A gui allows you to interact with objects on the screen such as icons and buttons true or false

A user complains that the pointer movement on his laptop computer is very erratic what should you do?

Nadusha1986 [10]

Answer:

just go to mouse settings click additional mouse settings and you can go to a speed tab to chance pointer speed. its probably set to a higher setting than what he normally uses

Explanation:

8 0

3 years ago

Think of some local businesses that have websites. Look online and identify two different websites for businesses or services in

garri49 [273]

I don't think we can answer this question because we don't know businesses in your area...

3 0

3 years ago

First write a method to calculate the greatest common divisor (GCD) of two positive integers using Euclid’s algorithm (also know

MAVERICK [17]

Answer:

import java.util.Scanner;

public class Gcd

{

public static void main(String[] args) {

Scanner input = new Scanner(System.in);

System.out.print("Enter two numbers to find their GCD: ");

int number1 = input.nextInt();

int number2 = input.nextInt();

if(number1 > 0 && number2 > 0)

System.out.println("GCD of " + number1 +" and " + number2 +" is: " + findGCD(number1, number2));

else

System.out.println("Invalid Input!");

}

public static int findGCD(int number1, int number2) {

if(number2 == 0){

return number1;

}

return findGCD(number2, number1 % number2);

}

Explanation:

To be able to find GCD, we need to factorize the numbers and multiply common factors.

- Inside the function:

- Recursively divide number1 by number2 until number2 is equal to zero.

- Return number1 when the number2 is equal to zero. That means we found all the divisors of number1

- Inside the main:

- Ask user to input numbers

- Check if both numbers are greater than zero

- If they satisfy the condition, call the method and print the result.

7 0

3 years ago

Write a C++ program to count even and odd numbers in array. The array size is 50. The array elements will be entered by the user

vlabodo [156]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

int main(){

int numbers[50];

int evekount = 0, odkount = 0;

for(int i = 0; i<50;i++){

cin>>numbers[i];

if(numbers[i]%2==0){ evekount++; }

else{ odkount++; }

}

cout<<"Even Count: "<<evekount<<endl;

cout<<"Odd Count: "<<odkount<<endl;

return 0;

}

Explanation:

This declares the integer array of number

int numbers[50];

This initializes the even count and odd count to 0

int evekount = 0, odkount = 0;

This iterates from 1 to 50

for(int i = 0; i<50;i++){

This gets input for the array

cin>>numbers[i];

This checks for even

if(numbers[i]%2==0){ evekount++; }

This checks for odd

else{ odkount++; }

}

This prints the even count

cout<<"Even Count: "<<evekount<<endl;

This prints the odd count

cout<<"Odd Count: "<<odkount<<endl;

3 0

3 years ago

A hard drive cannot be partitioned until the device is set.

PilotLPTM [1.2K]

Yeah that true I asked my grandpa

3 0

3 years ago