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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).
How do you add colors in c?
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;
What pivots on c2 and lacks a body?
The Atlas - C1.
Add the contents of cells C1and C2 together?
type into a given cell =c1+c2 If you mean words, use this =c1&c2 which means join the string values.
Which island is locates in grid squares c1 and c2?
Alexander Island.
2 capacitances are in parallel gives 6 micro farad and in series gives 25 what will be the value of individual capacitance?
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
Why is an intervertebral disc not present between C1 and C2?
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".
What if you break your c1 and c2 vertebre?
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.
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
What is a quadratic curve?
A quadratic curve has the form C2X2+C1X1+C0 where (C2,C1,C0) are coefficients. If C2=0, it degrades to the equation for line. C1 or C0 may also =0
The first cervical vertebra is called the axis?
Yes it is called the atlas
How do you know when an equation has infinite answers?
say system of equation is as follows a1x + b1y = c1 a2x + b2y = c2 if a1/a2 = b1/b2 = c1/c2 equation will have infinite answers
What vertebra allows you to shake your head?
C1 (atlas) and C2 (axis). C1 allows for nodding yes (flexion and extension) while C2 allows for shaking head no (lateral rotation).
