GingerAle

MAD is often very funny, but who is Justin Bieber?

Well, Mr. Blarney, they have shown their work. Now all you need is a brain – an active one, of course. Having a petrified pile of oatmeal in its place won’t do.

Here is an easier way using logarithms:

Write 85184^85184 as a power of 1936

.

\(\dfrac{85184*log(85184)}{log(1936)} =127776\\ \;\\ 85184^{85184} = 1936^{1227776}\)

Normally I wouldn’t bother presenting this for you, because it’s like trying to teach a pìg to sing, and doing that is a waste time and annoys the pìg. However, there may be a bright eleven-year-old who will understand this. It’s not that difficult, really--not for a bright eleven-year-old.

Hear! Hear!

Guest, would you create an account and message me, if you please?  I may very well be acquainted with your “Troubleshooter”

EP, why did you use a slide rule?  Scientific Calculators were very common and cheap then, and a few of the programmable calculators cost about the same as an expensive text book. 

Did you go to an Amish school?

Comments from SevenUp

Time to test my genetic enhancement.

\(\text {Velocity of hammer at impact with pile}\\ V_f= \sqrt{2ad} \hspace{1em}|\sqrt{2*9.81 m/s^2*5m} = 9.904568m/s\\ \)

\(m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\leftarrow \text {conservation of momentum}\\ 150kg (9.905m/s) = \underbrace{(150 + 400)kg}_{\text{mass of pile & driver}} (v)\leftarrow \text {velocity of combined masses}\\ \hspace {7em} v=\dfrac{150kg (9.905 m/s)}{(150 + 400)kg} = 2.70136m/s\\\)

\(\text {Energies:}\\ \text{kinetic energy} = \frac{1}{2}(550 kg)(0 - 2.70136 m/s)^2 = - 2006.77 J\\ \text{potential energy} = (550 kg) (9.81 m/s^2)(0-0.075 m ) = -404.66 J \\\)

\(\text {Energy to work formula:}\\ \text{(-F)}*(0.075 m) = -2411.43 J \hspace{1em} | (-F) = \text{negative Force} \\ \text{(-F)}=\dfrac{-2411.43J}{0.075m}= 32152.40 \ newtons\\\)

\(\text{Average resistance of the ground }= 32,152.40 \ newtons\)

No figs. Just newtons on this ground. 

This concludes the test of my genetic enhancement.

I think I passed.

In the post above mine, Sir CPhill typed the “Roman Zero” just as Morgan Tud described it. I reasoned Sir CPhill found it in the bolder of Sisyphus and chose a subtle path to announce it.

Here's a start.

This is Classic Zegroes and Rosala with guest star Rose98 (a.k.a. “Rose not La”)  

Top of the morning to you, Sir CPhill:

LancelotLink and Naus were both persuaded by Morgan Tud’s prophecy that you would indeed, one day, intercept Sisyphus’s bolder and extract the Roman Zero and display the precious Roman knowledge of nothing.

Ah yes . . .  the ever changing baseline of moral codes.  I wonder what the formula is.  Whatever it is, it probably follows an infinite descent model. Mine does. A few years ago, I would never have been a consort to a troll.  Now, it’s hard to imagine not being one. 

One thing about Lancelot, his liquor is always high quality. You can’t get a better banana daiquiri than one made by a genetically enhanced chimp.