Which method can help you prevent RSI while using a keyboard?Keep the left-hand’s fingers on J, K, L, and ; keys.
Rest the left-hand’s fingers on A,S, D, and F keys.
Rest the right-hand’s fingers on A,S, D, and F keys.
Rest the right-hand’s fingers on Q,W, E, and R keys.

Answer:

The program written in Python is as follows:

See Explanation section for line by line explanation

for n in range(100,1000):

     isum = 0

     for d in range(1,n):

           if n%d == 0:

                 isum += d

     if isum == n * 2:

           print(n)

Explanation:

The program only considers 3 digit numbers. hence the range of n is from 100 to 999

for n in range(100,1000):

This line initializes sum to 0

     isum = 0

This line is an iteration that stands as the divisor

     for d in range(1,n):

This line checks if a number, d can evenly divide n

           if n%d == 0:

If yes, the sum is updated

                 isum += d

This line checks if the current number n is a double-perfect number

     if isum == n * 2:

If yes, n is printed

           print(n)

When the program is run, the displayed output is 120 and 672

Answer:

Let f be a function

a) f(n) = n²

b) f(n) = n/2

c) f(n) = 0

Explanation:

a) f(n) = n²

This function is one-to-one function because the square of two different or distinct natural numbers cannot be equal.

Let a and b are two elements both belong to N i.e. a ∈ N and b ∈ N. Then:

                                f(a) = f(b) ⇒ a² = b² ⇒ a = b

The function f(n)= n² is not an onto function because not every natural number is a square of a natural number. This means that there is no other natural number that can be squared to result in that natural number.  For example 2 is a natural numbers but not a perfect square and also 24 is a natural number but not a perfect square.

b) f(n) = n/2

The above function example is an onto function because every natural number, lets say n is a natural number that belongs to N, is the image of 2n. For example:

                                f(2n) = [2n/2] = n

The above function is not one-to-one function because there are certain different natural numbers that have the same value or image. For example:

When the value of n=1, then    

                                   n/2 = [1/2] = [0.5] = 1

When the value of n=2 then

                                    n/2 = [2/2] = [1] = 1

c) f(n) = 0

The above function is neither one-to-one nor onto. In order to depict that a function is not one-to-one there should be two elements in N having same image and the above example is not one to one because every integer has the same image.  The above function example is also not an onto function because every positive integer is not an image of any natural number.

Answer:

The output is 20

Explanation:

This line divides the value of x by userVal

tmpVal = x / userVal;

i.e.

tmpVal = 100/5

tmpVal = 20

This line then prints the value of tmpVal

System.out.print(tmpVal);

i.e 20

Hence, The output is 20

Final answer:

The provided Java code checks if the variable userVal does not equal 0. As this is the case when userVal is 5, another variable tmpVal is assigned the value of another variable x (which is 100) divided by userVal. Hence, the output of the code would be 20.

Explanation:

In the given piece of code, the variable userVal is assigned a value of 5. The program also contains a variable x which is assigned a value of 100. An if statement checks whether userVal does not equal 0 - since 5 != 0, the condition is true. A new variable tmpVal is then declared and assigned the value of x divided by userVal, so tmpVal equals 100 / 5, which is 20. So, the output of this code when userVal is 5 would be 20.

Learn more about Java Programming here:

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Answer:

The system cannot find the path specified.

Explanation:

please give brain thx! :D

good luck!

Answer:

It's pie chart because it shows the percentage of multiple things.

Explanation:

And I did an exam aced it and got this question right.

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: