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BabaBlast [244]
3 years ago
10

What are two ways to access the options for scaling and page orientation? click the home tab, then click alignment, or click the

file tab. click the file tab, then click print, or click the page layout tab. click the page layout tab, or click the review tab. click the review tab, or click the home tab?
Computers and Technology
1 answer:
inessss [21]3 years ago
4 0

The answer is Click the file tab, then click print, or click the page layout tab.

Sometimes, you may want certain pages or sections to have a different orientation (vertical or landscape layout) from the rest of the content inside. To achieve this, you could either go to the file tab then click print or click the page layout tab and click on the drop down arrow to select the options available.  

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_______ is a way to minimize technical problems with your computer.
ikadub [295]
Hi,
The answer should be letter A.
7 0
3 years ago
Read 2 more answers
To move or copy a range of cells, select the correct order:
Karo-lina-s [1.5K]

Answer:

Correct Order

2. Select the cell or range you want to move or copy.

1. Move the pointer over the border of the selection until the pointer changes shape.

3. To move the range, click the border and drag the selection to a new location, or to copy the range, hold down the Ctrl key and drag the selection to a new location.

Explanation:

To move or copy range of cells in MS Excel, You first select the cell/range you want to move or copy, hover the mouse pointer and take note when it changes shape, then finally click the border (when you noticed the change of shape of the pointer) and hold down the ctrl key and drag it to the destination location.

6 0
3 years ago
Make a ladtract class that has 2 fields, one for the tractor's length and one for the width. The class should have a method that
Virty [35]

Explanation:

Below is the java code for the ladtract class :-

public class ladtract

{

private double length;  //length of tractor

private double width;   //width of tractor

public double calculateArea(){  

    return length*width;    //calculate and return the area of the tractor

}

public boolean equals(Object o) {

    if (o == this) {    //check if it's the same object

        return true;

    }

       if(o.length==this.length && o.width==this.width){   //check if the length and width are same for both objects

           return true;

       }

       return false;  

}

public String toString(){

    return "Area="+calculateArea(); //return the area of the tractor as a string

}

}

7 0
3 years ago
Write a method transpose, that takes as argument a 2D array of integers, myTable, and returns the transpose of myTable. The tran
Gelneren [198K]

Answer:

The program to this question can be given as:

Program:

import java.util.*; //import package

public class Main //defining class

{

public static int[][] transpose(int[][] a1) //defining function transpose

{

int i,j; //define variables

int[][] transpose = new int[a1[0].length][a1.length]; //define variable transpose.

//loop for transpose matrix.

for(i = 0; i < a1.length; i++)  

{

for(j = 0; j < a1[i].length; j++)

{

transpose [j][i] = a1[i][j]; //hold value in transpose variable

}

}

return transpose ; //return transpose matrix.  

}

public static void main(String[] args) //main method.

{

int row,col,i,j; //defining variable.

Scanner obj = new Scanner(System.in); //creating Scanner class Object

System.out.print("Enter number of rows: ");//message

row = obj.nextInt(); //input rows form user.

System.out.print("Enter number of columns: ");//message

col = obj.nextInt(); //input columns form user.

int[][] a1 = new int[row][col]; //defining array.

System.out.println("Enter matrix"); //message

// loop for input array elements from user

for(i=0; i<a1.length;i++)  

{

for(j=0; j<a1[i].length;j++)  

{

a1[i][j]=obj.nextInt(); //input in array variable.

}

}

System.out.println("Matrix you insert :"); //message

// loop for display array.

for(i=0; i<a1.length;i++)  

{

for(j=0; j<a1[i].length;j++)  

{

System.out.print(a1[i][j]+" "); //print matrix

}

System.out.print("\n");

}

int[][] result = transpose(a1); //define result array variable that holds function value.

System.out.println("Transposed matrix is :");

//use loop for print Transpose matrix.

for(i = 0;i < result.length;i++)

{

for (j = 0; j < result[i].length;j++)

{

System.out.print(result[i][j] + " "); //print matrix

}

System.out.print("\n");

}

}

}

Output:

Enter number of rows: 2

Enter number of columns: 3

Enter matrix

2

3

4

9

8

7

Matrix you insert :

2 3 4  

9 8 7  

Transposed matrix is :

2 9  

3 8  

4 7  

Explanation:

The description of the above program can be given as:

  • In the above program firstly import a package, then declare a class that is "Main" inside a class defining a transpose() function this function returns a matrix in transpose form.
  • In the main class, we define a variable and matrix that is "row, col, i, j and a1[][]".  Then we Creating a scanner class object, which is used by row, col and a1[][] variables for user input.  
  • The row and col variables are used to declare matrix size and i, j variable is used in the loop for insert and display matrix.  
  • a1[][] matrix is used for input matrix elements and passes into transpose() function as an argument that returns a value that is held in the result variable. This is used as a loop for the print transpose matrix.

5 0
3 years ago
Compute the present value of a $5,500 deposit in year 1, and another $5,000 deposit at the end of year 4 using an 8 percent inte
marusya05 [52]

Answer:

Explanation:

Present value is the value in the present of a sum of money, in contrast to some future value it will have when it has been invested at compound interest.

It can be calculated using future value formula below

A = P(1+r/100)^n

where

A = Future value

P = Present value

r = Rate of interest

n = time period

Present Value = Value at Year 1 + Value at Year 4

Calculating Value at Year 1

A = $5,500

r = 8%

n = 1

From A = P(1+r/100)^n ; Make P the subject of formula

P = A ÷ (1 + r/100)^n

Substitute in values

P = $5,500 ÷ (1 + 8/100)^1

P = $5,500 ÷ (1 + 0.08)

P = $5,500/1.08

P = $5092.5925926

P = $5092.59 ----

Present Value at Year 1 = $5092.59

Calculating Value at Year 4

A = $5,000

r = 8%

n = 1

From A = P(1+r/100)^n ; Make P the subject of formula

P = A ÷ (1 + r/100)^n

Substitute in values

P = $5,000 ÷ (1 + 8/100)⁴

P = $5,000 ÷ (1 + 0.08)⁴

P = $5,000/1.08⁴

P = $3675.149263982267

P = $3675.15 ----

Present Value at Year 4 = $3675.15

Present Value = Value at Year 1 + Value at Year 4

Substitute each value

Present Value = $5092.59 + $3675.15

Present Value = $8,767.74

8 0
3 years ago
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