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3 years ago

Two parter:

Computers and Technology

1 answer:

sashaice [31]3 years ago

6 0

Answer:

This is an infinite loop.

Explanation:

The while loop while never end since the last statement inside the while loop will continue to decrement the value of num by 1 and the condition num < 9 will always be true.

So either change the condition of the while loop like: num > -9

or start incrementing the variable num in the last statement of the while loop like:

num = num + 1

Note: Don't make both changes at the same time.

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What is the mean of 9, 25, and 53?

Alex17521 [72]

Answer:

i think its 29

- add all the numbers up and divide by how many numbers there are.

8 0

3 years ago

Integers and booleans. Write a program RightTriangle that takes three int command-line arguments and determines whether they con

icang [17]

Answer:

The programming language is not stated;

I'll answer using C++

#include<iostream>

#include<cmath>

using namespace std;

int main()

{

int side1, side2, side3;

cout<<"Enter the three sides of the triangle: "<<endl;

cin>>side1>>side2>>side3;

if(side1<=0 || side2 <= 0 || side3 <= 0) {

cout<<"Invalid Inputs";

}

else {

if(abs(pow(side1,2) - (pow(side2,2) + pow(side3, 2)))<0.001) {

cout<<"Right Angled";

}

else if(abs(pow(side2,2) - (pow(side1,2) + pow(side3, 2)))<0.001) {

cout<<"Right Angled";

}

else if(abs(pow(side3,2) - (pow(side2,2) + pow(side1, 2)))<0.001) {

cout<<"Right Angled";

}

else {

cout<<"Not Right Angled";

}

return 0;

}

Explanation:

The following line declares the three variables

int side1, side2, side3;

The next line prompts user for input of the three sides

cout<<"Enter the three sides of the triangle: "<<endl;

The next line gets user input

cin>>side1>>side2>>side3;

The following if condition checks if any of user input is negative or 0

if(side1<=0 || side2 <= 0 || side3 <= 0) {

cout<<"Invalid Inputs";

}

If otherwise

else {

The following if condition assumes that side1 is the largest and test using Pythagoras Theorem

if(abs(pow(side1,2) - (pow(side2,2) + pow(side3, 2)))<0.001) {

cout<<"Right Angled";

}

The following if condition assumes that side2 is the largest and test using Pythagoras Theorem

else if(abs(pow(side2,2) - (pow(side1,2) + pow(side3, 2)))<0.001) {

cout<<"Right Angled";

}

The following if condition assumes that side3 is the largest and test using Pythagoras Theorem

else if(abs(pow(side3,2) - (pow(side2,2) + pow(side1, 2)))<0.001) {

cout<<"Right Angled";

}

If none of the above conditions is true, then the triangle is not a right angles triangle

else {

cout<<"Not Right Angled";

}

}

return 0;

Download cpp

4 0

4 years ago

HTML5 is______because it works on tablet or smartphone,notebooks

jeyben [28]

Answer:

Wonderful and easy language

Explanation:

Hope this helps

3 0

3 years ago

Read 2 more answers

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

String time1 = "18:00:00";

String time2 = "7:30:50";

// Creating a SimpleDateFormat object

// to parse time in the format HH:MM:SS

SimpleDateFormat simpleDateFormat

= new SimpleDateFormat("HH:mm:ss");

// Parsing the Time Period

Date date1 = simpleDateFormat.parse(time1);

Date date2 = simpleDateFormat.parse(time2);

// Calculating the difference in milliseconds

long differenceInMilliSeconds

= Math.abs(date2.getTime() - date1.getTime());

// 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

Write pseudocode to solve the following problem: You are given an array A[1 . . . n] whose each element is a point of the plane

bixtya [17]

Answer:

Answer explained below

Explanation:

void bubbleSort(int X[], int Y[], int n)

{

int i, j;

for (i = 0; i < n-1; i++)

// Last i elements are already in place

for (j = 0; j < n-i-1; j++)

if (X[j] > X[j+1])

{

swap(X[j],X[j+1])

swap(Y[j],Y[j+1]);

}

if (X[j] == X[j+1]&&Y[j]<Y[j+1])

{

swap(X[j],X[j+1])

swap(Y[j],Y[j+1]);

}

Since the above algorithm contains 2 nested loops over n.

So, it is O(n^2)

5 0

4 years ago