If x = y and y = z then x = z
#include <iostream> using namespace std; int main() { int x, y, z; cout << "Enter 3 numbers: \n"; cin >> x; cin >> y; cin >> z; if(x < y && x < z) { cout << x << " "; if(y < z) { cout << y << " " << z; } else if(z < y) { cout << z << " " << y; } } else if(y < x && y < z) { cout << y << " "; if(x < z) { cout << x << " " << z; } else if(z < x) { cout << z << " " << x; } } else if(z < y && z < x) { cout << z << " "; if(y < x) { cout << y << " " << x; } else if(x < y) { cout << x << " " << y; } } char wait; cin >> wait; return 0; }
3 out of 4. 8 possiableaties------------------ coins 1--- 2--- 3--- 4--- 5--- 6--- 7--- 8 x y--- x--- x--- x--- x--- y--- y---- y--- y x z--- x--- x--- z--- z--- x--- x----z--- z y z--- y----z---y--- z--- y--- z----y-----z There are 8 possiabilities for the three coins to land, you count the matches, there 6 out of 8 that match.
(x + y)/z = x/z + y/z where z is non-zero.
3x + y + z = 63x - y + 2z = 9y + z = 3y + z = 3y = 3 - z (substitute 3 - z for y into the first equation of the system)3x + y + z = 63x + (3 - z) + z = 63x + 3 = 63x = 3x = 1 (substitute 3 - z for y and 1 for x into the second equation of the system)3x - y + 2z = 93(1) - (3 - z) + 2z = 93 - 3 + z + 2z = 93z = 9z = 3 (which yields y = 0)y = 3 - z = 3 - 3 = 0So that solution of the system of the equations is x = 1, y = 0, and z = 3.
xy + y = z xy = z - y (xy)/y = (z - y)/y x = (z - y)/y
The expression y^4 - z^4 can be factored using the difference of squares formula. It can be written as (y^2 + z^2)(y^2 - z^2), which can be further simplified to (y^2 + z^2)(y + z)(y - z).
If x = y and y = z then x = z
x=abs(y+z) x=+(y+z)=y+z x=-(y+z)=-y-z
-zy or y(-z) and viceversa or y x -z and viceversa
There are 8 different subsets. The null set. {x} {y} {z} {x y} {x z} {y z} {x y z}
If x y and y z, which statement is true
Commutative x + y = y + x x . y = y . x Associative x+(y+z) = (x+y)+z = x+y+z x.(y.z) = (x.y).z = x.y.z Distributive x.(y+z) = x.y + x.z (w+x)(y+z) = wy + xy + wz + xz x + xy = x x + x'y = x + y where, x & y & z are inputs.
(x - y)2 - z2 is a difference of two squares (DOTS), those of (x-y) and z. So the factorisation is [(x - y) + z]*[(x - y) - z] = (x - y + z)*(x - y - z)
#include <iostream> using namespace std; int main() { int x, y, z; cout << "Enter 3 numbers: \n"; cin >> x; cin >> y; cin >> z; if(x < y && x < z) { cout << x << " "; if(y < z) { cout << y << " " << z; } else if(z < y) { cout << z << " " << y; } } else if(y < x && y < z) { cout << y << " "; if(x < z) { cout << x << " " << z; } else if(z < x) { cout << z << " " << x; } } else if(z < y && z < x) { cout << z << " "; if(y < x) { cout << y << " " << x; } else if(x < y) { cout << x << " " << y; } } char wait; cin >> wait; return 0; }
It is not easy to illustrate this in a table since it would need to be a 3-d table. The browser which we are required to use is bad enough for ordinary tables. So here goes:X = True, Y = True, Z = True then (X or Y or Z) = TrueX = True, Y = True, Z = False then (X or Y or Z) = TrueX = True, Y = False, Z = True then (X or Y or Z) = TrueX = True, Y = False, Z = False then (X or Y or Z) = TrueX = True, Y = True, Z = True then (X or Y or Z) = TrueX = True, Y = True, Z = False then (X or Y or Z) = TrueX = True, Y = False, Z = True then (X or Y or Z) = TrueX = True, Y = False, Z = False then (X or Y or Z) = FalseThe simple way to remember is that the OR gate gives False only when each input is False.
If it is required, the apostrophe would come after the Z but before the S. "This is Mr. Buzz's stinger!"