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bezimeni [28]
3 years ago
5

When a new version of a system is installed, the prior release is _____, or stored.​?

Computers and Technology
1 answer:
Firlakuza [10]3 years ago
7 0

<span><span>Version control is the process of tracking system releases or versions</span>.
When a new version of a system is installed, the prior release is archived, or stored.​</span>
<span> Version Archive is a repository of all entries which have ever appeared .
</span>Each change is documented and installed as a new version of the system called a maintenance release .

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/* Returns the sum of series */

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int sum = 0;

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

sum += x * x * x;

return sum;

}

// Driver Function

public static void main(String[] args)

{

int n = 5;

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}

// Code Contributed by Mohit Gupta_OMG <(0_o)>

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Superclass in python explanation
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Answer:

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Explanation:

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Read 2 more answers
Exercise1 : Defining Circle type and Creating a Driver Circle class private members double radius double xPos double yPos public
asambeis [7]

Answer:

Aee explaination

Explanation:

1) class_circle.h

#include<iostream>

#include<cmath>

#include<iomanip>

#define PI 3.14159

using namespace std;

class circle

{

private:

double radius,xPos,yPos;

public:

double diameter();

double area();

double circumference();

double getRadius();

double getX();

double getY();

void setX(double x);

void setY(double y);

double distanceToOrigin();

bool intersect(const circle& otherCircle);

bool setRadius(double r);

};

class CircleDriver:public circle

{

private:

circle circ1;

circle circ2;

void obtainCircles();

void printCircleInfo();

public:

void run();

};

2) class_circle.cpp

#include "class_circle.h"

double circle::diameter()

{

return (radius * 2);

}

double circle::area()

{

return PI * radius * radius;

}

double circle::circumference()

{

return 2 * PI * radius;

}

double circle::distanceToOrigin()

{

return sqrt((xPos * xPos) + (yPos * yPos));

}

double circle::getRadius()

{

return radius;

}

double circle::getX()

{

return xPos;

}

double circle::getY()

{

return yPos;

}

void circle::setX(double x)

{

xPos = x;

}

void circle::setY(double y)

{

yPos = y;

}

bool circle::setRadius(double r)

{

if(r > 0)

{

radius = r;

return true;

}

else

{

radius = 0;

return false;

}

}

bool circle::intersect(const circle& otherCircle)

{

double dist1 = (radius - otherCircle.radius) * (radius - otherCircle.radius);

double dist2 = ((xPos - otherCircle.xPos) * (xPos - otherCircle.xPos)) + ((yPos - otherCircle.yPos)*(yPos - otherCircle.yPos));

double dist3 = (radius + otherCircle.radius) * (radius + otherCircle.radius);

if((dist1 <= dist2 ) && (dist2 <= dist3) && (dist1 < dist3))

return true;

else

return false;

}

void CircleDriver::obtainCircles()

{

double x,y,rad;

cout<<"Enter xPos yPos circle 1 :";

cin>>x>>y;

do

{

cout<<"\n\nEnter radius for circle 1 :";

cin>>rad;

}while(rad <= 0);

circ1.setX(x);

circ1.setY(y);

circ1.setRadius(rad);

cout<<"\n\nEnter xPos yPos circle 2 :";

cin>>x>>y;

do

{

cout<<"\n\nEnter radius for circle 2 :";

cin>>rad;

}while(rad <= 0);

circ2.setX(x);

circ2.setY(y);

circ2.setRadius(rad);

}

void CircleDriver::printCircleInfo()

{

cout<<setprecision(6)<<endl;

cout<<"\n\nInformation for Circle 1:"<<endl;

cout<<"location:( "<<circ1.getX()<<" ,"<<circ1.getY()<<" )"<<endl;

cout<<"diameter: "<<circ1.diameter()<<endl;

cout<<"area: "<<circ1.area()<<endl;

cout<<"circumference: "<<circ1.circumference()<<endl;

cout<<"distance from the origin: "<<circ1.distanceToOrigin()<<endl;

cout<<"\n\nInformation for Circle 2:"<<endl;

cout<<"location:( "<<circ2.getX()<<" ,"<<circ2.getY()<<" )"<<endl;

cout<<"diameter: "<<circ2.diameter()<<endl;

cout<<"area: "<<circ2.area()<<endl;

cout<<"circumference: "<<circ2.circumference()<<endl;

cout<<"distance from the origin: "<<circ2.distanceToOrigin()<<endl<<endl;

if(circ1.intersect(circ2))

cout<<"The circles intersect."<<endl;

else

cout<<"The circles does not intersect."<<endl;

}

void CircleDriver::run()

{

obtainCircles();

printCircleInfo();

}

3) main.cpp

#include "class_circle.h"

int main()

{

CircleDriver myDriver;

myDriver.run();

return(0);

}

Please refer below output for reference

Enter xPos yPos circle 1 :0.0 50.0

Enter radius for circle 1 :100.0

Enter xPos yPos circle 2 :50.0 0.0

Enter radius for circle 2 :100.0

Information for Circle 1:

location:( 0 ,50 )

diameter: 200

area: 31415.9

circumference: 628.318

distance from the origin: 50

Information for Circle 2:

location:( 50 ,0 )

diameter: 200

area: 31415.9

circumference: 628.318

distance from the origin: 50

The circles intersect.

Process returned 0 (0x0) execution time : 18.245 s

Press any key to continue.

8 0
4 years ago
What is the different between image processing and machine vision? Please explain with an example.
HACTEHA [7]

Answer:

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For example: Television images.

Machine vision is defined as the technology in which it provide various image based applications like system control, automatic inspections and also used in various industries. It basically refers to the hardware, software and integration system in the computer science.

For example: Analysis of various types of electronic components.

4 0
4 years ago
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