Q: What is a example of disrupting a threat C2 system?

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12c

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 â‰¥ mc + c1 and y â‰¤ mx + c2

a2 + b2 = c2. 62 + 82=c2. 36+64=c2. 100=c2. sqrt(100)=sqrt(c2). c=10. So the diagonal is 10 m. long.

(c2 + 11)(c2 - 11)

#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(); }

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Citroen C2 can be found most easily on the Autotrader website. The Citroen C2 is a car manufactured in France, and its engine utilizes a hybrid system.

use Pythagorean theorem (A2 + B2 = C2).A is a side touching the right angle, B is the other side and C is the hypotenuse. example: A = 3, B = 4, C = ?, A2 + B2 = C2, 32 + 42 = c2, 9 + 16 = c2, 25 = c2 5 = c

Pythagorean Theorem

Here~ D2 D2 D2 B D2 C2 B A C2 C2 C2 A C2 B A G D2 D2 D2 G G A B C2 C2 C2 C2 D2 C2 B A G D2 D2 D2 B D2 D2 D2 B D2 D2 D2 E2 D2 D2 B C2 C2 C2 A C2 C2 C2 A C2 C2 C2 B2 C2 B A G

12c

A c2 c2 c2

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

The pythagorean theory or pythagorean theorem is a formula to find the leg or the hypotenuse for a right triangle. There are three parts to a triangle, The legs(A2) and (B2). The hypotenuse (C2). The hypotenuse is always the longest side of the triangle it is always adjacent to the 900 angle of the right triangle. The actual pythagorean theorem is A2 + B2 = C2. Example: A=2 B= 4 C=? A2 + B2 =C2 22 + 42 =C2 4 + 16= C2 20=C2 Now you find the square root for the two numbers you just added 4.4 = C

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 â‰¥ mc + c1 and y â‰¤ mx + c2