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
I think it's Alexander island
a1 = b1 = c1 a2 = b2 = c2
#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(); }
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).
Yes it is called the atlas
Eg: typedef struct RGB { unsigned char R, G, B; } c1, c2, csum; ... if ((int)c1.R + (int)c2.R > 255) csum.R = 255; else csum.R = c1.R + c2.R; if ((int)c1.G + (int)c2.G > 255) csum.G = 255; else csum.G = c1.G + c2.G; if ((int)c1.B + (int)c2.B > 255) csum.B = 255; else csum.B = c1.B + c2.B;
The Atlas - C1.
C1 is called the atlas and C2 is called the axis
type into a given cell =c1+c2 If you mean words, use this =c1&c2 which means join the string values.
There are only 2 cervical vertebrae that have common names: the atlas and the axis, they act as the pivot that allows you to turn your head.
Alexander Island.
Let K1 & K2 be the equivalent capacitence in series and parallel resp. if c1 and c2 b the values of capacitor we have 1/c1+1/c2=1/6 c1+c2=25 solving we get c1=10 MF c2 =15 MF or vice cersa
yes. the atlas and axis, or C1 and C2, do not have an intervertebral disc, nor an intervertebral foramen, between them. C1 looks like an oval. it has two lateral masses (no vertebral body) where it makes contact with the occiput and C2. the inferior articular facets of the C1 and the superior articular facets of C2 form 2 joints, one on each side. there is also a third joint formed by the dens, or odontoid process, of C2 and the interior of the anterior arch of C1. this is the joint you use to shake your head "no".
when you break C2 and C1 worst case senario is when your in a ventialator beacuase you usually cant breath on your own and your sometimes paralyzed from the neck down.
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
The name of the joint that is between C1 & C2 is the atlantoaxial joint. C1 is the atlas and C2 is the axis. The type of joint is called a pivot joint, and often referred to as the no-no joint, because it is used to rotate your head left and right, as in saying 'no'.