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Veseljchak [2.6K]
4 years ago
5

A ____ is a synonym for a virtualization application.

Computers and Technology
1 answer:
forsale [732]4 years ago
3 0

hypervisor is the answer

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Savatey [412]

Answer:

TELL IT TO YOUR TEACHERS HOW

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3 years ago
Suppose that the code below is the body of some loop. Given variables x and y write some code that reads a value into the variab
ra1l [238]

Answer:

I will code in JAVA.

import java.util.Scanner;

public class Main {

public static void main(String[] args) {

     float x;

     float y;

     float result;

     Scanner input = new Scanner(System.in);

     while(true){ <em>//this is an infinite loop</em>

       x = input.nextFloat(); <em>//wait the input for x</em>

       y = input.nextFloat(); <em>//wait the input for y</em>

       if(y == 0) {

         continue; <em>//next iteration without print.</em>

       } else{

         result = x/y;

         System.out.print(result); <em>//print the result.</em>

       }

     }

 }

}

Explanation:

To get the numbers, you need to import Scanner class, and use the method nextFloat() to admit floats and integers.

If y is 0 then, executes the continue statement, if not prints the division between x and y.

4 0
4 years ago
Describe the effect of a pull up resistor
Vesna [10]
With a pull-up resistor, the input pin will read a high state when the button is not pressed. In other words, a small amount of current is flowing between VCC and the input pin (not to ground), thus the input pin reads close to VCC. When the button is pressed, it connects the input pin directly to ground.
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Which of the following are best practices that should be done in Photoshop before placing images in InDesign?
SSSSS [86.1K]
I’m pretty sure it’s C good luck luv
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Explain what the problems with the implementation of the function are, and show a way to fix them.// return true if two C string
Nataly [62]

Answer:

The code is not dereferencing the pointers. You have to place an asterisk in front of the pointer to read the value the pointer points to.

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So "if (str1 != str2)" must be "if (*str1 != *str2)".

likewise:

   while (*str1 != 0 && *str2 != 0)

and

     result = (*str1 == *str2);

7 0
3 years ago
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