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Tamiku [17]
2 years ago
15

One should take to prevent ulcers in the gastrointestinal tract?​

Computers and Technology
1 answer:
Andrew [12]2 years ago
7 0

Answer:

food??

Explanation:

if you don't eat food, the acid produced in the stomach will break down the stomach walls instead of the food, giving you ulcers

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says the answer in the question

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System Architecture: Describe the system architecture. Specifically, be sure to address the corporate organization and culture,
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Answer:

Explanation:

In information technology, architecture plays a major role in the aspects of business modernization, IT transformation, software development, as well as other major initiatives within the enterprise. IT architecture is used to implement an efficient, flexible, and high quality technology solution for a business problem, and is classified into three different categories: enterprise architecture, solution architecture and system architecture. Each of these classifications varies in their implementation and design, depending on the contextual business scope, organization structure, and corporate culture.

Architecture Level

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Before making decision regarding system architecture the designer must consider the following points:

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Compare Fibonacci (recursion vs. bottom up)
ipn [44]

Answer:

C++ code explained below

Explanation:

#include<bits/stdc++.h>

#include <iostream>

using namespace std;

int FiboNR(int n)

{

int max=n+1;

int F[max];

F[0]=0;F[1]=1;

for(int i=2;i<=n;i++)

{

F[i]=F[i-1]+F[i-2];

}

return (F[n]);

}

int FiboR(int n)

{

if(n==0||n==1)

return n;

else

return (FiboR(n-1)+FiboR(n-2));

}

int main()

{

long long int i,f;

double t1,t2;

int n[]={1,5,10,15,20,25,30,35,40,45,50,55,60,65,70,75};

cout<<"Fibonacci time analysis ( recursive vs. non-recursive "<<endl;

cout<<"Integer FiboR(seconds) FiboNR(seconds) Fibo-value"<<endl;

for(i=0;i<16;i++)

{

clock_t begin = clock();

f=FiboR(n[i]);

clock_t end = clock();

t1=double(end-begin); // elapsed time in milli secons

begin = clock();

f=FiboNR(n[i]);

end = clock();

t2=double(end-begin);

cout<<n[i]<<" "<<t1*1.0/CLOCKS_PER_SEC <<" "<<t2*1.0/CLOCKS_PER_SEC <<" "<<f<<endl; //elapsed time in seconds

}

return 0;

}

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2 years ago
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