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Funny how memory plays tricks on you. I just loooove Crow! Maybe that's why.

Nope. I was wrong. I went back and had another look. It was "Tom Servo" who said it voiced by Kevin Murphy. It was from mst3k s03e19.

War of the Colossal Beast mst3k version mst3k can be annoying at first but it gets more enjoyable the more you watch it. Like when the collossal beast (giant man) litereraly picks up a bus and the mst3k crew interrupt saying "Hold the bus." I thought that was funny. Can't remember but I think...

Cambridge y11 Ex15C Q14a,15

Well if you really must use projections, you can use the formula \begin{aligned}d&=|{\bf p}-{\bf a}-\text{Proj}_{\bf n}({\bf p}-{\bf a})|\\&=\textstyle|-{\bf i}-4{\bf j}-6{\bf k}-\frac{(-{\bf i}-4{\bf j}-6{\bf k})\boldsymbol{\cdot}(2{\bf i}+{\bf j}+{\bf k})}{|2{\bf i}+{\bf j}+{\bf k}|^2}(2{\bf...

\text{Well }6\lambda^2+24\lambda+53\text{ is a quadratic } \text{For }a>0\text{ the quadratic }a\lambda^2+b\lambda+c\text{ is minimum when }\lambda=\frac{-b}{2a}

I wouldn't use projections. Method 1 (without finding the minimum distance) \text{If }d=\text{minimum distance then} d^2=(1+2\lambda)^2+(4+\lambda)^2+(6+\lambda)^2=6\lambda^2+24\lambda+53 \text{and }\lambda=\frac{-b}{2a}=\frac{-24}{2\times6}=-2 \text{Point is...

There is yet a third even quicker way using the hyperbolic cotangent method, which is outside the syllabus, but actually much easier: (z+1)^{2n}+(z-1)^{2n}=0\Rightarrow(z+1)^{2n}=-(z-1)^{2n}\text{ and since }z\ne1, \textstyle\left(\frac{z+1}{z-1}\right)^{2n}=-1=e^{i(2k+1)\pi}\text{ for integers...

Rather than explain the rather long solution to 15a, here is a shorter solution instead: (z+1)^{2n}+(z-1)^{2n}=0\Rightarrow(z+1)^{2n}=-(z-1)^{2n}\text{ and since }z\ne1, \textstyle\left(\frac{z+1}{z-1}\right)^{2n}=-1=e^{i(2k+1)\pi}\text{ for integers }k.\...

These are Cambridge Ext 2 Ex 3C Q13-15 Here the solutions to 13b,c,14c,15a. I included 13b too because the solution to 13c refers to part of the solution to 13b .

It is cambridge y11 Ex14D Q5

So there's this dude from Harvard University, Lars Ahlfors He is like the coolest dude ever for complex numbers - like the GOAT of complex numbers who wrote the book Complex Analysis - probably the one which has been used more than any other for that subject. My very short solution...

Well here is the trailer which gives you a glimpse of what I just endured for the last hour: Unfortunately it doesn't show Spock singing but you get the drift.

Oh my goodness what is star trek coming to? Spock singing? etc. Really? Strange new world indeed!

There is a new resource for the fx-8200 AU Kissane, Barry (2023) Scientific calculators and irrational numbers. Australian Mathematics Education Journal, 5(1), 4-9. attached

\textstyle\frac{1}{4}\ge\frac{x}{x^2+4} \textstyle\int_0^\alpha\frac{1}{4}dx\ge\int_0^\alpha\frac{x}{x^2+4}dx \textstyle[\frac{1}{4}x]_0^\alpha\ge\frac{1}{2}\int_0^\alpha\frac{2x}{x^2+4}dx=\frac{1}{2}[\ln(x^2+4)]_0^\alpha \textstyle\frac{1}{4}\alpha\ge\frac{1}{2}\ln\frac{\alpha^2+4}{4}...

It is yr 11 cambridge 14C Q5

I'm watching barbie now and wishing oppenheimer goes into barbieland to blow it up with an atom bomb!:newburn:

Not as famous, but better is a documentary which came out a few days before the Oppenheimer movie called "To End All War: Oppenheimer & the Atomic Bomb" I thought that was better than the movie

These are yr 11 cambridge 14A Q10-12