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WITCHER [35]
3 years ago
12

The​ ___________________ is the most commonly used methodology for building information systems.

Computers and Technology
1 answer:
spin [16.1K]3 years ago
5 0
The answer to this question is "traditional system life cycles". The traditional system life cycles are the most commonly and oldest used methodology for building information system. This methodology is a little bit costly.Other methods used are outsourcing and prototyping.
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Explanation:

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¿Qué diferencia existe entre un virus biológico y virus informático?
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3 years ago
Your ipad has an application that will not stop running. what feature/tool can you use to stop it? kill treason task manager for
saveliy_v [14]

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Read 2 more answers
In which of the following locations can you edit all of the Properties of a PowerPoint file?
Lapatulllka [165]
Access the File<span> menu, choose </span>Info Pane<span> to get to </span>Backstage view, you can see Properties on t<span>he area on the right side </span>of the current PowerPoint presentation.  <span>Within the </span>Properties<span> pane click the </span>Show All Properties<span>  option , T</span><span>his will displays properties such as </span>Size<span>, the number of </span>Slides<span>,  </span>Hidden Slides<span>, the number of </span>Multimedia Clips, etc.  Some of the entries are editable w<span>ithin the  </span>Properties pane, and some are not. Just move your mouse cursor over any detail of a property. The editable sections will change the cursor into edit mode.  
6 0
3 years ago
Write an algorithm that prints out all the subsets of 3 elements of a set of n elements. The elements of the set are stored in a
Ksenya-84 [330]

void Print_3_Element_Subsets(int n,const element_type S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

print("No subset found");

}

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

for(int k =j+1; k < n ; k++)

{

print(S[i],S[j],S[k]);

}

}

}

}

C++ Implementation;

#include<cstdlib>

#include<time.h>

using namespace std;

int S1[5]= {1,2,3,4,5};

int n1 = 5;

int S2[2]={1,2};

int n2 = 2;

void Print_3_Element_Subsets(int n,const int S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

printf("No subset found\n");

}

printf("3 Subsets are\n");

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

for(int k =j+1; k < n ; k++)

{

cout<<S[i]<<" "<<S[j]<<" "<<S[k]<<endl;

}

}

}

}

int main()

{

cout<<"Case 1"<<endl;

Print_3_Element_Subsets(n1,S1);

cout<<endl<<"Case 2"<<endl;

Print_3_Element_Subsets(n2,S2);

}

OUTPUT

Case 1

3 Subsets are

1 2 3

1 2 4

1 2 5

1 3 4

1 3 5

1 4 5

2 3 4

2 3 5

2 4 5

3 4 5

Case 2

No subset found

3 Subsets are

--------------------------------

Process exited after 0.0137 seconds with return value 0

Press any key to continue . . .

7 0
3 years ago
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