I've skipped and skimmed.
My pron name is Smokey Princess. Damn.
Sail, did you say that you have an all-over itch with no other symptoms and no specific cause? I had that. Zyrtec was the only thing that stopped it. My dermatologist also warned me against using those dryer sheets that stop static cling. Apparently they leave a residue that some people are allergic to.
Steph, I have friends who swear by Airborne. Vitamin C and zinc have been shown to shorten the duration of colds; echinacea has not so far been proven to do squat. And, congrats on the gym and the healthy dinner! You are now my rolemodel.
So many pretty clothes! I need more. Perhaps this time I'll buy them in sizes that actually fit me.
Happy Anniversary, Nora & Tom!
~ma to everyone who needs it! Too many Buffistas having sad times, lately.
I have one picture of us at the wedding this past weekend. I look like a basketball in a black dress.
The boy crashed out on the chair tonight.
Oh thank God! I've spent the better part of the past two nights wrestling with integrating the equation for a parabola (mostly wrestling with the arithmetic -- going from -b^3+2b^2sqrt(b^2-4ac)-b^3+4abc+.... to something slightly more compact just to get 1/6th of the equation written) and it has, despite all expectations, come out right (that is to say, I either had no stray dropped negatives in there, or I had enough that I could cancel them out!). It's taken up about five sheets of notebook paper... now to type it up semi-concisely. But if any of you need some kind of proof about the relationship between the area of a parabolic segment and the triangle inscribed therein, I'm your girl!
Oh frell. I just realized that if I was going to assume it crossed the x-axis, I could just as well have assumed one endpoint was at 0, and then I would only have had to calculate from (-b+sqrt(b^2-4ac))/2a to 0, rather than from that to (-b-sqrt(b^2-4ac))/2a, but no. Instead I did twice the work, because I am a mow-ron.
(Sorry folks. Ignore the math and it'll go away. Just needed to vent.)
With the black dress and black tux we blend together like some freakish conjoined twins.
Gorgeous couple, cutiehead sleeping toddler
Annabel is mowing her way through a slice of pepperoni pizza. I never know what to expect when I give her foods these days.
The "pining for the fjords" sketch is on Monty Python on BBC America.
I hope I can get through the conference this weekend without my head exploding. In a possible bout of insanity, I volunteered to do the same job next year....
How do you inscribe a triangle within a parabola? One point in the peak, one at each x-axis crossing? (assuming it's a parabola with exactly two real solutions)
Heh. Math geekery.
Oh, sorry, did I say that? It's meant to be a parabolic segment and its inscribed triangle, so I've said, basically, wherever the line is which cuts off the segment, move the segment so that line is at the x-axis and the segment is above it (so yes, the points of the triangle will be at the x crossings and at the vertex).
