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Ierofanga [76]
2 years ago
13

Can someone please help me! It’s due Thursday!

Computers and Technology
1 answer:
Eva8 [605]2 years ago
7 0

Answer:

if the input is zero the out put is 1

Explanation:

because if you think about it if the input was 1 the output would be zero

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Write a routine to interchange the mth and nth elements of a singly-linked list. You may assume that the ranks m and n are passe
IRINA_888 [86]

Answer:

//The routine to swap m th and nth ranked elements

void swapmAndn(int m, int n)

{

  //Set the linked list head node

  Linked_Node **node_head

  //To find the m ranked node

  //set previous node  

  Linked_Node *node_prev = NULL;

 

  //set node to store m ranked node as head node

  Linked_Node *node_current_m = *node_head;

 

  //set the rank as rm=1

  rm =1

  //traverse to find the m ranked node

  while (node_current_m && rm<m)

  {    

      node_prev = node_current_m;

      node_current_m = node_current_m->next;

      rm++;

  }

 

  //To find the n ranked node

  //set previous node  

  Linked_Node *node_prev_n = NULL;

 

  //set node to store n ranked node as head node

  Linked_Node *node_current_n = *node_head;

 

  //set the rank as rn=1

  rn =1

  //traverse to find the n ranked node

  while (node_current_n && rn<n)

  {    

      node_prev_n= node_current_n;

      node_current_n = node_current_n->next;

      rn++;

  }

 

  //if m is not first ranked node

  if (node_prev != NULL)

      //set node_current_n as previous node

      node_prev->next = node_current_n;

 

  //otherwise Set the node_current_n

  else

      *node_head = node_current_n;

 

  // If n is not first ranked node

  if (node_prev_n!= NULL)

      node_prev_n->next = node_current_m;

  else

      *node_head = node_current_m;

 

  //now swap the next pointers also

  //to make the swap process complete  

  Linked_Node *temp = node_current_n->next;

  node_current_n->next = node_current_m->next;

  node_current_m->next = temp;

}

Explanation:

3 0
3 years ago
Which of the following is a valid formula?<br> (C2-D16)<br> O A2/F2<br> O=B5/10<br> O =E10
dmitriy555 [2]

Answer:

=B5/10

=E10

Explanation:

The above two formulas are the correct options. Remember for a valid formula in Excel, we must have the = sign first, and this should be followed by a valid formula. Saying that C5 should be E10 by writing =E10 at C5 is a formula certainly. And storing B5/10 at C5, through =B5/10 is also a valid formula. Hence, two formula out of four, and as mentioned above are the correct ones.

4 0
3 years ago
Explain why computer users should use the cloud?
Arlecino [84]

Answer:

Explanation:

Developing in the cloud enables users to get their applications to market quickly. Hardware failures do not result in data loss because of networked backups. Cloud computing uses remote resources, saving organizations the cost of servers and other equipment.

hope it helps.

3 0
3 years ago
When should students practice netiquette in an online course? Check all that apply.
Elden [556K]

Answer:

tienesrazon  

Explanation:

6 0
3 years ago
Write a program that asks the user to enter a number of seconds. There are 60 seconds in a minute. If the number of seconds ente
frez [133]

Answer:

// here is code in c.

#include <stdio.h>

// main function

int main()

{

// variable to store seconds

long long int second;

printf("enter seconds:");

// read the seconds

scanf("%lld",&second);

// if seconds is in between 60 and 3600

if(second>=60&& second<3600)

{

// find the minutes

int min=second/60;

printf("there are %d minutes in %lld seconds.",min,second);

}

// if seconds is in between 3600 and 86400

else if(second>=3600&&second<86400)

{

// find the hours

int hours=second/3600;

printf("there are %d minutes in %lld seconds.",hours,second);

}

// if seconds is greater than 86400

else if(second>86400)

{

// find the days

int days=second/86400;

printf("there are %d minutes in %lld seconds.",days,second);

}

return 0;

}

Explanation:

Read the seconds from user.If the seconds is in between 60 and 3600 then find the minutes by dividing seconds with 60 and print it.If seconds if in between 3600 and 86400 then find the hours by dividing second with 3600 and print it. If the seconds is greater than 86400 then find the days by dividing it with 86400 and print it.

Output:

enter seconds:89

there are 1 minutes in 89 seconds.

enter seconds:890000

there are 10 days in 890000 seconds.

8 0
3 years ago
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