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ivolga24 [154]
3 years ago
6

Tips for staying safe using social media. Using in your own words.

Computers and Technology
1 answer:
OLga [1]3 years ago
7 0

Whatever you post on social media, it's now public. Yes there are ways to make posts private and such, but glitches do happen. Not only that, but hackers may be able to turn privacy features off. Also, sometimes the social media company changes policy on some aspects, which may turn some private features into public ones. That means you should be careful not to post anything too personal and also don't post anything such as addresses, bank account numbers, passwords, etc. This is probably common sense to you, but it's a good thing to keep in mind regardless.

Also, if there are people who verbally threaten or harass you, then you should either report them or block them. Preferably both, but blocking is probably more effective. If someone friend requests you, then it's ideal to only accept their request if you know who they are. Though you can relax this if you aren't worried about who sees your profile page.

In addition, you should be careful as to what you read on social media. Not everything is true. This is because practically anyone can post a news story even if it's completely made up. Lately there have been some efforts to do fact checking, but there's still a lot of fictional stuff out there. Be sure to avoid clicking on any links you may not recognized as they could be phishing links, or may lead to harmful malware. It's probably just best to get an antivirus of some kind.

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Overnight Delivery Service delivers a package to Pam. At the request of Overnight's delivery person, to acknowledge receipt Pam
SSSSS [86.1K]

Answer:

Option b (a digitized handwritten signature) would be the right option.

Explanation:

  • Another photograph of such a handwritten signature was used to digitally sign transcripts that would be perceived to have become a "digitized signature."
  • Those same kinds of signature verification may take a glance official, but they don't protect against widespread fraud, a vital component of every other internet signature.

The latter available options weren’t connected to the type of situation in question. So the response above would be the correct one.

3 0
3 years ago
Danny is editing the text content of a single page on a website he created a year ago. What part of the tag would he have to upd
Verdich [7]
<h2>"<lastmod>" is the right choice for the given scenario</h2>

Explanation:

<update> This tag does not belong to <URL> tag of the site map. Hence this option goes invalid.

<lastmod> This gives details of the date and time when the website content was changed. It will be in "W3C Date time format". This may omit time at times. This is the write answer.

<loc> This contains the URL of the web page.

<changefreq>: How frequently the "web page is likely to get modified". The  values which it accepts are: "monthly, always , daily , weekly ,hourly, yearly , never"

5 0
3 years ago
Given an array A of size N, and a number K. Task is to find out if it is possible to partition the array A into K contiguous sub
Dennis_Churaev [7]

Answer: First lets solve the Prerequisite part

Lets say we have an input array of N numbers {3,2,5,0,5}. We have to  find number of ways to divide this array into 3 contiguous parts having equal sum. So the output for the above input array will be 2 as there are 2 ways to divide the array. One is (3,2),(5),(0,5) and the other is (3,2),(5,0),(5).

Following are the steps to achieve the above outcome.

  • Let p and q point to the index of array such that sum of array elements from 0 to p-1 is equal to sum of array elements from p to q which is equal to the sum of array elements from q+1 to N-1.  
  • If we see the array we can tell that the sum of 3 contiguous parts is 5. So the condition would be that sum of all array elements should be equal to 5 or sum of each contiguous part is equal to sum of all array elements divided by 5.
  • Now create 2 arrays prefix and postfix of size of input array. Index p of prefix array carries sum of input array elements from index 0 to index p. Index q of postfix array carries sum of input array elements from index p to index N-1
  • Next move through prefix array suppose at the index p of prefix array : value of prefix array == (sum of all input array elements)/5.
  • Search the postfix array for p index found above. Search it starting from p+2 index. Increment the count variable by 1 when the value of postfix array =(sum of all input array elements)/5 and push that index of postfix array into a new array. Use searching algorithm on new array to calculate number of values in postfix array.

Now lets solve the main task

We have an array A of size N and a number K. where A[]= {1,6,3,4,7} N=5 and K=3. We have to find if its possible to partition A into 3 contiguous subarrays such that sum of elements in each subarray is the same. It is possible in this example. Here we have 3 partitions (1,6),(3,4),(7) and sum of each of subarrays is same (1+6) (3+4) (7) which is 7.

Following are the steps to achieve the above outcome.

  • In order create K contiguous subarrays where each subarray has equal sum, first check the condition that sum of all elements in the given array should be divisible by K. Lets name another array as arrsum that will be the size of array A. Traverse A from first to  last index and keep adding current element of A with previous value in arrsum. Example A contains (1,6,3,4,7} and arrsum has {1,7,10,14,21}
  • If the above condition holds, now check the condition that each subarray or partition has equal sum. Suppose we represent sum1 to sum of all element in given array and sum2 of sum of each partition then: sum2 = sum2 / K.
  • Compare arrsum to subarray, begining from index 0 and when it becomes equal to sum2 this means that end of one subarray is reached. Lets say index q is pointing to that subarray.
  • Now from q+1 index find p index in which following condition holds: (arrsum[p] - arrsum[q])=sum2
  • Continue the above step untill K contigous subarrays are found. This loop will break if, at some index, sum2 of any subarray gets greater than required sum2 (as we know that every contiguous subarray should contain equal sum).

A easier function Partition for this task:

int Partition(int A[], int N, int k) // A arra y of size N and number k

{      int sum = 0;    int count = 0;  //variables initialization    

   for(int j = 0; j < N; j++)  //Loop that calculates sum of A

  sum = sum + A[j];        

  if(sum % k != 0) //checks condition that sum of all elements of A should be //divisible by k

   return 0;        

   sum = sum / k;  

   int sum2 = 0;  //represents sum of subarray

  for(int j = 0; j < N; j++) // Loop on subarrays

  {      sum2=sum2 + A[j];  

   if(sum2 == sum)    { //these lines locates subarrays and sum of elements //of subarrays should be equal

       sum2 = 0;  

       count++;  }  }  

/*calculate count of subarrays whose

sum is equal to (sum of complete array/ k.)

if count == k print Yes else print No*/

if(count == k)    

return 1;  

   else

   return 0;  }

6 0
3 years ago
Instructions: Use the tables below to answer the questions that follow:
Svet_ta [14]

Answer:

same as above....................

3 0
3 years ago
WILL GIVE BRAINLIEST PLEASE HELP / urgent
Alex Ar [27]
Web design is awesome! Alright, so -

If you want to call some attention to text, you need to focus on the basic essentials. 

You want your text to be brief and split up. If someone goes on your site and see's walls of text, they'll be overwhelmed and leave.

So, to call attention - make it brief, and split it up into nice paragraphs.

Another way to call attention to text is to have a <em>really </em>good colour scheme. Having text easy on the eyes attracts the reader more, and encourages them to dive deeper.

If your text is unattractive and hard to look at it, it'll certainly get their attention - but not the attention you want.

Finally, another way to call attention is with visual adjustments such as making text bold, making it <em>italicised, </em>making it ALL CAPS, <em>or just GOING CRAZY WITH UNNECESSARY TEXT ATTENTION ATTRACTING POWER!!!!!!!

</em>Ahem... Anyways, these are just a few ways to get their attention. =) If you need any other help, private message me because I love web development! =)<em>
</em>
4 0
3 years ago
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