_______ allows you to specify how a photo is positioned in the text of your document.

JAVA

butalik [34]

Answer:

import java.util.*;

public class MyClass {

public static void main(String args[]) {

Scanner input = new Scanner(System.in);

System.out.print("Input a word: ");

String userinput = input.nextLine();

for(int i =0;i<userinput.length();i+=2) {

System.out.print(userinput.charAt(i));

}

Explanation:

This line prompts user for input

System.out.print("Input a word: ");

This declares a string variable named userinput and also gets input from the user

String userinput = input.nextLine();

The following iterates through every other character of userinput from the first using iteration variable i and i is incremented by 2

for(int i =0;i<userinput.length();i+=2) {

This prints characters at i-th position

System.out.print(userinput.charAt(i));

5 0

3 years ago

5)What are the differences in the function calls between the four member functions of the Shape class below?void Shape::member(S

Stells [14]

Answer:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

Explanation:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

The s1 and s2 objects are passed by value as there is no * or & sign with them. If any change is made to s1 or s2 object, there will not be any change to the original object.

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

The s1 and s2 objects are passed by pointer as there is a * sign and not & sign with them. If any change is made to s1 or s2 object, there will be a change to the original object.

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

The s1 and s2 objects are passed by reference as there is a & sign and not * sign with them. If any change is made to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

The s1 and s2 objects are passed by reference as there is a & sign and not * sign with them. The major change is the usage of const keyword here. Const keyword restricts us so we cannot make any change to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

The s1 and s2 objects are passed by reference as there is a & sign and not * sign with them. const keyword restricts us so we cannot make any change to s1 or s2 object as well as the Shape function itself.

5 0

3 years ago

How many channels can the 2.4 GHz band be divided into?

wlad13 [49]

Answer:Fourteen channels

Explanation:

8 0

3 years ago

Technician A says that generic scan tools must be able to read all generic OBD-II codes. Technician B says that all generic scan

Likurg_2 [28]

Answer:

A. Technicians A and B

Explanation:

When we're talking about generic scanners and about all OBD-II codes, in this case, both technician A and B is the correct answer. Because we can scan all OBD-II codes with a generic scan.

But the technician A just says generic tools must be able to read all generic OBD-II codes and technicians B just says generic scan tools must be able to read manufacture OBD-II code, both are correct.

4 0

3 years ago

What project started the development of inter-network connections using tcp/ip that has evolved into the internet today.?

rosijanka [135]

Dial Up, help start it

3 0

3 years ago