Answer:
// here is code in c++ to find the approx value of "e".
#include <bits/stdc++.h>
using namespace std;
// function to find factorial of a number
double fact(int n){
double f =1.0;
// if n=0 then return 1
if(n==0)
return 1;
for(int a=1;a<=n;++a)
f = f *a;
// return the factorial of number
return f;
}
// driver function
int main()
{
// variable
int n;
double sum=0;
cout<<"enter n:";
// read the value of n
cin>>n;
// Calculate the sum of the series
for (int x = 0; x <= n; x++)
{
sum += 1.0/fact(x);
}
// print the approx value of "e"
cout<<"Approx Value of e is: "<<sum<<endl;
return 0;
}
Explanation:
Read the value of "n" from user. Declare and initialize variable "sum" to store the sum of series.Create a function to Calculate the factorial of a given number. Calculate the sum of all the term of the series 1+1/1!+1/2!.....+1/n!.This will be the approx value of "e".
Output:
enter n:12
Approx Value of e is: 2.71828
The code that remove duplicate is as follows:
def remove_duplicate(mylist):
mylist = list(dict.fromkeys(mylist))
return mylist
print(remove_duplicate([1, 1, 2, 3, 3, 5, 6, 7]))
<h3>Code explanation</h3>
The code is written in python.
- we defined a function named "remove_duplicate" and it accept the parameter "mylist".
- The variable "mylist" is used to store the new value after the duplicate vallues has been removed.
- Then, wed returned mylist.
- Finally, we call the function with the print statement . The function takes the required parameter.
learn more on python here: brainly.com/question/21126936
