Many filesystems attempt to logically group inodes together in an attempt to keep data close together on the disk. Consider a file system configuration of 1KB size blocks, 16384 blocks/group, and 4096 inodes/group, with each inode sized 128 bytes.
Answer:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
double side, height;
cout<<"Enter the length of one of the sides of the base: ";
cin>>side;
cout<<"Enter the height: ";
cin>>height;
double area = side * side;
double volume = area * height / 3;
cout<<"The area of the base of the pyramid is: "<<area<<endl;
cout<<"The height of the pyramid is: "<<height<<endl;
cout<<"The volume of the pyramid is: "<<fixed<<setprecision(2)<<volume<<endl;
return 0;
}
<u>Pseudocode:</u>
Declare side, height
Get side
Get height
Set area = side * side
Set volume = area * height / 3
Print area
Print height
Print volume
Explanation:
Include <iomanip> to have two decimal places
Declare the side and height
Ask the user to enter side and height
Calculate the base area, multiply side by side
Calculate the volume using the given formula
Print area, height and volume (use <em>fixed</em> and <em>setprecision(2)</em> to have two decimal places)
Answer:
x= -2
Explanation:
3=2x+7
subtract the 7 on both sides of the equal sign
-4 =2x
divide by 2 on both sides of the equal sign
-2=x
B. Generally backdoors are logins/access to something without permission and without admins knowing.
Answer:
Explanation:
Since the array is not provided, I created a Python function that takes in the array and loops through it counting all of the words that are longer than 5. Then it returns the variable longer_than_five. To test this function I created an array of words based on the synapse of Pride and Prejudice. The output can be seen in the attached picture below.
def countWords(p_and_p_words):
longer_than_five = 0
for word in p_and_p_words:
if len(word) > 5:
longer_than_five += 1
return longer_than_five