+4 find domain of sqrt(x^2+4x C 2X^2+3) {C is used 2 represent combination $${\frac{({\mathtt{X}}){!}}{\left((\left({\mathtt{X}}{\mathtt{\,-\,}}{\mathtt{Y}}\right)){!}{\mathtt{\,\times\,}}({\mathtt{Y}}){!}\right)}}$$ )
+118723 This is a bit confusing - I'll assume x and X are the same thing.
Actually I don't think this has a domain because there is an X and a Y and no equal sign.
Okay so I will just look at possible values of x
$$\sqrt{x^2+4x C 2X^2+3} \\\\
\sqrt{x^2+4x \frac{X!}{(x-y)!Y!} 2X^2+3} \\\\
\sqrt{x^2+8x^3 \frac{x!}{(x-y)!y!}+3} \\\\$$
Taking this interpretation, $$x\in Z\ge0, \qquad y\in Z\ge 0, \qquad x>y$$
so x must be a cardinal number (counting number 1,2,3,.... )
that's what I think anyway.
+118723 This is a bit confusing - I'll assume x and X are the same thing.
Actually I don't think this has a domain because there is an X and a Y and no equal sign.
Okay so I will just look at possible values of x
$$\sqrt{x^2+4x C 2X^2+3} \\\\
\sqrt{x^2+4x \frac{X!}{(x-y)!Y!} 2X^2+3} \\\\
\sqrt{x^2+8x^3 \frac{x!}{(x-y)!y!}+3} \\\\$$
Taking this interpretation, $$x\in Z\ge0, \qquad y\in Z\ge 0, \qquad x>y$$
so x must be a cardinal number (counting number 1,2,3,.... )
that's what I think anyway.
