 10 and 20 are 11, 13, 17 and 19. There is only one square number between 10 and 20: 16. Using the criteria given, the number that fits is 17.
Each term is a square or triangular number. In the context of the sequence of square numbers, the first term is the first square number, the second term is the second square number and so on.
there is no such method using string copy
To reverse a number, first convert the number to a string, then reverse the string. Given your number consists of alphanumeric characters, the number must already be a string so simply reverse the string: #include<string> using std::string; string reverse (const string& s) { string str {}; for (auto c : s) str.insert (str.begin(), c); return str; } int main () { std::cout << "Enter a number: "; string s {}; std::cin >> s; std::cout << "The number in reverse is: " << reverse (s); }
/*Java program to find out square root of a given number * without using any Built-In Functions */ public class SquareRootDemo2 { public static void main(String[] args) { //Number for which square root is to be found double number = -12; //This method finds out the square root findSquareRoot(number); } /*This method finds out the square root without using any built-in functions and displays it */ public static void findSquareRoot(double number) { boolean isPositiveNumber = true; double g1; //if the number given is a 0 if(number==0) { System.out.println("Square root of "+number+" = "+0); } //If the number given is a -ve number else if(number<0) { number=-number; isPositiveNumber = false; } //Proceeding to find out square root of the number double squareRoot = number/2; do { g1=squareRoot; squareRoot = (g1 + (number/g1))/2; } while((g1-squareRoot)!=0); //Displays square root in the case of a positive number if(isPositiveNumber) { System.out.println("Square roots of "+number+" are "); System.out.println("+"+squareRoot); System.out.println("-"+squareRoot); } //Displays square root in the case of a -ve number else { System.out.println("Square roots of -"+number+" are "); System.out.println("+"+squareRoot+" i"); System.out.println("-"+squareRoot+" i"); } } }
#include#includeusing std::cin;using std::cout;using std::endl;using std::string;int main(){string number = "0";cout number;cout
class Abc { public static void main(String[] args) { System.out.println("Hello World!"); } }
#include #include using std::cout;using std::cin;using std::endl;using std::string;int main(){string myNumber = "";cout myNumber;cout
construct a square that perfectly circumscribes (surrounds) a given square * * * * * What? The same square will do it - perfectly! It is, in fact, the challenge of constructing a square with the same area as a given circle by using only a finite number of steps with compass and straightedge. As pi (π) is a transcendental, rather than an algebraic irrational number, it cannot be done.
Take out one letter, using a random number. Then replace the string with the new string, without the extracted letter. Repeat, until the remaining string has a length of zero. The extracted letters get added to a new string.Take out one letter, using a random number. Then replace the string with the new string, without the extracted letter. Repeat, until the remaining string has a length of zero. The extracted letters get added to a new string.Take out one letter, using a random number. Then replace the string with the new string, without the extracted letter. Repeat, until the remaining string has a length of zero. The extracted letters get added to a new string.Take out one letter, using a random number. Then replace the string with the new string, without the extracted letter. Repeat, until the remaining string has a length of zero. The extracted letters get added to a new string.
As a numerical string, 9621.
to construct (using a compass and straight-edge) a square with the same area as a given circle using only a finite number of steps. "Squaring the circle" was an ancient problem that has been proved impossible to do.
to construct (using a compass and straight-edge) a square with the same area as a given circle using only a finite number of steps. "Squaring the circle" was an ancient problem that has been proved impossible to do.
Console.WriteLine("Please input a string:"); string str = Console.ReadLine(); Console.WriteLine("Number of characters: " + str.Length);