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Vikentia [17]
3 years ago
12

Mark often gets headaches while working on his computer in the office. What activity can Mark do in the office to prevent headac

hes?
stretch his neck
stretch his back
look away from the computer monitor to rest his eyes
sit straight with cushioning
Computers and Technology
1 answer:
svet-max [94.6K]3 years ago
4 0
Mark can prevent these headaches that he is getting on his computer by looking away from his monitor every now and again so he can rest his eyes.
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Which of the following is a type of monitor port?
Soloha48 [4]
B. all of those.

just to clarify,
HDMI is the port for monitor and sound.

VGA is for video/monitor port
and display port.... I don't know, but it's a type of monitor port.
7 0
3 years ago
Which statements describe the use of styles in Word? Check all that apply.
algol [13]

Answer:

can be used to make word docments look the same

Explanation:

4 0
3 years ago
Read 2 more answers
Finish and test the following two functions append and merge in the skeleton file:
avanturin [10]

Answer:

Explanation:

#include <iostream>

using namespace std;

int* append(int*,int,int*,int);

int* merge(int*,int,int*,int);

void print(int*,int);

int main()

{ int a[] = {11,33,55,77,99};

int b[] = {22,44,66,88};

print(a,5);

print(b,4);

int* c = append(a,5,b,4); // c points to the appended array=

print(c,9);

int* d = merge(a,5,b,4);

print(d,9);

}

void print(int* a, int n)

{ cout << "{" << a[0];

for (int i=1; i<n; i++)

cout << "," << a[i];

cout << "}\n";

}

int* append(int* a, int m, int* b, int n)

{

int * p= (int *)malloc(sizeof(int)*(m+n));

int i,index=0;

for(i=0;i<m;i++)

p[index++]=a[i];

for(i=0;i<n;i++)

p[index++]=b[i];

return p;

}

int* merge(int* a, int m, int* b, int n)

{

int i, j, k;

j = k = 0;

int *mergeRes = (int *)malloc(sizeof(int)*(m+n));

for (i = 0; i < m + n;) {

if (j < m && k < n) {

if (a[j] < b[k]) {

mergeRes[i] = a[j];

j++;

}

else {

mergeRes[i] = b[k];

k++;

}

i++;

}

// copying remaining elements from the b

else if (j == m) {

for (; i < m + n;) {

mergeRes[i] = b[k];

k++;

i++;

}

}

// copying remaining elements from the a

else {

for (; i < m + n;) {

mergeRes[i] = a[j];

j++;

i++;

}

}

}

return mergeRes;

}

4 0
3 years ago
Your ghost hunting group is recording the sound inside a haunted Stanford classroom for 20 hours as MP3 audio files. About how m
Mademuasel [1]

Answer:

1.152 GB if it will be at 128 kbps quality

Explanation:

So, let's say you choose your mp3 be 128 kbps (kilobits per second).

One byte is 8 bits. So 128 kbps = 16 kilobytes per second.

There is 60*60 seconds in one hour. So 16*60*60 = 57.6 megabytes per hour

Multiply by 20 hours to get your answer: 57.6*20 = 1152 MB = 1.152 GB.

3 0
3 years ago
The compare_strings function is supposed to compare just the alphanumeric content of two strings, ignoring upper vs lower case a
Korolek [52]

Answer:

There is a problem in the given code in the following statement:

Problem:

punctuation = r"[.?!,;:-']"

This produces the following error:

Error:

bad character range

Fix:

The hyphen - should be placed at the start or end of punctuation characters. Here the role of hyphen is to determine the range of characters. Another way is to escape the hyphen - using using backslash \ symbol.

So the above statement becomes:

punctuation = r"[-.?!,;:']"  

You can also do this:

punctuation = r"[.?!,;:'-]"  

You can also change this statement as:

punctuation = r"[.?!,;:\-']"

Explanation:

The complete program is as follows. I have added a print statement print('string1:',string1,'\nstring2:',string2) that prints the string1 and string2 followed by return string1 == string2  which either returns true or false. However you can omit this print('string1:',string1,'\nstring2:',string2) statement and the output will just display either true or false

import re  #to use regular expressions

def compare_strings(string1, string2):  #function compare_strings that takes two strings as argument and compares them

   string1 = string1.lower().strip()  # converts the string1 characters to lowercase using lower() method and removes trailing blanks

   string2 = string2.lower().strip()  # converts the string1 characters to lowercase using lower() method and removes trailing blanks

   punctuation = r"[-.?!,;:']"  #regular expression for punctuation characters

   string1 = re.sub(punctuation, r"", string1)  # specifies RE pattern i.e. punctuation in the 1st argument, new string r in 2nd argument, and a string to be handle i.e. string1 in the 3rd argument

   string2 = re.sub(punctuation, r"", string2)  # same as above statement but works on string2 as 3rd argument

   print('string1:',string1,'\nstring2:',string2)  #prints both the strings separated with a new line

   return string1 == string2  # compares strings and returns true if they matched else false

#function calls to test the working of the above function compare_strings

print(compare_strings("Have a Great Day!","Have a great day?")) # True

print(compare_strings("It's raining again.","its raining, again")) # True

print(compare_strings("Learn to count: 1, 2, 3.","Learn to count: one, two, three.")) # False

print(compare_strings("They found some body.","They found somebody.")) # False

The screenshot of the program along with its output is attached.

4 0
3 years ago
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