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1 year ago

What should you do if a headset plugged into your computer is not working properly?

2 answers:

8 0

If a headset plugged into computer is not working properly than one can go for updating the device driver. The correct option is B.

<h3>What is a device driver?</h3>

A device driver is a type of software application that allows one hardware device (such as a computer) to communicate with another hardware device (such as a printer). A device driver is indeed referred to as a software driver.

A driver, also known as a device driver, is a collection of files that instructs a piece of hardware on how to operate by communicating with a computer's operating system.

Every piece of hardware, from internal computer components like your graphics card to external peripherals like a printer, requires a driver.

If a headset plugged into a computer is not working properly, the device driver can be updated.

Thus, the correct option is B.

For more details regarding device driver, visit:

brainly.com/question/14054807

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jeka941 year ago

7 0

Answer:

You can unplug the headset and then if it's still nit working you need to replace the computer's power cord

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Which statement describes the relationship between science and technology?

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Answer: science is the study of the world, and technology changes the world to solve problems.

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3 years ago

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Nunya is a computer software company that employs highly intelligent, but somewhat unusual people. Every Friday, free lollipops,

mario62 [17]

Answer:

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3 years ago

match the type of secondary storage devices to their examples. CD, hard disk, flash memory. answer choices: magnetic storage dev

alisha [4.7K]

Hi pupil here's your answer ::

➡➡➡➡➡➡➡➡➡➡➡➡➡

Column A Column B

1》CD = Optical Storage Devices

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3 0

3 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

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3 0

2 years ago

100 POINTS FOR ANYONE WHO CAN DO THIS!

Nataly [62]

Let's check what can be modified

Before calling def we need adjective and conjunctions stored inside variables

Store them(You may change according to your choice)

$\tt adjectives=("foolish","bad","long","hard")$

$\tt conjunctions=("and","but","for","after")$

We have to make optional ,easy way ask the user to do instead of yourself .

$\tt con=input("Enter\: yes\: if \:you \:want \:to \:use \:conjunctions:")$

$\tt adj=input("Enter\:yes\:if\:you\:want\:to\:use\: adjectives:")$

If they click then we can proceed else no problem let the program run

$\tt def\: conjunctionPhrase():$

$\quad\tt if\: con=="yes":$

$\quad\quad\tt return\:random.choice(conjunctions)+"\:"+nounPhrase()$

$\quad\tt else:$

$\quad\quad\tt continue$

You may use pass also

$\tt def\: adjectivePhrase():$

$\quad\tt if\:adj=="yes":$

$\quad\quad\tt return\:random.choice(adjectives)+"\:"+nounPhrase()$

$\quad\tt else:$

$\quad\quad\tt continue$

4 0

2 years ago