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Dovator [93]
2 years ago
10

The radial gradient option creates a gradient that shades from the starting point to the end point in a _________ pattern.

Computers and Technology
1 answer:
rusak2 [61]2 years ago
7 0

Answer:

The correct answer is <u>Circular</u>

I hope this helps! ^-^

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In this question, you will experimentally verify the sensitivity of using a precise Pi to the accuracy of computing area. You ne
zhannawk [14.2K]

Answer:

Follows are the code to this question:

import math as x #import math package

#option a

radius = 10#defining radius variable  

print("radius = ", radius)#print radius value

realA = x.pi * radius * radius#calculate the area in realA variable

print("\nrealA = ", realA)#print realA value

#option b

a1 = 3.1  * radius * radius#calculate first area in a1 variable  

print("Area 1= ", a1)#print Area

print("Percentage difference= ", ((realA - a1)/realA) * 100) #print difference  

a2 = 3.14  * radius * radius#calculate first area in a2 variable                            

print("Area 2= ", a2)#print Area

print("Percentage difference= ", ((realA - a2)/realA) * 100)#print difference  

a3 = 3.141  * radius * radius#calculate first area in a2 variable                       print("Area 3= ", a3)#print Area

print("Percentage difference= ", ((realA - a3)/realA) * 100) #print difference  

Output:

please find the attached file.

Explanation:

In the given Python code, firstly we import the math package after importing the package a "radius" variable is defined, that holds a value 10, in the next step, a "realA" variable is defined that calculate the area value.

In the next step, the "a1, a2, and a3" variable is used, which holds three values, that is "3.1, 3.14, and 3.141", and use the print method to print its percentage difference value.  

4 0
3 years ago
What is the characteristics of a spear phishing message
Cerrena [4.2K]
It is targeted at one person specifically. More work for the attacker, but higher chances of obtaining sensitive information.
7 0
3 years ago
Read 2 more answers
Write a removeDuplicates() method for the LinkedList class we saw in lecture. The method will remove all duplicate elements from
oksano4ka [1.4K]

Answer:

removeDuplicates() function:-

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

     

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       

       //make ptr1 equals to head.

       ptr1 = head;

        //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

               

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

Explanation:

Complete Code:-

//Create Linked List Class.

class LinkedList {

       //Create head pointer.

       static ListNode head;

       //define structure of ListNode.

       //it has int val(data) and pointer to ListNode i.e, next.

   static class ListNode {

       int val;

       ListNode next;

       //constructor to  create and initialize a node.

       ListNode(int d) {

               val = d;

           next = null;

       }

   }

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

       

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       

       //make ptr1 equals to head.

       ptr1 = head;

       

       

       //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

               

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

   //display() function prints linked list.

   void display(ListNode node)

   {

       //run while loop till last node.

       while (node != null)

       {

               //print node value of current node.

           System.out.print(node.val + " ");

           

           //move node pointer by one node.

           node = node.next;

       }

   }

   public static void main(String[] args) {

       

       //Create object of Linked List class.

       LinkedList list = new LinkedList();

       

       //first we create nodes and connect them to form a linked list.

       //Create Linked List 1-> 2-> 3-> 2-> 4-> 2-> 5-> 2.

       

       //Create a Node having node data = 1 and assign head pointer to it.

       //As head is listNode of static type. so, we call head pointer using class Name instead of object name.

       LinkedList.head = new ListNode(1);

       

       //Create a Node having node data = 2 and assign head.next to it.

       LinkedList.head.next = new ListNode(2);

       LinkedList.head.next.next = new ListNode(3);

       LinkedList.head.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next = new ListNode(4);

       LinkedList.head.next.next.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next.next.next = new ListNode(5);

       LinkedList.head.next.next.next.next.next.next.next = new ListNode(2);

       //display linked list before Removing duplicates.

       System.out.println("Linked List before removing duplicates : ");

       list.display(head);

       //call removeDuplicates() function to remove duplicates from linked list.

       list.removeDuplicates();

       System.out.println("")

       //display linked list after Removing duplicates.

       System.out.println("Linked List after removing duplicates :  ");

       list.display(head);

   }

}

Output:-

6 0
3 years ago
(within 200 words) analyzing the importance of information and communication technology in personal life.
Volgvan

Explanation:

Information technology is inescapable in modern day life, turn on the coffee pot and there are tiny microprocessor inside. Start your automobile and every aspect of operation is controlled by dozens of computer control modules. The current mobile phone amazing power and speed for the user. Technology is transforming every aspect of our life. Nowever is this more apparent than our place of employment. Developing of it makes our family to live happy because it helps the members of family to spend their most time with their family. It helps us to communicate the family member who live far from their family. It is also used in house hold work. It's saves our time to do more better and better

6 0
4 years ago
You have received a Word memo from your supervisor outlining the five-year strategic plan for your company, which operates in a
Maksim231197 [3]

Answer:

The correct option to the following question is C.) Rights Management Services.

Explanation:

The Right Management Services is also referred to as RMS or AD(Active Directory) RMS.

It is the MS Windows security tool which provides us with persistent data protections by enforcing our data access policy.

Our documents are protected with the RMS and the AD RMS, it is an application by which the documents are associated with and be RMS-aware.

It actually called the Windows RMS but its name was changed in MS Windows 2008, RMS to AD RMS.

7 0
4 years ago
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