Vinculum

LMAO, I realised my mistake!!!!

I should have looked at the final answer before submitting it!!!

-Vinculum

It's 3, trust me

-Vinculum

The domain of the function in union of intervals is: \(\left(-\infty \:,\:-8\right)\cup \left(-8,\:\infty \:\right)\)

I would say the easiest way to do it is to just graph and find the distance from one side of the circle to the origin.

The distance from the left end of the circle to the center is 8.

Applying this to the circle area formula:

A=pi*r^{2}

A=pi*8^{2}

You can calculate the rest by yourself by entering pi*8^{2} in a calc.

-Vinculum

"was ist 4 mal 4"

4 mal 4 ist 16

-Vinculum

"pourquoi sa ne marche pas dès qu'on tape sur le étégale ?"

peut-être que ça ne marche pas, essayez de rafraîchir

-Vinculum

Ight so solve for z...

That gives \(z=12-16i\)

Now you are asking for \(|z|\) or in other words the magnitude of \(z\)

So convert \(12-16i\) to polar coords, which is \(20 e^{-i tan^{-1}{(4/3)}}\)

The magnitude is \(20\)

-Vinculum

\(\frac{3}{m}\le \:-\frac{7}{2}\)

\(\frac{6+7m}{m}\le \:0\)

\(-\frac{6}{7}\le \:m<0\)

I think you got it from here after looking at the intervals...

-Vinculum