Answer:
She should use her "notes pane" in the PowerPoint
Explanation:
PowerPoint is a very good graphics software program available in different versions as a result of upgrades, with in-built functions which aid presentations of information in different formats such as text, videos, graphs in slides. It also allows printing into a hard copy.
Notes pane is a section in the PowerPoint software program located below the slide pane, in other words, it is a box found at the bottom of the slide. You write information in that box when you don't want the audience to see the information when they are viewing the slides during a presentation, giving Beth, for instance, the ability to still add some things that could assist her presentation.
Answer:
Putting the entire class here would be large, so I've attached a class file
Explanation:
All the explanation is done inside the class file.
It's a pretty simple Java GUI Program in which users can input the x and y cordinates into some fields and then a button is pressed to print out the info for the point.
Answer:
Action buttons are built-in shapes you can add to a presentation and set to link to another slide, play a sound, or perform a similar action.
Explanation:
Answer:
return value =2.
Here the function f() returns the length of the substring we traversed before we find the same character at the equal index of two substrings.
Take the inputs s= “abcd” and t= “bccd”.
• Now, p1 points to s1, i.e., p1 points to the character ‘a’ of “abcd”. And similarly, p2 points to ‘b’ of “bccd”.
• Then we compare the values at p1 and p2, are not equal, so p1 and p2 both are incremented by 1.
• Now the characters ‘b’ and ‘c’ of “abcd” and “bccd” respectively are compared. They are not equal. So both p1 and p2 both are incremented by 1.
• Now, p1 points to ‘c’ of “ abcd” that is the element at index 2 of s. And p2 points to ‘c’ of “bccd” that is the element at index 2 of t. Here value at p1 and p2 becomes equal. So the break statement is executed. We stop moving forward.
• As p1 is pointing to index 2 and s is pointing to the base that is index 0, so p1-s = 2.
Explanation:
#include<stdio.h>
int f(char *s, char *t);
void main()
{
int k = f("abcd", "bccd");
printf("%d", k);
}
int f(char *s, char *t)
{
char *p1, *p2;
for(p1 = s, p2 = t; *p1 != '\0'&& *p2 != '\0'; p1++, p2++)
{
if (*p1 ==*p2)
break;
}
return (p1-s);
}
OUPUT is given as image
Actually, a lot, and it’s especially helpful when the answer is expert verified