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Ilia_Sergeevich [38]
3 years ago
13

Which statement is NOT a rule for the data source for the mail merge?

Computers and Technology
2 answers:
ad-work [718]3 years ago
7 0

Answer:

Field names should start with a letter.

Explanation:

Natali [406]3 years ago
4 0

Answer:

Field names should start with a letter  is not a rule for data source for a mail merge.

Explanation:

The best Microsoft tool to use for a data source in a Microsoft mail merge is Excel spreadsheet, a data source for a mail merge can also be a database or a table .

Mail merge is the merging/combining of mail and letters or labels meant for mailing in mass from a form letter. this simply means that Microsoft word documents can take in content from databases, spreadsheet.

A data source is used to define the connection between a data from its source to its recipient. it is the name used when querying a data .

every other options given is a rule because it makes the field name unique and easy to query the data source as well

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What process combines data from a list with the content of a document to provide personalized documents?
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Write a function analyze_text that receives a string as input. Your function should count the number of alphabetic characters (a
laiz [17]

Answer:

def analyze_text(sentence):

   count = 0

   e_count = 0

   for s in sentence:

       s = s.lower()

       if s.isalpha():

           count += 1

       if s == "e":

           e_count += 1

   return "The text contains " + str(count) + " alphabetic characters, of which " + str(e_count) + " (" + str(e_count*100/count) + "%) are ‘e’."

Explanation:

Create a function called analyze_text takes a string, sentence

Initialize the count and e_count variables as 0

Create a for loop that iterates through the sentence

Inside the loop, convert all letters to lowercase using lower() function. Check each character. If a character is a letter, increment the count by 1. If a character is "e", increment the e_count by 1.

Return the count and e_count in required format

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When an employee is told that his job will be outsourced within a year, he knows that his job will become unnecessary
Anvisha [2.4K]
False!! Hope this helps
8 0
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Integers and booleans. Write a program RightTriangle that takes three int command-line arguments and determines whether they con
icang [17]

Answer:

<em>The programming language is not stated;</em>

<em>I'll answer using C++</em>

#include<iostream>

#include<cmath>

using namespace std;

int main()

{

int side1, side2, side3;

cout<<"Enter the three sides of the triangle: "<<endl;

cin>>side1>>side2>>side3;

if(side1<=0 || side2 <= 0 || side3 <= 0) {

 cout<<"Invalid Inputs";

}

else {

 if(abs(pow(side1,2) - (pow(side2,2) + pow(side3, 2)))<0.001) {

  cout<<"Right Angled";

 }

 else if(abs(pow(side2,2) - (pow(side1,2) + pow(side3, 2)))<0.001) {

  cout<<"Right Angled";

 }

 else if(abs(pow(side3,2) - (pow(side2,2) + pow(side1, 2)))<0.001) {

  cout<<"Right Angled";

 }

 else {

  cout<<"Not Right Angled";

 }

}

return 0;

}

Explanation:

The following line declares the three variables

int side1, side2, side3;

The next line prompts user for input of the three sides

cout<<"Enter the three sides of the triangle: "<<endl;

The next line gets user input

cin>>side1>>side2>>side3;

The following if condition checks if any of user input is negative or 0

<em> if(side1<=0 || side2 <= 0 || side3 <= 0) { </em>

<em>  cout<<"Invalid Inputs"; </em>

<em> } </em>

If otherwise

else {

The following if condition assumes that side1 is the largest and test using Pythagoras Theorem

<em>if(abs(pow(side1,2) - (pow(side2,2) + pow(side3, 2)))<0.001) { </em>

<em>   cout<<"Right Angled"; </em>

<em>  } </em>

The following if condition assumes that side2 is the largest and test using Pythagoras Theorem

<em>  else if(abs(pow(side2,2) - (pow(side1,2) + pow(side3, 2)))<0.001) { </em>

<em>   cout<<"Right Angled"; </em>

<em>  } </em>

The following if condition assumes that side3 is the largest and test using Pythagoras Theorem

<em>  else if(abs(pow(side3,2) - (pow(side2,2) + pow(side1, 2)))<0.001) { </em>

<em>   cout<<"Right Angled"; </em>

<em>  } </em>

If none of the above conditions is true, then the triangle is not a right angles triangle

<em>  else { </em>

<em>   cout<<"Not Right Angled"; </em>

<em>  } </em>

}

return 0;

Download cpp
4 0
3 years ago
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