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Ahat [919]
3 years ago
12

There are 10 girls and 8 boys at a party. A cartoonist want to sketch a picture of each boy with each girl. How many sketches ar

e required?​
Computers and Technology
1 answer:
kykrilka [37]3 years ago
3 0
80 sketches will be required. It’s just multiplication. Let x=boys and y=girls. If x=8 and y=10, then one boy has to be with one girl. In total 1 boy will be with 10 girls, so that’s 10 sketches. You do the same for boy #2 and so on. It adds up to 80.
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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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Answer:

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Explanation:

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