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jolli1 [7]
3 years ago
15

What is a special type of variable used in subroutines that refers to a piece of data?

Computers and Technology
1 answer:
Fed [463]3 years ago
8 0

Answer:parameter

Explanation:

I got it right

You might be interested in
Write a C program that has the following statements: int a, b; a = 10; b = a + fun(); printf("With the function call on the righ
mart [117]

Answer:Following is the C program:-

#include <stdio.h>

int fun()//function fun of return type int and it returns value 6.

{

   return 6;

}

int main() {

  int a, b;

  a = 10;

  b = a + fun();//adds 6 to a.

  printf("With the function call on the right, ");

  printf("\n%d ",b);//printing b..

return 0;

}

Output:-

With the function call on the right,  

16

Explanation:

The function fun return the value 6 so it adds 6 to a and stores the result in b.

6 0
4 years ago
How many transponders are contained within a typical satellite?
Ilia_Sergeevich [38]

Answer:

d.

24 to 32

Explanation:

Satellite programmers broadcast, signals to the satellite which they own or lease the channel space from. Uplinked signals sent by the programmers are received by the transponder located on satellite. It is a device which receives signals and transmits them back to Earth after converting the signals to frequency which could be received by the ground-based antenna. <u>There are typically 24 to 32 transponders on a satellite. </u>

8 0
3 years ago
Write a Java program that prompts the user to enter integer values for the sides of a triangle and then displays the values and
satela [25.4K]

Answer:

import java.util.Scanner;

public class triangle {

       public static void main (String [] args) {

           int sideOne, sideTwo, sideThree;

           Scanner in = new Scanner (System.in);

           System.out.println("Enter the first side of the triangle");

           sideOne = in.nextInt();

           System.out.println("Enter the secon side of the triangle");

           sideTwo = in.nextInt();

           System.out.println("Enter the third side of the triangle");

           sideThree = in.nextInt();

           if ( sideOne<=0||sideTwo<=0|| sideThree<=0){

               System.out.println(" The Values enter are "+sideOne+ "," +sideThree+ ","+sideThree+ " These values don't make a valid triangle");

           }

           else if ((sideOne + sideTwo> sideThree) || (sideOne+sideThree > sideTwo) || (sideThree+sideTwo > sideOne))

           {

               System.out.println ("The triangle is Valid");

           }

           else {

               System.out.println(" The Values enter are "+sideOne+ "," +sideTwo+ ","+sideThree+ " These values don't make a valid triangle");

           }

       }

Explanation:

for a Triangle to be Valid one of the three sides of the triangle must greater than the other two sides.  The code above enforces this condition using an if statement in combination with the Or Operator

The following conditions where enforced

side one, side two and side three != 0

Side One + Side Three > Side Two

Side Three + Side Two> Side One

Side Two + Side One > Side Three

7 0
3 years ago
c++ Consider this data sequence: "3 11 5 5 5 2 4 6 6 7 3 -8". Any value that is the same as the immediately preceding value is c
Ksivusya [100]

<u>Answer:</u>

<em>int fNumber,scndNumber = -1,  </em>

<em>dup = 0; </em>

<em>do { </em>

<em>cin >> fNumber; </em>

<em>if ( scndNumber == -1) { </em>

<em>scndNumber = fNumber; </em>

<em>} </em>

<em>else { </em>

<em>if ( scndNumber == fNumber ) </em>

<em>duplicates++; </em>

<em>else </em>

<em>scndNumber = fNumber; </em>

<em>} </em>

<em>} while(fNumber > 0 );  </em>

<em>cout << dup; </em>

<u>Explanation:</u>

Here three variables are declared to hold the first number which is used obtain all the inputs given by the user, second number to hold the value of <em>last encountered number and “dup” variable to count the number of duplicate values.</em>

<em>“Do-while”</em> loop help us to get the input check whether it is same as previous input if yes then it <em>adds to the duplicate</em> value otherwise the new previous value if stored.

4 0
3 years ago
Read 2 more answers
Two machines can finish a job in StartFraction 20 Over 9 EndFraction hours. Working​ alone, one machine would take one hour long
Alborosie

<em><u>Answer</u></em>

5 hours

<em><u>Explanation</u></em>

The two working together can finish a job in

\frac{20}{9}  \: hours

Also, working alone, one machine would take one hour longer than the other to complete the same job.

Let the slower machine working alone take x hours. Then the faster machine takes x-1 hours to complete the same task working alone.

Their combined rate in terms of x is

\frac{1}{x}    +  \frac{1}{x - 1}

This should be equal to 20/9 hours.

\frac{1}{x}  +  \frac{1}{x - 1}  =  \frac{9}{20}

Multiply through by;

20x(x - 1) \times \frac{1}{x}  +20x(x - 1) \times   \frac{1}{x - 1}  =  20x(x - 1) \times \frac{9}{0}

20(x - 1)  +20x = 9x(x - 1)

20x - 20+20x = 9{x}^{2}  - 9x

9{x}^{2}  - 9x - 20x - 20x + 20= 0

9{x}^{2}  - 49x  + 20= 0

Factor to get:

(9x - 4)(x - 5) = 0

x =  \frac{4}{9}  \: or \: x = 5

It is not feasible for the slower machine to complete the work alone in 4/9 hours if the two will finish in 20/9 hours.

Therefore the slower finish in 5 hours.

8 0
3 years ago
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