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Snezhnost [94]
3 years ago
15

Peter works on a huge database of numerical figures in a worksheet ranging from cell A1 to cell I50. He has to print the workshe

et and carry it to a client meeting. Which options should he select to fit the page to a standard A4 page for printing?
A. Left or right page orientation
B. Top and bottom page orientation
C. Landscape or vertical page orientation
D. Horizontal or vertical page orientation
E. Middle and side page orientation
Computers and Technology
2 answers:
bekas [8.4K]3 years ago
8 0

Answer:

Landscape or vertical page orientation

Explanation:

In landscape  he can mange the number of cell that are required to print on A4 page size.

On landscape orientation of the page we can add more columns of the sheet. That is the reason, we should choose landscape or horizontal page orientation to adjust all columns on the single page.

OLEGan [10]3 years ago
6 0

Answer:

Landscape or vertical page orientation.

Explanation:

By choosing this option, Peter will be able to fit the huge table into an A4 sheet. If he chooses a landscape orientation, he will be able to fit a larger number of columns. On the other hand, if he chooses vertical orientation, he will be able to fit more rows. Looking at this option will allow Peter to better adjust the information he will present to a design that he is happy with.

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You don't

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In general, use no more than _____ font types in a worksheet.
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The answer is a.two
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He should identify his various
joja [24]

Answer:

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Explanation:

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3 years ago
Write a function with this prototype:
sp2606 [1]

Answer:

Following are the code to this question:

#include <iostream> //defining header file  

using namespace std;

void numbers(ostream &outs, const string& prefix, unsigned int levels); // method declaration

void numbers(ostream &outs, const string& prefix, unsigned int levels) //defining method number

{

string s; //defining string variable

if(levels == 0) //defining condition statement that check levels value is equal to 0

{

outs << prefix << endl;  //use value

}

else //define else part

{

for(char c = '1'; c <= '9'; c++) //define loop that calls numbers method

{

s = prefix + c + '.'; // holding value in s variable  

numbers(outs, s, levels-1); //call method numbers

}

}

}

int main() //defining main method

{

numbers(cout, "THERBLIG", 2); //call method numbers method that accepts value

return 0;

}

Output:

please find the attachment.

Explanation:

Program description:

  • In the given program, a method number is declared, that accepts three arguments in its parameter that are "outs, prefix, levels", and all the variable uses the address operator to hold its value.
  • Inside the method a conditional statement is used in which string variable s and a conditional statement is used, in if the block it checks level variable value is equal to 0. if it is false it will go to else block that uses the loop to call method.
  • In the main method we call the number method and pass the value in its parameter.  

5 0
3 years ago
The ArrayList class contains a trim method that resizes the internal array to exactly the capacity. The trim method is intended
andreev551 [17]

Answer:

public void trimToSize() {

modCount++;

if (size < elementData.length) {

elementData = (size == 0)

? EMPTY_ELEMENTDATA

: Arrays.copyOf(elementData, size);

}

}

Now, the running time for copyOf() function is O(N) where N is the size/capacity of the ArrayList. Hence, the time complexity of trimToSize() function is O(N).

Hence, the running time of building an N-item ArrayList is O(N^2).

Please find the sample code below.

CODE

==================

import java.util.ArrayList;

public class Driver {

  public static void main(String[] args) throws Exception{

      int N = 100000;

      ArrayList<Integer> arr = new ArrayList<>(N);

     

      long startTime = System.currentTimeMillis();

      for(int i=0; i<N; i++) {

          arr.add(i);

          arr.trimToSize();

      }

      long endTime = System.currentTimeMillis();

      double time = (endTime - startTime)/1000;

      System.out.println("Total time taken = " + time + " seconds.");

  }

}

Explanation:

5 0
3 years ago
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