Answer:
The right approach will be "PSH".
Explanation:
- The PSH flag seems to be an acceptable solution through TCP which always encourages the submit implementation to immediately transfer data but only whenever the buffer isn't complete.
- The implementation helps to formulate this same flag towards true for something like the connection, as well as TCP will instantly begin transmitting the relevant information.
import java.util.Scanner;
public class InchesToFeetInteractive
{
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
final int INCHES_IN_FOOT = 12;
int inches = scan.nextInt();
int feet;
int inchesLeft;
feet = inches / INCHES_IN_FOOT;
inchesLeft = inches % INCHES_IN_FOOT;
System.out.println(inches + " inches is " +
feet + " feet and " + inchesLeft + " inches");
}
}
We import the Scanner class and then initialize a new Scanner named scan. We then get an integer representation of inches from the user and calculate the feet and inches from the value entered by the user.
Answer:
The correct answer to the following question will be "Taxonomy
".
Explanation:
A location taxonomy seems to be the direction it helped organize the information into subgroups as well as subsections, often shown in a network map, in comparison to internet sites and access points.
- Is a very well-planned approach that implies that somehow the navigation features on the website become straightforward to do this and feature-packed
- The scientific method of categorizing, or classifying, items predicated on a default framework.
Just- you don’t wanna know alr? Alr :)
Answer:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
double side, height;
cout<<"Enter the length of one of the sides of the base: ";
cin>>side;
cout<<"Enter the height: ";
cin>>height;
double area = side * side;
double volume = area * height / 3;
cout<<"The area of the base of the pyramid is: "<<area<<endl;
cout<<"The height of the pyramid is: "<<height<<endl;
cout<<"The volume of the pyramid is: "<<fixed<<setprecision(2)<<volume<<endl;
return 0;
}
<u>Pseudocode:</u>
Declare side, height
Get side
Get height
Set area = side * side
Set volume = area * height / 3
Print area
Print height
Print volume
Explanation:
Include <iomanip> to have two decimal places
Declare the side and height
Ask the user to enter side and height
Calculate the base area, multiply side by side
Calculate the volume using the given formula
Print area, height and volume (use <em>fixed</em> and <em>setprecision(2)</em> to have two decimal places)
