Homework Help Thread

Discuss literature (e.g. books, newspapers), educational studies (getting help or opinions on homework or an essay), and philosophy.

Unread postby Seven at One Stroke » Thu Oct 09, 2003 2:26 pm

Jiang Xun wrote:Hmmmm.....i never really thought 'bout it that way....i was thinking more in the context of the lack of distinguishing details...the fact that they have no names, the fact that their story can be transplanted to any time during a confilct.


That is certainly true. But IMO, since the story is titled "War" and not "at XXX on the XXX train" means that the author is talking about wars in general, even though there is one mention of the "King". You're right of course, about the story being able to be transplanted to any time, but the fact that it is detached from any time period to start out with (I know nothing about the author) makes it pretty obvious.

Whatever your arguments are, just make sure they support your thesis, that is all. The rest you can write however you want. I can see how you can use the lack of distinguishing detail to build a strong thesis but I can't think of one right now. Sorry.
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Unread postby Justin » Thu Oct 09, 2003 5:07 pm

If anyone has experience in Computer Theory and Finite Automata, Regular Expressions and the like I could use a study partner per say. This class is killing me and I could use a person to go over things with.
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Unread postby Harimau » Tue Oct 14, 2003 11:12 am

Some harder geometry for anyone interested.

Consider an equilateral triangle ABC, of sides a units. P is a point on BC. P is projected to a point P1 on AB, such that PP1 is perpendicular to AB. Similarly, P1 is projected to a point P2 on AC such that P1P2 is perpendicular to AC. Again, The point P2 is projected to a point P3 on BC such that P2P3 is perpendicular to BC. (Note that P3 does not equal P) This pattern is then repeated on to infinity.

Now as n approaches infinity, what do the shapes of the points P(n-1), P(n) and P(n+1) tend towards? Find the length of the sides of this shape.
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Unread postby Seven at One Stroke » Tue Oct 14, 2003 7:07 pm

Harimau wrote:Some harder geometry for anyone interested.

Consider an equilateral triangle ABC, of sides a units. P is a point on BC. P is projected to a point P1 on AB, such that PP1 is perpendicular to AB. Similarly, P1 is projected to a point P2 on AC such that P1P2 is perpendicular to AC. Again, The point P2 is projected to a point P3 on BC such that P2P3 is perpendicular to BC. (Note that P3 does not equal P) This pattern is then repeated on to infinity.

Now as n approaches infinity, what do the shapes of the points P(n-1), P(n) and P(n+1) tend towards? Find the length of the sides of this shape.


Sorry, I've never seen this type of problems before (never took any geometry higher than than high school geometry) so I can only give you a rough sketch of what I think is the answer.

First of all, there can only be one answer if there is a shape as n->lazy 8, and that is an equilateral triangle inside the big one. Otherwise there is no definitive shape to speak of. Further more, the projection of the points make 3 nice 1-2-root3 triangle for every three points. So if indeed the shape approaches that triangle, then the total length of the three perpendicular projections will be a×root3.

Now let's try and show that the shape does tend toward that triangle we think it does. we'll consider a set of three projections at one time. Let the original length (which is also the hypotenus of the first triangle drawn) be x0, then because of the 60degree right triangle, the second hypotenus will be a-x0/2, and the third one will be a-x1/2, and the fourth, will be a-x2/2.

Since we assume that there is a difference between the length x0 and x3 (which we call z), then it's easy to calculate z=(6a-9x0)/8. If we plug in 2a/3 as x0, then the small equilateral triangle simply repeats itself as z=0. But if we consider x0<2a/3 or x0>2a/3, then we can see that, if x0<2a/3, then z is positive, meaning that the x0 for the next set of 3 projections will be closer to 2a/3; while if x0>2a/3, then z becomes negative, meaning the x0 for the next set will be smaller, and closer to 2a/3 as well.

There's still the problem of how much the increment z is, and we have to comfirm that it actually makes the x0 for the next set closer to 2a/3 in the case that x0+z jumps from less than 2a/3 to greater...but I'm not that far yet. Hope it helps (which I doubt, since I've never seen problems like this before).
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Unread postby Jiang Xun » Mon Oct 20, 2003 6:47 pm

This is not a homework question, but a question i've had for like...two or three years, and nobody has been able to answer it for me...

Ok...Humans breath air, which is Nitrogen and Oxygen, we use the Oxygen, but what happens to the Nitrogen? If it dissolves in the blood and doesn't go anywhere wouldn't you get a form of the "Bends"? And if you exhaled it, wouldn't you be able to create a larger flame if you blew on it? By larger i mean almost like using hair spray, not fanning the flame...

Just wondering....
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Unread postby Seven at One Stroke » Mon Oct 20, 2003 9:43 pm

Jiang Xun wrote:This is not a homework question, but a question i've had for like...two or three years, and nobody has been able to answer it for me...

Ok...Humans breath air, which is Nitrogen and Oxygen, we use the Oxygen, but what happens to the Nitrogen? If it dissolves in the blood and doesn't go anywhere wouldn't you get a form of the "Bends"? And if you exhaled it, wouldn't you be able to create a larger flame if you blew on it? By larger i mean almost like using hair spray, not fanning the flame...


Gaseous nitrogen has the formula :N≡N:, which is an extremely stable molecule, mostly inert and doesn't participate in many chemical reactions. To put it simply, the nitrogen that goes in goes out when you breathe out.

Humans normally have the same amount of pressure inside and out, so if one liter of gas comes into your body, it stays at one liter inside your body, whether it's dissolved or not. It's like if you eat a quarter pound steak, your body weight increases by a quarter pound, no more no less. When you're breathing in nitrogen, only a small fraction of the nitrogen is dissolved in your blood; the rest goes out as you breathe out. If you don't, then your lungs are full and no gas exchange can take place. Because the pressure doesn't change inside and outside your body, the volume of air stays constant, so there is no "bloating."

For a more comprehensive discussion of the bends with real scientific principles behind it, this is a very informative link here

Gaseous nitrogen is not flammable and doesn't combust very well (due to the strong triple bond). The requirements for a flame (as told by my local firefighter) are oxygen (flame is a form of oxidation), fuel, and heat. Nitrogen doesn't figure into any of the three. Whatever's in the hairspray is probably a nitrogen derivative (such as nitrate or amines etc) and much more reactive than the diatomic nitrogen in the air.
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Unread postby Rhiannon » Sun Oct 26, 2003 11:05 pm

Ech, I can't help you Alacard, I read those words and went "wha?". At least in the context I know they're in.

I could use an explanation here, myself, for my Critical Reasoning class, regarding the syntax of logic.

There are two logical operators, "If ... then" ( -> ) and "If and only If" (<->). The first is the conditional, the second is the biconditional. These I understand on their own.

However, in translating arguments in english into logical syntax, I am confused with the following statement:

"P only if Q."

According to my book, "only if" is the same as a conditional, but in reverse order. In other words, they say that argument translates into:

Q -> P.

This makes very little sense to me, and I don't understand why it would be that way.

But to complicate matters, when doing practice problems, I ran into an argument which the first premise in english was just this, P only if Q. I could not figure out how to prove the argument with propositional calculus, because I was using Q -> P as the book said 'only if' translated into. I glanced back and found the argument translation in the back of the book, and it was NOT Q -> P, as they had told me prior, but rather, P <-> Q. I grumbled, and assumed that they had made a typo (in typing 'only if' instead of 'if and only if'.)

It's only made matters worse, however. Now, not only do I not understand how 'only if' translates into a conditional and not a biconditional, but I now also have doubts as to the book's veracity on this. Someone... please. :D
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Unread postby Seven at One Stroke » Mon Oct 27, 2003 2:39 am

P Q (P only if Q)
t t t
f t t
t f f
f f t

from the truth table: P=>Q

Reasoning: P only if Q implies that: When P is true, Q must be true, but if P is not true, Q can be either true or false.

Truth be told I got it wrong too (until I opened up my $120 book) :(

By the way, which version of book do you have?
I'm using Douglas Smith's Transition to Advanced Mathematics, 5th edition.
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Unread postby Rhiannon » Mon Oct 27, 2003 5:07 pm

Hmm, I suppose that makes sense... so am I right in supposing their sample problem with "p only if q" translating into p<->q was indeed a typo?

And I'm using Schaum's Complete Outline of Logic, 4th edition (title may be off there, I don't have it with me today.)
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Unread postby Seven at One Stroke » Mon Oct 27, 2003 9:49 pm

Wild-Eyes wrote:Hmm, I suppose that makes sense... so am I right in supposing their sample problem with "p only if q" translating into p<->q was indeed a typo?


That is probably the case, but go and check with your professor or TA and show them your reasoning.

[Edit]

I just realized something funny. "P only if q" does indeed translate to "p=>q", but "p if only q" translates into "p<=>q". The reasoning goes like this:

if only means two things, if p then q, and if ~p then ~q, which, by the law of contrapositives, implies q=>p, therefore it is a biconditional.
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