Answer:
Check the explanation
Explanation:
The above question can be sovled in the below step by step way:
15 can be written as 16-1 = 24 -1
Therefore, 15 *13 = (24 -1)*13
= 13*24 - 13
Now, when we left shift a number we multiply it by 2k .Where k is no of times we left shifted it.
Therefore, 13*24 = shift 13 to left 4 times
15 * 13 = shift 13 to left 4 times and subtract 13 from it.
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;
}
The answer is C.54.45.43.54
<span>The schema will have to accommodate to make the person more easily able to perform the new task. Accommodation allows the new information to be made a part of a schema without changing the overall concepts in the schema. The schema itself stays unchanged for the most part, but the new information is more of a "tweak" to the schema than a full-on update.</span>
In a relational database application, a <u>foreign key</u> is used to link one table with another.
<h3>What is a
foreign key?</h3>
In database management system (DBMS), a foreign key can be defined as a column whose value provides a link between the data that are stored in a table or relational database.
This ultimately implies that, a <u>foreign key</u> is used to link one table with another in a relational database application.
Read more on keys here: brainly.com/question/8131854
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