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Wittaler [7]
3 years ago
7

Text that does not fit in a cell

Computers and Technology
1 answer:
Aneli [31]3 years ago
3 0
To make all wrapped text visibly, do the following to manually adjust the row height:<span>Select the cell or range for which you want to adjust the row height.On the Home tab, in the Cells group, click Format.<span>Under Cell Size , do one of the following: To automatically adjust the row height, click AutoFit Row Height.</span></span>
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Cations - Office 2016 A
Vika [28.1K]

Answer:

1345

Explanation:

6 0
3 years ago
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Write a recursive method to return the number of uppercase letters in a String. You need to define the following two methods. Th
satela [25.4K]

Answer:

<u>Recursive function with two parameters that return the number of uppercase letters in a String</u>

public static int count(String str,int h)//Defining function

{

      if(str.charAt(0)>='A' && str.charAt(0)<='Z')//Checking the characters from A to Z

{

          h++; //incrementing the counter

          if(str.length()>=2){

              return count(str.substring(1),h);//recalling function

          }

      }

      if(str.length()>=2){

              return count(str.substring(1),h); //recalling function

      }

      return h;

  }

This is the recursive function with the name count of return type integer,having parameters str of string type and h of integer type.In this we are checking the characters at a particular position from A to Z.

<u>Recursive function with one parameter that return the number of uppercase letters in a String</u>

public static int count(String str)//Defining function

{

      if(str.charAt(0)>='A' && str.charAt(0)<='Z')//Checking the characters from A to Z

{

          count++; //incrementing the counter

          if(str.length()>=2){

              return count(str.substring(1));//recalling function

          }

      }

      if(str.length()>=2){

              return count(str.substring(1)); //recalling function

      }

      return count;

  }

This is the recursive function with the name count of return type integer,having parameters str of string type .In this we are checking the characters at a particular position from A to Z.

<u>Java program that return the number of uppercase letters in a String</u>

import java.util.*;

public class Myjava{

static int count =0;//Defining globally  

 

public static int count(String str,int h)//Defining function

{

      if(str.charAt(0)>='A' && str.charAt(0)<='Z')//Checking the characters from A to Z

{

          h++;

//incrementing the counter

          if(str.length()>=2){

              return count(str.substring(1),h);//recalling function

          }

      }

      if(str.length()>=2){

              return count(str.substring(1),h);

//recalling function

      }

      return h;

 

  }

  public static void main(String[] args)//driver function

  {

      System.out.println("Enter a string");//taking input

      Scanner scan = new Scanner(System.in);

      String s = scan.nextLine();

      int h =0;

      System.out.println("Counting the Uppercase letters: "+count(s,h));

  }

}

<u>Output</u>

Enter a string  WolFIE

Counting the Uppercase letters: 4

4 0
3 years ago
Write a C++ program to find if a given array of integers is sorted in a descending order. The program should print "SORTED" if t
LenKa [72]

Answer:

The cpp program for the given scenario is as follows.

#include <iostream>

#include <iterator>

using namespace std;

int main()

{

   //boolean variable declared and initialized  

   bool sorted=true;

   //integer array declared and initialized

   int arr[] = {1, 2, 3, 4, 5};

   //integer variables declared and initialized to length of the array, arr

   int len = std::end(arr) - std::begin(arr);

       //array tested for being sorted

    for(int idx=0; idx<len-1; idx++)

    {

        if(arr[idx] < arr[idx+1])

           {

               sorted=false;

            break;

           }

    }

    //message displayed  

    if(sorted == false)

     cout<< "UNSORTED" <<endl;

 else

    cout<< "UNSORTED" <<endl;

return 0;

}

OUTPUT

UNSORTED

Explanation:

1. An integer array, arr, is declared and initialized as shown.

int arr[] = {1, 2, 3, 4, 5};

2. An integer variable, len, is declared and initialized to the size of the array created above.

int len = std::end(arr) - std::begin(arr);

3. A Boolean variable, sorted, is declared and initialized to true.

bool sorted=true;  

4. All the variables and array are declared inside main().

5. Inside for loop, the array, arr, is tested for being sorted or unsorted.  The for loop executes over an integer variable, idx, which ranges from 0 till the length of the array, arr.

6. The array is assumed to be sorted if all the elements of the array are in descending order.

7. If the elements of the array are in ascending order, the Boolean variable, sorted, is assigned the value false and for loop is exited using break keyword. The testing is done using if statement.

8. Based on the value of the Boolean variable, sorted, a message is displayed to the user.

9. The program can be tested for any size of the array and for any order of the elements, ascending or descending. The program can also be tested for array of other numeric data type including float and double.

10. All the code is written inside the main().

11. The length of the array is found using begin() and end() methods as shown previously. For this, the iterator header file is included in the program.

4 0
3 years ago
Ask the user to input a country name. Display the output the message “I
Sever21 [200]

Answer:

Explanation:

The following code is written in Python, it asks the user for an input saves it to a variable called country, and then prints out the sentence example in the question using the user's input. The output can be seen in the picture attached below

country = input("input country")

print('I would love to go to ' + country)

4 0
3 years ago
All linear programming problems have all of the following properties EXCEPT
lyudmila [28]

Answer:

alternative optimal solutions

Explanation:

8 0
3 years ago
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