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