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timurjin [86]
3 years ago
11

What is the forging of the return address on an email so that the email message appears to come from someone other than the actu

al sender? Multiple Choice Hoaxes Spoofing Malicious code Sniffer
Computers and Technology
1 answer:
Anestetic [448]3 years ago
5 0

Answer:

The correct option is: B. Spoofing

Explanation:

In network security, a spoofing attack is defined as a technique by which an individual or a program impersonates as another by deliberately falsifying information or data to spread malware, steal information, or bypass access controls. The common types of spoofing are email spoofing, IP spoofing, DNS spoofing.

E-mail address spoofing is used by the hackers or spammers to hide the source or origin of the e-mails and is known as email spoofing. It involves creation of an email with a forged sender address.

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save as a word document (.docx) to keep editing offline then copy and paste into the google doc

5 0
3 years ago
Design and implement a class dayType that implements the day of the week in a program. The class dayType should store the day, s
Afina-wow [57]

The code is implemented based on the given operations.

Explanation:

#include <iostream>

#include <string>

using namespace std;

class dayType

{ private:

 string day[7];

 string presentDay;

 int numofDays;

public:

 void setDay(string freshDay);

 void printDay() const;

 int showDay(int &day);

 int nextDay(int day);

 int prevDay(int day) const;

 int calcDay(int day, int numofDays);    

 dayType()

 {

  day[0] = "Sunday";

  day[1] = "Monday";

  day[2] = "Tuesday";

  day[3] = "Wednesday";

  day[4] = "Thursday";

  day[5] = "Friday";

  day[6] = "Saturday";

  presentDay = day[0];

  numofDays = 0;

 };

 ~dayType();

};

#endif

#include "dayType.h"

void dayType::setDay(string freshDay)

{

  presentDay = freshDay;

}

void dayType::printDay()

{

  cout << "Day chosen is " << presentDay << endl;

}

int dayType::showDay(int& day)

{

  return day;

}

int dayType::nextDay(int day)

{

day = day++;

if (day > 6)

 day = day % 7;

switch (day)

{

case 0: cout << "The successive day is Sunday";

 break;

case 1: cout << "The successive day is Monday";

 break;

case 2: cout << "The successive day is Tuesday";

 break;

case 3: cout << "The successive day is Wednesday";

 break;

case 4: cout << "The successive day is Thursday";

 break;

case 5: cout << "The successive day is Friday";

 break;

case 6: cout << "The successive day is Saturday";

 break;

}

cout << endl;

return day;

}

 

int dayType::prevDay(int day)

{

day = day--;

switch (day)

{

case -1: cout << "The before day is Saturday.";

 break;

case 0: cout << "The before day is Saturday.";

 break;

case 1: cout << "The before day is Saturday.";

 break;

case 2: cout << "The before day is Saturday.";

 break;

case 3: cout << "The before day is Saturday.";

 break;

case 4: cout << "The before day is Saturday.";

 break;

case 5: cout << "The before day is Saturday.";

 break;

default: cout << "The before day is Saturday.";

}

cout << endl;

return day;

}

int dayType::calcDay(int addDays, int numofDays)

{

addDay = addDays + numofDays;

if (addDay > 6)

 addDay = addDay % 7;

switch(addDay)

{

case 0: cout << "The processed day is Sunday.";

 break;

case 1: cout << "The processedday is Monday.";

 break;

case 2: cout << "The processedday is Tuesday.";

 break;

case 3: cout << "The processedday is Wednesday.";

 break;

case 4: cout << "The processedday is Thursday.";

 break;

case 5: cout << "The processedday is Friday.";

 break;

case 6: cout << "The processedday is Saturday.";

 break;

default: cout << "Not valid choice.";

}

cout << endl;

return addDays;

}

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Which of the following was not important in the development of the internet?
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The help desk received a call from a user who cannot get any print jobs to print on the locally shared printer. While questionin
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Answer:

a. Power cycle the printer.

Explanation:

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The faster an object is moving, the ________ the shutter speed needs to be in order to freeze motion.
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Answer:

jkdsdjdshj,dfh.jhdfbhjf

Explanation:

bchSDCMHCXZ NHCXHBDSVCHDH,KC NBDBSDMJCBDBFD,JHCDSMNBBNCSCBFDNJCFJKC FMNSDNMSDFCH ĐS,CJDBS,CBSJBV,FJNDBFDSFDVBFĐDVFBVJFDCDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDHCD

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