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Svetradugi [14.3K]
2 years ago
5

What is folded card publishing?​

Computers and Technology
2 answers:
Oduvanchick [21]2 years ago
6 0
Folded card This layout type is for creating
greeting cards by printing pages on a sheet and then folding the sheet to make the card. If you choose the Folded card layout, then sheet fold options are displayed. Select an option in the list to specify how you will fold your publication
Alexeev081 [22]2 years ago
3 0

Answer:

Folded card This layout type is for creating greeting cards by printing pages on a sheet and then folding the sheet to make the card. If you choose the Folded card layout, then sheet fold options are displayed. Select an option in the list to specify how you will fold your publication

Explanation:

You might be interested in
Slack space most commonly contains visible data. True or false?
Ivanshal [37]
That is so True because slack space mostly has visible dada
7 0
3 years ago
Given two complex numbers, find the sum of the complex numbers using operator overloading.Write an operator overloading function
Inessa05 [86]

Answer:

I am writing the program in C++ programming language.  

#include<iostream>  // to use input output functions

using namespace std;   // to identify objects like cin cout

class ProblemSolution {  //class name

private:  

// private data members that only be accessed within ProblemSolution class

int real, imag;

 // private variables for real and imaginary part of complex numbers

public:  

// constructor ProblemSolution() to initialize values of real and imaginary numbers to 0. r is for real part and i for imaginary part of complex number

ProblemSolution(int r = 0, int i =0) {  

real = r; imag = i; }  

/* print() method that displays real and imaginary part in output of the sum of complex numbers */

void print(){  

//prints real and imaginary part of complex number with a space between //them

cout<<real<<" "<<imag;}  

// computes the sum of complex numbers using operator overloading

ProblemSolution operator + (ProblemSolution const &P){  //pass by ref

          ProblemSolution sum;  // object of ProblemSolution

          sum.real = real + P.real;  // adds the real part of the  complex nos.

          sum.imag = imag + P.imag;  //adds imaginary parts of  complex nos.

//returns the resulting object

          return sum;       }  //returns the sum of complex numbers

};   //end of the class ProblemSolution

int main(){    //start of the main() function body

int real,imag;  //declare variables for real and imaginary part of complex nos

//reads values of real and imaginary part of first input complex no.1

cin>>real>>imag;  

//creates object problemSolution1 for first complex number

ProblemSolution problemSolution1(real, imag);  //creates object

//reads values of real and imaginary part of first input complex no.2

cin>>real>>imag;

//creates object problemSolution2 for second complex number

ProblemSolution problemSolution2(real,imag);

//creates object problemSolution2 to store the addition of two complex nos.

ProblemSolution problemSolution3 = problemSolution1 + problemSolution2;

problemSolution3.print();} //calls print() method to display the result of the //sum with real and imaginary part of the sum displayed with a space

Explanation:

The program is well explained in the comments mentioned with each statement of the program. The program has a class named ProblemSolution which has two data members real and imag to hold the values for the real and imaginary parts of the complex number. A default constructor ProblemSolution() which initializes an the objects for complex numbers 1 and 2 automatically when they are created.

ProblemSolution operator + (ProblemSolution const &P) is the operator overloading function. This performs the overloading of a binary operator + operating on two operands. This is used here to add two complex numbers.  In order to use a binary operator one of the operands should be passed as argument to the operator function. Here one argument const &P is passed. This is call by reference. Object sum is created to add the two complex numbers with real and imaginary parts and then the resulting object which is the sum of these two complex numbers is returned.  

In the main() method, three objects of type ProblemSolution are created and user is prompted to enter real and imaginary parts for two complex numbers. These are stored in objects problemSolution1 and problemSolution2.  Then statement ProblemSolution problemSolution3 = problemSolution1 + problemSolution2;  creates another object problemSolution3 to hold the result of the addition. When this statement is executed it invokes the operator function ProblemSolution operator + (ProblemSolution const &P). This function returns the resultant complex number (object) to main() function and print() function is called which is used to display the output of the addition.

4 0
4 years ago
Thomas is getting ready for a presentation. His slides are set, and he's
PIT_PIT [208]
B copy the web page of the video that you want to use
5 0
4 years ago
Why do people still use Intel HD graphics for gaming?
pogonyaev
People usually do that because they can't afford a gpu or because they don't know anything about a gpu.
8 0
3 years ago
Consider the following recursive method: public int someFun(int n) { if (n &lt;= 0) return 2; else return someFun(n-1) * someFun
Stella [2.4K]

Answer:

(a) someFunc(3) will be called 4 times.

(b) For non negative number n someFunc method calculates 2^2^n.

Explanation:

When you call someFunc(5) it will call someFunc(4) two time.

So now we have two someFunc(4) now each someFunc(4) will call someFunc(3) two times.Hence the call to someFun(3) is 4 times.

someFunc(n) calculates someFunc(n-1) two times and calculates it's product.

someFunc(n) = someFunc(n-1)^2..........(1)

someFunc(n-1)=someFunc(n-2)^2..........(2)

substituting the value form eq2 to eq 1.

someFunc(n)=someFunc(n-2)^2^2

       .

       .

       .

       .

= someFunc(n-n)^2^n.

=2^2^n

2 raised to the power 2 raised to the power n.

3 0
3 years ago
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