Melody

This net might help you visualise the cylinder

Hi Atoombii        

I think that you are saying that there are 2 peices of paper.

Each one is 8.5 cm and 11cm   and they are each wrapped to make cylinders.

One is 8.5cm high and a circumference of 11cm

and the other is 11 cm high wirh a circumference of 8.5cm

Well, since they both use the same sized peice of paper the surface are of the curved surfaces must be identical.

So if you are refering to the SA of  the open cylinders then they are identical.

If you put ends on your cylinder then the one with the biggest circumference will have the biggest 'end' 

So  the short squat one will have the biggest surface area.     

this is not a quadratic equation.

Perhaps you want to rationalize the denominator?

$$\\\frac{4}{\sqrt3-5}\\\\
=\frac{4}{\sqrt3-5}\times \frac{\sqrt3+5}{\sqrt3+5}\\\\
=\frac{4(\sqrt3+5)}{3-25}\\\\
=\frac{4(\sqrt3+5)}{-22}\\\\
=\frac{-2(\sqrt3+5)}{11}\\\\
=\frac{-2\sqrt3-10}{11}\\\\$$

So it seems you have rationalized the denominator correctly :)

   [2nd][acos]

$$\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{\mathtt{6.5}} = {\frac{{\mathtt{13}}}{{\mathtt{6}}}} = {\mathtt{2.166\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

Yes I will second MG.

Good one Rosala :)

Yes, .............     thank you :))

$${{\mathtt{\pi}}}^{{\mathtt{x}}} = {\mathtt{41}} \Rightarrow {\mathtt{x}} = {\frac{{ln}{\left({\mathtt{41}}\right)}}{{ln}{\left({\mathtt{\pi}}\right)}}} \Rightarrow {\mathtt{x}} = {\mathtt{3.244\: \!059\: \!679\: \!588\: \!781\: \!6}}$$

yes, it works too :)

BrittanyJ, on the web2 calc it is    [2nd] [atan]  

inverse tan is he same as arc tan :)

I'll give you he extra 3 points Geno