Answer:
if ( name1 > name2) {
first = name1;
} else {
first = name2;
}
Explanation:
First you need take a decision for that reason you need and if - else structure to decide if asign the variable name1 or the variable name2 to the result. Later you can use an > or < to compare two strings because it use the ASCII code to compare wich one is larger than another.
if ( name1 > name2) {
first = name1;
} else {
first = name2;
}
10.......................................
Answer:
local Area Network, Wide Local Area Network and Personal Area Network
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:
> we8thereRatings[1:5,]
Food Service Value Atmosphere Overall
1 5 5 5 5 5
2 5 5 5 5 5
5 5 5 4 4 5
11 5 5 5 5 5
12 5 5 5 4 5
