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insens350 [35]
2 years ago
13

Which of the following internet protocols is used to request and send pages and files on the World Wide Web

Computers and Technology
1 answer:
Marianna [84]2 years ago
7 0

it is . org Explanation:

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In this lesson, you learned how to create reports and how to display them in a Web application. You will use the same chinook da
Nimfa-mama [501]

create a report that displays back tge total sales by country

6 0
2 years ago
What type of firewall works on the Session layer that creates a connection and allows packets to flow between the two hosts with
valkas [14]

Answer:

The correct answer is "circuit-level gateway"

Explanation:

A circuit-level gateway is a type of firewall that works at the session layer between the transport and application layer.

8 0
3 years ago
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;

}

7 0
3 years ago
Which of the following is not a web browser? A. safari B. explorer C. chrome D. google​
Brums [2.3K]

Answer:

D) Google

Explanation:

Google on it's own is a search engine. Safari, Explorer, and Chrome are web browsers.

I hope it helps :)))

7 0
3 years ago
Read 2 more answers
Int[] myArray1= new int[24]; int[] myArray2= new int[10]; float[] myArray3= new float[8]; int index1, index =0 for(int index1 =0
anastassius [24]

Answer:

The value of myArray2[index2] when index1 = 12 is 30

Explanation:

In the source code, the formula for myArray2[index2] is;

   myArray2[index2] = index2 + index3 + myArray1[index1],

   myArray1[index1] = index1 * 2,

   index2 = index % 10 (equal to the remainder) and

   index3  = index % 8

When index1 increases to 12 in the for-loop statement, the "myArray1[index1]" is equal to 24, index2 is equal to 2 and index3 is 4. The total sum is equal to 30 and assigned to "myArray2[index2]".

3 0
3 years ago
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