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notka56 [123]
3 years ago
14

Write a program whose input is two integers. Output the first integer and subsequent increments of 10 as long as the value is le

ss than or equal to the second integer.
Computers and Technology
1 answer:
Lubov Fominskaja [6]3 years ago
3 0

Answer:

import java.util.Scanner;

public class TestClock {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter two integer numbers");

       int num1 = in.nextInt();

       int num2 = in.nextInt();

       int newSum=num1+10;

       System.out.println("The first number is "+num1);

       do{

          System.out.println(newSum);

          newSum +=10;

       }while (newSum <=num2);

   }

}

Explanation:

Using Java Programming language

  1. Prompt user for the two inputs and save them as num1 and num2(Using the scanner class)
  2. Create a new Variable newSum = num1+10
  3. Create a do...while loop to continually print the value of newSum, and increment it by 10 while it is less or equal to num2
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Dmitry_Shevchenko [17]

Answer:

There is no direct return on investment in building security systems.

Security systems are detrimental to usability and can make IT systems less functional, and therefore less attractive to the customer.

There is pressure to reduce the time it takes to get a new IT product or system onto the market, so security systems are sacrificed in order to reduce the time-to-market.

Explanation:

Cyber security has always been challenging for the organizations. There have been groups which are always looking for loop holes in cyber security and hacks the details and asks the ransom to restore. IT systems have now been more complex in this era. Users are increasing every new day and network accounts security is more demanding. Computer connection are more complex and require special attention to control them. The obstacles in IT systems are of concern as there is need for dynamic IT solution to counter the challenging hackers.  New programs and customized demand of IT systems need customized IT security systems.

4 0
3 years ago
Mix ‘em Let s1 and s2 be 2 c-strings of the same length. Write a function char* mixem(char*s1, char* s2) which returns a c-strin
Shkiper50 [21]

Answer:

See Explaination

Explanation:

#include <iostream>

#include <string.h>

using namespace std;

char *mixem(char *s1, char *s2);

int main() {

cout << mixem("abc", "123") << endl;

cout << mixem("def", "456") << endl;

return 0;

}

char *mixem(char *s1, char *s2) {

char *result = new char[1 + strlen(s1) + strlen(s2)];

char *p1 = s1;

char *p2 = s2;

char *p = result;

while (*p1 || *p2) {

if (*p1) {

*p = *p1;

p1++;

p++;

}

if (*p2) {

*p = *p2;

p2++;

p++;

}

}

*p = '\0';

return result;

}

5 0
3 years ago
Which of the following does not reflect the second step of effective communication?
Ilya [14]

Answer:

c

Explanation:

3 0
3 years ago
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Mario was surprised that the box that was supposed to tell him how many words he used in his document did not appear after the s
Nookie1986 [14]

Answer:

Click the Display tab

Explanation:

Choose the Word count option to make it visible, within the Display tab dropdown

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3 years ago
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Write a program in Java programming language that given two clock times prints out the absolute number of minutes between them
Triss [41]

Answer:

// Java Program to Find the difference

// between Two Time Periods

// Importing the Date Class from the util package

import java.util.*;

// Importing the SimpleDateFormat

// Class from the text package

import java.text.*;

public class GFG {

public static void main(String[] args) throws Exception

{

 // Dates to be parsed

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 String time2 = "7:30:50";

 // Creating a SimpleDateFormat object

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 SimpleDateFormat simpleDateFormat

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 Date date2 = simpleDateFormat.parse(time2);

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 long differenceInMilliSeconds

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 // Calculating the difference in Hours

 long differenceInHours

  = (differenceInMilliSeconds / (60 * 60 * 1000))

  % 24;

 // Calculating the difference in Minutes

 long differenceInMinutes

  = (differenceInMilliSeconds / (60 * 1000)) % 60;

 // Calculating the difference in Seconds

 long differenceInSeconds

  = (differenceInMilliSeconds / 1000) % 60;

 // Printing the answer

 System.out.println(

  "Difference is " + differenceInHours + " hours "

  + differenceInMinutes + " minutes "

  + differenceInSeconds + " Seconds. ");

}

}

7 0
2 years ago
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