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3 years ago

How many junctions are in a typical house

Engineering

2 answers:

masya89 [10]3 years ago

5 0

Explanation:

there are no limits of the number of junction boxes in any particular circuit.

hjlf3 years ago

3 0

No limits whatsoever

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Write the implementation (.cpp file) of the Player class from the previous exercise. Again, the class contains:

hoa [83]

Answer:

//Define the header file

#ifndef PLAYER_H

#define PLAYER_H

//header file.

#include <string>

//Use the standard namespace.

using namespace std;

//Define the class Player.

class Player

{

//Declare the required data members.

string name;

int score;

public:

//Declare the required

//member functions.

void setName(string par_name);

void setScore(int par_score);

string getName();

int getScore();

}

//End the definition

//of the header file.

#endif

Player.cpp:

//Include the "Player.h" header file,

#include "Player.h"

//Define the setName() function.

void Player::setName(string par_name)

{

name = par_name;

}

//Define the setScore() function.

void Player::setScore(int par_score)

{

score = par_score;

}

//Define the getName() function.

string Player::getName()

{

return name;

}

//Define the getScore() function.

int Player::getScore()

{

return score;

}

7 0

3 years ago

Where is the Volkswagen super factory located? how is this locate relevant to us?

Blizzard [7]

It’s in Wolfsburg Germany

4 0

3 years ago

ferry boat steamed along at 8 mph through calm seas, passenger casey exercised by walking the perimeter of the rectangular deck,

olga_2 [115]

Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember

6 0

3 years ago

Simplify the following expressions, then implement them using digital logic gates. (a) f = A + AB + AC (b) f = AB + AC + BC (c)

Tema [17]

<u>Explanation</u>:

(a)

$f=A+A B+A C\\=A[1+B+C]=A \quad[1+x=1]\\F=A$

No gate is required to implement this function

(b)

$\begin{aligned}&\ f=A B+\bar{A} C+B C\\&\therefore f=A B+A C\end{aligned}$ $\begin{array}{l}(A B+\bar{A} C+B E=A B+\bar{A} C \\B C \text { is redendant })\end{array}$

Note: Refer the first image.

(c)

$\begin{aligned}f &=\overline{A+B}+A \bar{B}+B \bar{C} \\&=(\bar{A} \bar{B})+A \bar{B}+B \bar{C} \\&=\bar{B}[A+\bar{A}]+B \bar{C} \\& F=\bar{B}+B \bar{C} =\bar{B}+\bar{C}\end{aligned}$

Note: Refer the second image

(d)

$\begin{aligned}f=& A B \bar{c}+\overline{A+\bar{c}} \\=& A B \bar{c}+\bar{A} \bar{c}=\bar{A} B \bar{c}+\bar{A} c \\f=& \bar{A}[c+B \bar{c}] . \\& f=\bar{A} B+\bar{A} c=\bar{A}(B+c)\end{aligned}$