Volume of Pyramid = A*h/3 where A is the area of the base of the pyramid and h is the height of the pyramid. Write a C++ program that asks the user to enter the necessary information about the pyramid (note that the user would not know the area of the base of the pyramid, you need to ask them for the length of one of the sides of the base and then calculate the area of the base). Using the information the user input, calculate the volume of the pyramid. Next display the results (rounded to two decimal places). Example Output (output will change depending on user input): The area of the base of the pyramid is: 25.00 The height of the pyramid is: 5.00 The volume of the pyramid is: 41.67 *Pseudocode IS required for this program and is worth 1 point. The program IS auto-graded.

Answer:

-11001

Explanation:

The following steps are performed in order to perform subtraction of the given binary numbers:

Step 1:

Find 2’s complement of the subtrahend. The subtrahend here is 100000

100000

First take 1's complement of 100000

1's complement is taken by inverting 100000

1's complement of 100000 = 011111

Now takes 2's complement by adding 1 to the result of 1's complement:

011111 + 1 = 100000

2's complement of 100000 = 100000

Step 2:

Add the 2's complement of the subtrahend to the minuend.

The number of bits in the minuend is less than that of subtrahend. Make the number of bits in the minuend equal to that of subtrahend by placing 0s in before minuend. So the minuend 111 becomes:

000111

Now add the 2's complement of 100000 to 000111

   0 0 0 1  1  1

+   1 0 0 0 0 0

    1 0 0 1  1  1

The result of the addition is :

 1 0 0 1  1  1

Step 3:

Since there is no carry over the next step is to take 2's complement of the sum and place negative sign with the result as the result is negative.

sum =   1 0 0 1  1  1

2's complement of sum:

First take 1's complement of 1 0 0 1  1  1

1's complement is taken by inverting 1 0 0 1  1  1

1's complement of 1 0 0 1  1  1 = 0 1 1 0 0 0

Now takes 2's complement by adding 1 to the result of 1's complement:

0 1 1 0 0 0 + 1 = 0 1 1 0 0 1

Now place the minus sign with the result of 2's complement:

- 0 1 1 0 0 1

Hence the subtraction of two binary numbers (100000)₂ and (111)₂  is

(-011001)₂

This can also be written as:

(-11001)₂

Answer:

Here the code is  given as,

Explanation:

Code:

#include <math.h>

#include <cmath>  

#include <iostream>

using namespace std;

int main() {

int v_stop = 0,count = 0 ;

int x;

double y;

int t_count [100];

double p_item [100];

double Total_rev = 0.0;

double cost_trx[100];

double Largest_element , Smallest_element;

double unit_sold = 0.0;

for( int a = 1; a < 100 && v_stop != -99 ; a = a + 1 )

  {

     cout << "Transaction # " << a << " : " ;

     cin >> x >> y;

  t_count[a] = x;

  p_item [a] = y;

  cost_trx[a] = x*y;

  v_stop = x;

  count = count + 1;

  }

  for( int a = 1; a < count; a = a + 1 )

  {

   Total_rev = Total_rev + cost_trx[a];

   unit_sold = unit_sold + t_count[a];

  }

  Largest_element = cost_trx[1];

  for(int i = 2;i < count - 1; ++i)

   {

      // Change < to > if you want to find the smallest element

      if(Largest_element < cost_trx[i])

          Largest_element = cost_trx[i];

   }

Smallest_element = cost_trx[1];

  for(int i = 2;i < count - 1; ++i)

   {

      // Change < to > if you want to find the smallest element

      if(Smallest_element > cost_trx[i])

          Smallest_element = cost_trx[i];

   }

  cout << "TRANSACTION PROCESSING REPORT     " << endl;

  cout << "Transaction Processed :           " << count-1 << endl;

  cout << "Uints Sold:                       " << unit_sold << endl;

  cout << "Average Units per order:          " << unit_sold/(count - 1) << endl;

  cout << "Largest Transaction:              " << Largest_element << endl;

  cout << "Smallest Transaction:             " << Smallest_element << endl;

  cout << "Total Revenue:               $    " << Total_rev << endl;

  cout << "Average Revenue :            $    " << Total_rev/(count - 1) << endl;

  return 0;

}

Output:

Answer:

The correct option is D

Explanation:

Generally for a virtual circuit   both analog and digital networks take advantage of the virtual circuit concept

Generally virtual circuits require the assignment of dedicated physical links during network connection

 Generally regarding the efficient use of physical links,  virtual circuits are NOT more efficient than analog circuits

Answer:

Human visual behavior often includes searching, scanning, and monitoring.

While it should be possible to moderate the performance of these tasks based on the dimensions used in the table, it is often useful to analyze these situations using Signal Detection Theory (SDT).              

Three suggestions for my favourite search engine that includes search are:

1.  ensure that search results are graded according to the following categories using different colours:  

• Most accurate and relevant results;
• Accurate and slightly relevant results
• Fairly accurate and fairly relevant results

These can help the users identify more easily the results to spend more time on. This categorisation can be done using colours. This is because of the way the eye functions. Unlike a camera snapshot, for example, the eye does not capture everything in a scene equally, but selectively picks out salient objects and features from the current context, and focuses on them so they can be processed in more detail.

2.  Another suggestion is that attention needs to be paid to where people look most of the time. It is not out of place for people to instinctively avoid search results that have dollar signs or currency signs especially when they are not searching for a commercial item.  

3. Lastly in displaying results that have images, it best to use sharp images. The theory suggests with respect to contrast, clarity and brightness that sharper and more distinct objects appear to be nearer, and duller objects appear to be farther away. To elicit the interest of the reader, targeted results or information from search engines must make use of the factors of contrast, clarity and brightness.

Cheers!

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