Changing the configuration of a database falls under which category of databaseâ administration?
d. operation
Answer:
It allows the designer to focus on main logic without being distracted by programming languages syntax.
* Since it is language independent, it can be translated to any computer language code.
* It allows designer to express logic in plain natural language.
* It is easier to write actual code using pseudocode.
* Unlike algorithms, pseudocodes are concise so pseudocodes are more readable and easier to modify.
Explanation:
Answer:
Storage Spaces
Explanation:
The feature that would best accommodate his needs would be the Storage Spaces feature that has been implemented in the Windows and Windows Server systems. This feature basically takes various hard drives and combines them together, this ultimately acts as copies of each other which protect the data of each one in the unfortunate case that one of the drives tends to fail. This feature also allows you to add more drives to the existing ones with very minimal effort without restricting configuration. Which would solve the needs that are needed in this scenario.
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;
}
Answer:
A
Explanation:
The value will be the original stored in the int variables. This is because the method swap(int xp, int yp) accepts xp and yp parameters by value (They are passed by value and not by reference). Consider the implementation below:
public class Agbas {
public static void main(String[] args) {
int xp = 2;
int yp = xp;
swap(xp,yp); //will swap and print the same values
System.out.println(xp+", "+yp); // prints the original in values 2,2
int xp = 2;
int yp = 5;
swap(xp,yp); // will swap and print 5,2
System.out.println(xp+", "+yp); // prints the original in values 2,5
}
public static void swap(int xp, int yp){
int temp = xp;
xp = yp;
yp = temp;
System.out.println(xp+", "+yp);
}
}