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zloy xaker [14]
2 years ago
8

Your computer has gradually slowed down. What's the most likely reason?

Computers and Technology
1 answer:
sweet-ann [11.9K]2 years ago
8 0
It could be old ... could be your connection to the internet... could be a virus. You can go to walmart or your super store and buy a usb for your computer. There are ones that can speed up your computer. I had to buy it for mine and it worked. It might be over 20 dollars depends on where you go.
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Mrs. Johnson is here today to receive an intercostal nerve block to mitigate the debilitating pain of her malignancy. Her cancer
Furkat [3]

Answer:

The correct answer is:

a. M54.6, C79.51, C80.1

Explanation:

-  M54.6 Pain in thoracic spine. It is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. The 2020 edition of ICD-10-CM M54.

- C79.51: Secondary malignant neoplasm of bone, it is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes.

- G89. 3 is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. The 2020 edition of ICD-10-CM G89.

Malignant neoplasm of anus, unspecified

Neoplasm related pain (acute) (chronic)

Pain in thoracic spine. M54. 6 is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. The 2020 edition of ICD-10-CM M54.

Malignant (primary) neoplasm, unspecified

- C80. 1 is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. The 2020 edition of ICD-10-CM C80.

5 0
2 years ago
The major result of treating 1-butyne with 6M aqueous NaOH would be:_______.A. the production of an alkene.B. the production of
andriy [413]

Answer:

D. nothing, as the alkyne would not react to an appreciable extent.

Explanation:

Nothing, as the alkyne would not react to an appreciable extent.

6 0
1 year ago
c++ You are given an array A representing heights of students. All the students are asked to stand in rows. The students arrive
Lilit [14]

The below code will help you to solve the given problem and you can execute and cross verify with sample input and output.

#include<stdio.h>

#include<string.h>

 int* uniqueValue(int input1,int input2[])

 {

   int left, current;

   static int arr[4] = {0};

   int i      = 0;

     for(i=0;i<input1;i++)

      {

         current = input2[i];

         left    = 0;

         if(current > 0)

         left    = arr[(current-1)];

      if(left == 0 && arr[current] == 0)

       {

       arr[current] = input1-current;

       }

       else

   {

       for(int j=(i+1);j<input1;j++)

       {

           if(arr[j] == 0)

           {

               left = arr[(j-1)];

               arr[j] = left - 1;

           }

       }

   }

}

return arr;

}

4 0
2 years ago
Allison wants to use equations to simplify the process of explaining something to the Sales team, but the default eq
yarga [219]

Answer:

Search for something that fits her needs

Explanation:

There are several approaches to problem solving, the approach that will be employed depends on our definition of the problem. Not all processes can be modeled by mathematical equations, some processes are simply theoretical and can be explained by simple logic.

If Allison does not find the needed equations among the available default equations, she can come up with an equation that addresses the process at hand. If she cannot come up with any, she can use other approaches that fit her needs to explain the required concept to the team.

7 0
2 years ago
Read 2 more answers
The groups_per_user function receives a dictionary, which contains group names with the list of users. Users can belong to multi
Akimi4 [234]

The groups_per_user function receives a dictionary, which contains group names with the list of users.

Explanation:

The blanks to return a dictionary with the users as keys and a list of their groups as values is shown below :

def groups_per_user(group_dictionary):

   user_groups = {}

   # Go through group_dictionary

   for group,users in group_dictionary.items():

       # Now go through the users in the group

       for user in users:

       # Now add the group to the the list of

         # groups for this user, creating the entry

         # in the dictionary if necessary

         user_groups[user] = user_groups.get(user,[]) + [group]

   return(user_groups)

print(groups_per_user({"local": ["admin", "userA"],

       "public":  ["admin", "userB"],

       "administrator": ["admin"] }))

3 0
2 years ago
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