Answer:
Following are the program in the Java Programming Language.
public class Main // declared class
{
public void hopscotch(int x) // function definition
{
int count=0; // variable declaration
for (int j=0; j<x; j++) // iterating over the loop
{
System.out.println(" " + (++count)); // print the value of count
System.out.println((++count) + " " + (++count));
}
System.out.println(" " + (++count));
}
public static void main(String[] args) // main method
{
Main ob=new Main(); // creating object
ob.hopscotch(3); // calling hopscotch method
}
}
<u>Output</u>:
1
2 3
4
5 6
7
8 9
10
Explanation:
Here, we define a class "Main" and inside the class, we define a function "hopscotch()" and pass an integer type argument in its parentheses and inside the function.
- We set an integer variable "count" and assign value to 0.
- Set the for loop which starts from 0 and end at "x" then, print space and value inside the loop, again print value then space then value, then repeat the 1st print space then value.
- Finally, set the main function "main()" and inside the main function, we create the object of the class "ob" then, call the function " hopscotch()" through the object and pass the value 3 in its parentheses.
I believe the answer is bloody purple poop that is the size of an extra large grilled cheese
Answer:
I'm pretty sure its B.
Explanation:
Input: Force is applied to the pedals by the rider's feet then..
Process: the chain and gear system convert the energy to cause...
Output: the rear wheels to turn and make the bike go foward
Answer:
Check the explanation
Explanation:
We can utilize the above algorithm with a little in modification. If in each of the iteration, we discover a node with no inward edges, then we we’re expected succeed in creating a topological ordering.
If in a number of iteration, it becomes apparent that each of the node has a minimum of one inward edge, then there must be a presence of cycle in the graph.
So our algorithm in finding the cycle is this: continually follow an edge into the node we’re presently at (which is by choosing the first one on the adjacency list of inward edges to decrease the running time).
Since the entire node has an inward edge, we can do this continually or constantly until we revisit a node v for the first time.
The set of nodes that we will come across among these two successive visits is a cycle (which is traversed in the reverse direction).
