The above statement is FALSE.
The layout gallery displays 9 slide layouts NOT 10.
These 9 layouts have various<span> placeholders to define text and content positioning and formatting.
</span>1 - Title slide
2 - Title and content
3 - Section Header
4 - Two content
5 - Comparison
6 - Title only
7 - Blank
8 - Content with Caption
9 - Picture with Caption
Answer:
In Java:
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String isbn;
System.out.print("First 1:2 digits: ");
isbn = input.nextLine();
if(isbn.length()==12){
int chksum = 0;
for(int i = 0; i<12;i++){
if((i+1)%2==0){ chksum+= 3 * Character.getNumericValue(isbn.charAt(i)); }
else{ chksum+=Character.getNumericValue(isbn.charAt(i)); } }
chksum%=10;
chksum=10-chksum;
if(chksum==10){
System.out.print("The ISBN-13 number is "+isbn+"0");}
else{
System.out.print("The ISBN-13 number is "+isbn+""+chksum); } }
else{
System.out.print("Invalid Input");
} }}
Explanation:
See attachment for explanation where comments are used to explain each line
We can define a word as a group of characters without a space between them. To find the words of the input string , w can use split(delimiter) which returns a list of strings which had the defined delimiter between them in the input string.
def countWords(string):
words = string.split(" ")
count = len(words)
return count
Here we set the delimiter as the space character, and returned the length of the words list. I split each step into its own line for readability, however the function could be one line:
return len(string.split())
Here, no delimiter is specified. If one isn't given, it will default to split at any whitespace, including space.
The answer is d.mother board
