Which disk tool can be run on a magnetic hard drive to identify any critical errors on the drive?

A(n) _____ is a fake online persona created to promote a particular point of view, often in praise of a firm, product, or indivi

drek231 [11]

Answer:

Sock puppet

Explanation:

A sock puppet, a reference to a puppet created by placing a sock over one's hand, is a misleading or fake online identity created to promote a particular point of view, often in praise of a firm, organization, product, or individual; to manipulate public opinion or to argue, bully or review products. Regardless of the purpose it was created for, sock puppets are rarely welcome in online communities and forums.

4 0

3 years ago

Write essay about pokhara

myrzilka [38]

Explanation:

Pokhara is the second most visited city in Nepal, as well as one of the most popular tourist destinations. It is famous for its tranquil atmosphere and the beautiful surrounding countryside. Pokhara lies on the shores of the Phewa Lake. From Pokhara you can see three out of the ten highest mountains in the world (Dhaulagiri, Annapurna, Manasalu). The Machhapuchre (“Fishtail”) has become the icon of the city, thanks to its pointed peak.

As the base for trekkers who make the popular Annapurna Circuit, Pokhara offers many sightseeing opportunities. Lakeside is the area of the town located on the shores of Phewa Lake, and it is full of shops and restaurants. On the south-end of town, visitors can catch a boat to the historic Barahi temple, which is located on an island in the Phewa Lake.

On a hill overlooking the southern end of Phewa Lake, you will find the World Peace Stupa. From here, you can see a spectacular view of the lake, of Pokhara and Annapurna Himalayas. Sarangkot is a hill on the southern side of the lake which offers a wonderful dawn panorama, with green valleys framed by the snow-capped Annapurnas. The Seti Gandaki River has created spectacular gorges in and around the city. At places it is only a few meters wide and runs so far below that it is not visible from the top

5 0

3 years ago

You received a call form a user who brought her own device to the office but cannot find or connect to the company WLAN. The lap

lapo4ka [179]

Answer:

Company WAP has SSID broadcast disable

Explanation:

Disabling SSID broadcast will make your WLAN network name invisible to other users. However, this only hides the name, not the network itself. that's why the user cannot find the company WLAN on her computer but yet can connected to the company WLAN and can browse to multiple websites with no problem from their company-assigned laptops.

8 0

3 years ago

How can solve a Boolean algebra problem? Ex: JK+(~J+~K)L+JK

kicyunya [14]

I am not sure what you mean by that but a boolean is either true or false. So it is kind of like a true or false question. 4+4=7 is false. 2*4!=48 is true.

8 0

2 years ago

a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv

Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

// (if you want better performance: use array instead of unordered_map)

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]

bool helper(vector<string>& words, string& result, int digit, int widx, int sum) {

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

// if summation at digit position complete, validate it with result[digit].

if (widx == words.size()) {

if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {

if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

bool tmp = helper(words, result, digit+1, 0, sum/10);

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

} else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

return helper(words, result, digit+1, 0, sum/10);

} else {

return false;

}

// if word[widx] length less than digit, ignore and go to next word

if (digit >= words[widx].length()) {

return helper(words, result, digit, widx+1, sum);

}

// if word[widx][digit] already mapped to a value

if (c2i.count(words[widx][digit])) {

if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0)

return false;

return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);

}

// if word[widx][digit] not mapped to a value yet

for (int i = 0; i < 10; i++) {

if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

bool tmp = helper(words, result, digit, widx+1, sum+i);

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

void isSolvable(vector<string>& words, string result) {

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

vector<string> words={"GREEN" , "BLUE"} ;

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0

1 year ago