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Hatshy [7]
2 years ago
15

Question is attached as image, please help :>

Computers and Technology
1 answer:
harkovskaia [24]2 years ago
3 0

Answer:

Proved

Explanation:

Given

X = (a.\bar b)+(\bar a.b)

(a + b)\ .  \frac{}{a.b}

Required

Find out why they represent the same

To do this, we simplify either (a + b)\ .  \frac{}{a.b} or X = (a.\bar b)+(\bar a.b)

In this question, I will simplify (a + b)\ .  \frac{}{a.b}

Apply de morgan's law

(a + b)\ .  \frac{}{a.b} = (a + b) . (\bar a + \bar b)

Apply distribution property

(a + b)\ .  \frac{}{a.b} = a.\bar a + a.\bar b + \bar a . b + b . \bar b

To simplify, we apply the following rules:

a.\bar a = 0 --- Inverse law

b.\bar b = 0 --- Inverse law

So, the expression becomes

(a + b)\ .  \frac{}{a.b} = 0 + a.\bar b + \bar a.b + 0

(a + b)\ .  \frac{}{a.b}  = a.\bar b + \bar a.b

Rewrite as:

(a + b)\ .  \frac{}{a.b}  = (a.\bar b)+(\bar a.b)

From the given parameters:

X = (a.\bar b)+(\bar a.b)

This implies that:

(a + b)\ .  \frac{}{a.b} when simplified is X or (a.\bar b)+(\bar a.b)

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Answer:

B. operating tractors and machinery

D. maintaining vehicles and equipment

E. operating irrigation (watering) equipment

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Explanation:

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Imagine you were going to use a dedicated workstation for an animation job rather than a personal PC or the all-purpose PCs you
xeze [42]

Answer:

Explanation:

The three most notable differences that any user will notice between these two would be the following

  1. High-Resolution Display
  2. Specific Hardware (graphics tablet, camera, etc)
  3. Increased Speed

An individual that is working in a career in Animation needs a workstation that is capable of rendering animations which take lots of processing power. Therefore, the workstation would have to be high-end meaning that it will be much faster than a normal personal PC. Animation and Digital Design require a very High-Resolution display and specific hardware to allow you to bring your creations to life and make sure they look as beautiful as possible. All of this is not found in a normal personal PC either.

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3 years ago
Write a program that reads two fractions such as 1/2 and 1/4 and computes and stores the sum, the difference, the product and th
Vesnalui [34]

Answer:

1: #include <iostream>

2: using namespace std;

3: int main()

4: {

5: int a,b,c,d;

6: cout<<"n1: ";

7: cin>>a;

8: cout<<"d1: ";

9: cin>>b;

10: cout<<"n2: ";

11: cin>>c;

12: cout<<"d2: ";

13: cin>>d;

14: int top = (a*d)+(b*c);

15: int bottom = b*d;

16: cout<<"Sum: "<<top<<"/"<<bottom<<"\n";

17: top = (a*d)-(b*c);

18: cout<<"Difference: "<<top<<"/"<<bottom<<"\n";

19: top = a*c;

20: cout<<"Product: "<<top<<"/"<<bottom<<"\n";

21: top = a*d;

22: bottom = b*c;

23: cout<<"Quotient: "<<top<<"/"<<bottom<<"\n";

24: return 0;

25: }

Explanation:

The Program is written in C++ programming language

The Program is left numbered

Line 5 was used for variable declaration.

Variables a,b represents the numerator and denominator of the first fraction

While variables c,d represent the numerator and denominator of the second fraction

Line 6 prints "n1" without the quotes which represents the numerator of the first fraction

Line 7 accepts input for the numerator of the first fraction.

Line 8 prints "d1" without the quotes which represents the denominator of the first fraction

Line 9 accepts input for the denominator of the first fraction.

Line 10 prints "n2" without the quotes which represents the numerator of the second fraction

Line 11 accepts input for the numerator of the second fraction.

Line 12 prints "d2" without the quotes which represents the denominator of the second fraction

Line 13 accepts input for the denominator of the second fraction.

Line 14 and 15 calculate the sum of the fractions which is then printed on line 16

Line 17 calculates the difference of the fractions which is then printed on line 18

Line 19 calculates the product of the fractions which is then printed on line 20

Line 21 and 22 calculates the quotient of the fractions which is then printed on line 23

5 0
3 years ago
An output device is any device that displays, prints, or plays content stored on a computer. from the list below, identify the d
inysia [295]
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3 years ago
Given an array as follows
slava [35]

Answer:

1) Method calcTotal:

  1. public static long calcTotal(long [][] arr2D){
  2.        long total = 0;
  3.        
  4.        for(int i = 0; i < arr2D.length; i++ )
  5.        {
  6.            for(int j = 0; j < arr2D[i].length; j++)
  7.            {
  8.                total = total + arr2D[i][j];
  9.            }
  10.        }
  11.        
  12.        return total;
  13.    }

Explanation:

Line 1: Define a public method <em>calcTotal</em> and this method accept a two-dimensional array

Line 2: Declare a variable, total, and initialize it with zero.

Line 4: An outer for-loop to iterate through every rows of the two-dimensional array

Line 6: An inner  for-loop to iterate though every columns within a particular row.

Line 8: Within the inner for-loop, use current row and column index, i and j, to repeatedly extract the value of each element in the array and add it to the variable total.

Line 12: Return the final total of all the element values as an output

Answer:

2) Method calcAverage:

  1. public static double calcAverage(long [][] arr2D){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            for(int j = 0; j < arr2D[i].length; j++)
  8.            {
  9.                total = total + arr2D[i][j];
  10.                count++;
  11.            }
  12.            
  13.        }
  14.        
  15.        double average = total / count;
  16.        
  17.        return average;
  18.    }

Explanation:

The code in method <em>calcAverage</em> is quite similar to method <em>calcTotal</em>. We just need to add a counter and use that counter as a divisor of total values to obtain an average.

Line 4: Declare a variable, count, as an counter and initialize it to zero.

Line 11: Whenever an element of the 2D array is added to the total, the count is incremented by one. By doing so, we can get the total number of elements that exist in the array.

Line 16: Use the count as a divisor to the total to get average

Line 18: Return the average of all the values in the array as an output.

Answer:

3) calcRowAverage:

  1. public static double calcRowAverage(long [][] arr2D, int row){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            if(i == row)
  8.            {
  9.                for(int j = 0; j < arr2D[i].length; j++)
  10.                {
  11.                    total = total + arr2D[i][j];
  12.                    count++;
  13.                }
  14.            }
  15.            
  16.        }
  17.        
  18.        double average = total / count;
  19.        
  20.        return average;
  21.    }

Explanation:

By using method <em>calcAverage </em>as a foundation, add one more parameter, row, in the method <em>calcRowAverage</em>. The row number is used as an conditional checking criteria to ensure only that particular row of elements will be summed up and divided by the counter to get an average of that row.

Line 1: Add one more parameter, row,

Line 8-15: Check if current row index, i, is equal to the target row number, proceed to sum up the array element in that particular row and increment the counter.

5 0
3 years ago
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