Answer:
4 subnets
Explanation:
According to the specifications provided, it seems that you would need a total of 4 subnets for the entire server structure. This is because each web server needs 1 public subnet and each database needs one private subnet. If there are a total of 2 web servers and 2 database servers then each one will ultimately have 2 subnets meaning a total of 4 subnets. These would be 2 public subnets and 2 private subnets for the different availability zones, which would allow the server structure to maintain high availability.
B. The number of sharks decreases. Explanation: hope it's help i learned about it and you too
Answer:
1. The tools panel
2. The actions panel
Explanation:
The tools panel contains properties that allow for text and object creation. The actions panel contains properties that allow for the alignment and distribution of titles.
Under the tools panel, we can find properties like; line, arc, an arrow for selection, different shapes like rectangle and the clipped corner, vertical type, vertical area type, vertical path type, etc. Under the actions panel, properties like; align, center, and distribute can be found.
The answer is bash
. The bash command opens a Bourne-again sheel (bash) session. It is the standard shell used in most Linux computers and it uses commands similar to a UNIX shell. Bash includes features such as:
1) Command completion when pressing the tab key.
2) Command history.
3) Improved arithmetic functions.
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
