Answer:no thank you ima goody
Explanation:
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
Answer:
B is the answer.. pls mark me as brainliest
ctrl+shift+down arrow
Explanation:
Answer:
The answer to this question is given below in the explanation section.
Explanation:
The for-loop given in the question is:
for ( j = 0; j < 10; j++ )
{
appendItem (myList, aNumber); //this loop append a number to a list myList
}
This loop starts from J variable's value zero and when J's value is less than 10, the loop iterate through its body until J's value becomes greater or equal to 10. As J's value exceed nine, the loop will get terminated.
So this loop repeats 10 times its loop body, at the 11th time, the condition becomes false and the loop will get terminated.
