After he selects the Push transition, He should select from Left option from the Effect Options drop-down menu.
<h3>What is push transition in PowerPoint?</h3>
The Push Transition Effect is known to be a function in a PowerPoint that helps to pushes one slide to show the other.
Note that when this is applied to any presentation, it moves the current slide upwards to reveal the next slide and so on. A person can add use this kind of effect to present a scrolling effect to one's PowerPoint work.
Learn more about presentation from
brainly.com/question/24653274
Answer:
Adding extra horizontal scroll, Blocking mobile devices from viewing, Eliminating extra links, Resizing content to fit a screen.
Explanation:
Answer:
False
Explanation:
The private member of a class is not accessible by using the Dot notation ,however the private member are those which are not accessible inside the class they are accessible outside the class .The public member are accessible inside the class so they are accessible by using the dot operator .
<u>Following are the example is given below in C++ Language </u>
#include<iostream> // header file
using namespace std;
class Rectangle
{
private:
double r; // private member
public:
double area()
{ return 3.14*r*r;
}
};
int main()
{
Rectangle r1;// creating the object
r1.r = 3.5;
double t= r1.area(); // calling
cout<<" Area is:"<<t;
return 0;
}
Output:
compile time error is generated
<u>The correct program to access the private member of class is given below </u>
#include<iostream> // header file
using namespace std;
class Rectangle
{
private:
double r; // private member
public:
double area()
{
r1=r;
double t2=3.14*r2*r2;
return(t2); // return the value
}
};
int main()
{
Rectangle r1;// creating the object
r1.r = 1.5;
double t= r1.area(); // calling
cout<<" Area is:"<<t;
return 0;
}
Therefore the given statement is False
A hyperlink is a feature that takes you to a different page when you click on it. That page could be part of the same website or a totally different website.
Let me know if you have any questions.
