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
names = ["Kevin", "Joe", "Thor", "Adam", "Zoe"]
names.sort()
for x in names:
if x == "Thor":
break
else:
print(x)
I made up my own names for the sake of testing my code. I wrote my code in python 3.8. I hope this helps.
Answer:
c) 4.4
Explanation:
You can evaluate step by step the code. You know tha your input is 4.0
Step 1 Variables definitions:
double tax;
double total;
Step 2 Ask the user for the input:
System.out.print("Enter the cost of the item");
Step 3 read de input:
total = scan.nextDouble();
Step 4 evaluate the condition ( is the input greater or equal than 3.0 ? [True])
if ( total >= 3.0)
Step 5 get done the operations inside the condition and print it:
tax = 0.10;
System.out.println(total + (total * tax));
What's up!!! :D A Tech-Savvy here:
Answer:
Symmetric Digital Subscriber Line (SDSL)
Cheers,
Feel free to ask me anything regarding ICT/Tech.
Answer:
b. False
Explanation:
If you execute an infinite recursive function on a computer it will NOT execute forever.
