It will take 8 times more time to sort a 40-element list compared to the time spent on a 5-element list.
We can arrive at this answer as follows:
- We can see that if we divide a group of 40 elements into groups containing 5 elements, we will have 8 groups.
- In this case, the time it would take to sort the list of one group of 5 elements would be repeated 8 times so that we could sort all the groups and their elements.
Another way to do this is to divide the number 40 by 5. We would have the number 8 as a result, which indicates that we would need 8 times more time to sort a list of 40 elements, compared to a list of 5 elements.
You can get more information about lists at this link:
brainly.com/question/4757050
Answer:
Direct Selection Tool
Explanation:
The Direct Selection tool is a tool that allows the selection of a single object or a single path such that an object already grouped with other objects can be directly and moved to a desired location
The Direct Selection tool can be used to select a container's content including graphics which are imported and specific points or paths of a figure or text to allow for drawing, text editing or to edit paths.
Kinda of both. The processor, memory, hard drive and displays are all standard components and are provided by a variety computer competent manufacturers (except for the processors which are all supplied by Intel). Yet - while many components are standard - NO, the core, hardware components, like logic boards (motherboard), video cards, and other specialty components (some display connectors and displays, for example) are propriety Apple designs.
Answer:
#include <iostream>
using namespace std;
int main()
{
int sum=0;//taking an integer variable to store the sum with initial value 0..
for(int i=1;i<=10;i++)//looping from 1 to 10..
{
sum+=i;//adding each i in the sum..
}
cout<<sum<<endl;//printing the sum..
return 0;
}
Explanation:
output :- 55
1.I have taken an integer variable sum which is equal to 0.
2.Looping over first ten positive integers.
3.Adding every number to sum.
4.Printing the sum.