// this solves the area of a circle #include #define PI3.1415927 double Area(double r) double Circumference(double r); double Area (double r) { return(PI*r*r); double Circumference(double r) return(2*PI*r*r); intmain( ) double radius,a,c; cout<<"Enter radius); cin>>radius; a=Area(radius); c=Circumference(radius); cout<<"The circumference of the circle is "<

// this solves the area of a circle #include #define PI3.1415927 double Area(double r) double Circumference(double r); double Area (double r) { return(PIrr); double Circumference(double r) return(2PIr*r); intmain( ) double radius,a,c; cout<<"Enter radius); cin>>radius; a=Area(radius); c=Circumference(radius); cout<<"The circumference of the circle is "<

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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struct ID { int x; int y; }; struct name { ID r; }; int main() { name a[2]; for (int i=0;i<2;i++) { a[i).r.x=i; a[i].r.y=i+1; } for(int i=0;i<2;i++) cout<
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نقطة واحدة Let A = {a; b; c; d} and R= {(a; a); (b; c); (c; b); (d; d)} then R is Transitive Equivalent not transitive
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#include #include using namespace std; int main() { int x=4; for (int x=0; x++<10; x++) { cout<