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Is it possible for a system of two linear inequalities to have no solution and be parallel?
Yes, it is possible. Since their boundaries are parallel the relevant equations are of the form y = mx + c1 and y = mx + c2. Then if c1 > c2, the inequalities must be of the form y ≥ mc + c1 and y ≤ mx + c2
Which island is located in the grid squares c1 and c2?
I think it's Alexander island
What are the conditions of consistency for a pair of linear equations?
a1 = b1 = c1 a2 = b2 = c2
Write a program to find sum of two complex numbers?
#include<iostream.h> #include<conio.h> class complex { int a,b; public: void read() { cout<<"\n\nEnter the REAL PART : "; cin>>a; cout<<"\n\nEnter the IMAGINARY PART : "; cin>>b; } complex operator +(complex c2) { complex c3; c3.a=a+c2.a; c3.b=b+c2.b; return c3; } complex operator -(complex c2) { complex c3; c3.a=a-c2.a; c3.b=b-c2.b; return c3; } complex operator *(complex c2) { complex c3; c3.a=(a*c2.a)-(b*c2.b); c3.b=(b*c2.a)+(a*c2.b); return c3; } complex operator /(complex c2) { complex c3; c3.a=((a*c2.a)+(b*c2.b))/((c2.a*c2.a)+(c2.b*c2.b)); c3.b=((b*c2.a)-(a*c2.b))/((c2.a*c2.a)+(c2.b*c2.b)); return c3; } void display() { cout<<a<<"+"<<b<<"i"; } }; void main() { complex c1,c2,c3; int choice,cont; do { clrscr(); cout<<"\t\tCOMPLEX NUMBERS\n\n1.ADDITION\n\n2.SUBTRACTION\n\n3.MULTIPLICATION\n\n4.DIVISION"; cout<<"\n\nEnter your choice : "; cin>>choice; if(choice==1choice==2choice==3choice==4) { cout<<"\n\nEnter the First Complex Number"; c1.read(); cout<<"\n\nEnter the Second Complex Number"; c2.read(); } switch(choice) { case 1 : c3=c1+c2; cout<<"\n\nSUM = "; c3.display(); break; case 2 : c3=c1-c2; cout<<"\n\nResult = "; c3.display(); break; case 3 : c3=c1*c2; cout<<"\n\nPRODUCT = "; c3.display(); break; case 4 : c3=c1/c2; cout<<"\n\nQOUTIENT = "; c3.display(); break; default : cout<<"\n\nUndefined Choice"; } cout<<"\n\nDo You Want to Continue?(1-Y,0-N)"; cin>>cont; }while(cont==1); getch(); }
How do you find a point on a rhombus given the other three points?
Lets call the 4 points A, B, C and D, with D being the unknown point. Firstly, you would find the gradients of both the lines AB and BC. This will give you two values for "m" in the general equations for two different lines, in the form y = mx + c. Then, you should substitute the point that does not exist on each of the lines into it's equation. So, you would substitute C into y = mABx + c1 and A into y = mBCx + c2 This will allow you to find the values of c1 and c2. From this, two simultaneous equations are generated that should be fairly easy to solve by rearrangement and manipulation. Substitute for y in both of the equations and you are left with: mABx + c1 = mBCx + c2 This can be simplified to: ( c1 - c2 ) = (mBCx - mABx) ( c1 - c2 ) = x(mBC - mAB) ( c1 - c2 )/(mBC - mAB) = x Now that you have a value for x (The x-coordinate of the final vertex) you can find its corresponding y-coordinate, simply by substituting this x value into one of your two simultaneous equations. (Either y = mABx + c1 or y = mBCx + c2) From either of these, the y coordinate of the point can be obtained, and thus, the final step is to present the answer as a set of coordinates in the form (x,y).
