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The x, y, and z affair was an incident which french agents x, y, and z demanded payment before the U.S envoys could speak with French government. =P

Yush But Who Where X, Y, And Z!! That's Whut I Need To Nu!!!!! Eh, Oh, Sowwie People Lol

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Was that supposed to be XYZ? It's got something to do with Mr. X, Mr. Y, and Mr. Z. They did something bad.

Q: What was the x y and z affair?

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Y is for the letter Y. The letter Y, along with K & Z, were not part of the original Latin alphabet.

Commuting in algebra is often used for matrices. Say you have two matrices, A and B. These two matrices are commutative if A * B = B * A. This rule can also be used in regular binary operations(addition and multiplication). For example, if you have an X and Y. These two numbers would be commutative if X + Y = Y + X. The case is the same for X * Y = Y * X. There are operations like subtraction and division that are not commutative. These are referred to as noncommutative operations. Hope this helps!!

Henry knoX

It is almost always y vs. x. http://astro.uchicago.edu/cara/outreach/resources/other/howtograph.html

Your chromosome, if you have two x chromosome your a female, 1 x and 1 y your a male.

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Three French agents.

During the Quasi-War, the XYZ Affair damaged relations between America and France.

If x = y and y = z then x = z

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.

There are 8 different subsets. The null set. {x} {y} {z} {x y} {x z} {y z} {x y z}

x=abs(y+z) x=+(y+z)=y+z x=-(y+z)=-y-z

well on gamecube make a profile,exit,and on the main menu type in y,x,z,y,x,z,x,x,y,z,x,y for money or y,y,z,x,x,z,y,y,y,x,x,x for maximum reputation

(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)

If x y and y z, which statement is true

xy + y = z xy = z - y (xy)/y = (z - y)/y x = (z - y)/y

#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; }

x + 1 = y y + 3 = z z = y + 3 = (x + 1) + 3 = x + 4 Or: x = y - 1 = (z - 3) - 1 = z - 4 Which results in the same: x exceeds z by 4.