# Biscuits and gravy? Who the what?

1. Jun 18, 2006

### Pengwuino

I went to a restaurant this morning and had, among other things, biscuits and gravy. I then realized... what the hell is this? Does anyone else think biscuits and gravy seem very... out of place? It's like some kid just wanted something to eat, grabbed a biscuit and dunked it in a vat of gravy and said boom, staple. I'll never understand you people...

2. Jun 18, 2006

### Lisa!

Yeah, as we can never understand penguins too...

3. Jun 18, 2006

### wolram

I think the waiter saw that may be you were not ready for real solid food yet
and saw fit to serve you some thing your imature digestive system could handle :rofl:

4. Jun 18, 2006

### Staff: Mentor

Biscuits = flour, lard, and water

gravy = flour, lard, and water, just more water

5. Jun 18, 2006

### Cyrus

Evo, seriously.....lock this thread. It's utterly stupid. How about for every dumb thread he makes from now on, he has to solve a hard linear algebra problem

6. Jun 18, 2006

### Staff: Mentor

You know you have started a rash of members saying "thread locked" now? Right? :tongue:

Oh, and read my latest journal entry, I resisted the temptation to start another thread on my gardening woes.

And Cyrus, he admitted frying cheese in powdered sugar, surely that is worthy of a GOOBF card.

7. Jun 18, 2006

### Cyrus

Yeah, but at least your thread would have a point to it.

Oh man, because powered sugar looks soooooo much like flower....:uhh: The stuff doesnt even sift the same.......:uhh: Frying cheese in powdered sugar, wow.........im just speechless.

What's next, is he going to wipe his butt with sandpaper by accident?

Last edited: Jun 18, 2006
8. Jun 18, 2006

### Staff: Mentor

Awww, you're just jealous, come on...admit it.

9. Jun 18, 2006

### Danger

Ahem... regarding the original post...
I had absolutely never heard of the concept of biscuits and gravy until the first time that I went to Vegas to play pool. It's a bizarre concept, but it's pretty good if you overlook the fact that Yanks don't seem to have any idea of what gravy is.

Last edited: Jun 18, 2006
10. Jun 18, 2006

### Cyrus

Let T: V-> W be a linear transformation from a vector space V into a vector space W. Prove that the range of T is a subspace of W. [Hint: Typical elements of the range have the form T(X) and T(W) for some x,w in V.]

Have fun pengwuino....

Last edited: Jun 18, 2006
11. Jun 19, 2006

### Staff: Mentor

And now you're just being mean,

Found a sick kitten today, really sick. Bu it won't die. It's too weak to eat, distended bowels, should be dead. I finally found the painkiler my vet gave me for sick kittens. I just overdosed it. I'm in tears. I hope it goes soon. This sucks.

12. Jun 19, 2006

### Cyrus

You ended its suffering. There's nothing wrong with that. It was the only humane thing to do. Now, let's put some in pengwunio's food. You can always take it to the Vet. If he can't do anything about it they can put it to sleep.

13. Jun 19, 2006

### Staff: Mentor

Kitten just died, it's strong stuff, I hope it really stops the pain/

14. Jun 19, 2006

### Cyrus

City girl........moonbear woulda done it with a striaght face.......she's evilllllll...shhhhhh

15. Jun 19, 2006

### wolram

:yuck: Lard in gravy??? no oxo or meat juices

16. Jun 19, 2006

### Math Is Hard

Staff Emeritus
aw. I'm sorry you had to do that, Evo. Poor little critter.

17. Jun 19, 2006

### Math Is Hard

Staff Emeritus
As long as there's no powdered sugar in that gravy, I'm in. :tongue2:

18. Jun 19, 2006

### Staff: Mentor

Gravy on biscuits here is usually cream gravy, no meat.

19. Jun 19, 2006

### mattmns

Going easy on Pengwuion huh? This is a very easy proof

Proof:
The Proof is obvious and left as an exercise to the reader QED :tongue2:

OK (I will probably screw it up now that I said it is obvious :rofl:)
(i)let c be in R and let v',w' be in Range of T. That is there are some elements in V, say v and w, such that T(v) = v' and T(w) = w'. Since T is a linear transformation, T(cv + cw) = T(c(v+w)) = T(cv) + T(cw) = cv' + cw'

(ii) T is a linear transformation, so T(O) = O

Therefore the Range of T is a subspace of W QED.

edit... I am making some assumptions on a few words here. Specifically, when I took linear algebra we didn't use the word "Linear Transformation" (we used "Linear Map") and we didn't use "Range" (we used "image").

Last edited: Jun 19, 2006
20. Jun 19, 2006

### Cyrus

<shrug> I dont know I forgot all that crap two days after the class. :rofl:

I'm an engineer, I don't need to know that kinda crap.