Homework Help: Math

ocarinagirl · 02-24-2005, 04:57 PM

Now, this subject is the reason that I wanted to elect a different member to run. I don't like math. It is the bane of my very existance. I can't stand it. So it wouldn't make much sense for me to head a math topic, would it?

I'm open to suggestions as to who should be TEH MATHZ GURU!!!111 although in my mind, I'm thinking either TML or Jay would be excellent picks (you have to post sometime, Jay...mwahaha..)

Although I might be able to help with...Geometry. Or trig. A little. [img]tongue.gif[/img]

[sage] · 02-24-2005, 11:25 PM

Why do you need someone to be the guru, why not just let anyone who can answer a question answer.

ocarinagirl · 02-24-2005, 11:37 PM

I...said in my other topics that anyone can answer a question in a topic. The "guru" would just be the main person who helps. Other people can help if they want. [img]tongue.gif[/img]

But yeah, I guess I don't really NEED someone running a topic. I just thought it would be a good idea, but eh...=/

The Missing Link · 02-25-2005, 02:45 AM

I'll be happy to be math guru if you need one.

"There are some who call me... Link?"

"Carpe Gaium Domesticum!" (Seize the Cucco!)
Zelda: The Grand Adventures | Triforce MUCK
ザ行方不明リンク
悪いユウモアの賢人

Eagle Eye · 02-25-2005, 09:53 PM

SuperLuigi64 · 02-26-2005, 12:25 AM

I don't know bout Andre, but I can vouch for Team L being a math god ^_^

Swordmaster Link · 02-26-2005, 01:11 AM

Quote:

Originally posted by Marth:
Why do you need someone to be the guru, why not just let anyone who can answer a question answer?

Kukurio · 02-26-2005, 01:06 PM

Ok, I have a question. For homework, I'm supposed to prove that the zeros of the zeta function (which are not negative even integers) all have a real part equal to 1/2. Any tips?

ocarinagirl · 02-26-2005, 02:03 PM

Quote:

Originally posted by Swordmaster Link:
</font><blockquote>quote:</font><hr />Originally posted by Marth:
Why do you need someone to be the guru, why not just let anyone who can answer a question answer?

</font>[/quote]

Quote:

I...said in my other topics that anyone can answer a question in a topic. The "guru" would just be the main person who helps. Other people can help if they want.

Yaaaaaaay reading comprehension.

Swordmaster Link · 02-26-2005, 02:35 PM

I was just saying it to bust your chops, Andi. I saw your post. [img]tongue.gif[/img]

The Missing Link · 02-27-2005, 02:49 AM

Quote:

Originally posted by The Lurker:
Ok, I have a question. For homework, I'm supposed to prove that the zeros of the zeta function (which are not negative even integers) all have a real part equal to 1/2. Any tips?

You have to prove the Reimann Hypothesis for homework?? According to our friends at MathWorld, no proof of that has yet been constructed.

"There are some who call me... Link?"

"Carpe Gaium Domesticum!" (Seize the Cucco!)
Zelda: The Grand Adventures | Triforce MUCK
ザ行方不明リンク
悪いユウモアの賢人

imported_Andrew · 02-27-2005, 12:26 PM

I got a easy ass 10th grade problem I need help on, because I'm a dumbass (that and I can't remember the formula):

Given the parallelogram below, find the point of intersection of the diagonals.

Point locactions are, (-2, 5) (5,6) (-4,-3) (3,-2)

ready...go!

Swordmaster Link · 02-27-2005, 01:07 PM

Quote:

Originally posted by ShyGuy Wants Baseball Season:
I got a easy ass 10th grade problem I need help on, because I'm a dumbass (that and I can't remember the formula):

Given the parallelogram below, find the point of intersection of the diagonals.

Point locactions are, (-2, 5) (5,6) (-4,-3) (3,-2)

ready...go!

Easy indeed, now this I can solve.

Now, first you have to know that a diagonal is a line connecting two non-consecutive angles. (Or something, don't know the exact definition but you get it.) So essentially, what they're asking you for is the midpoint. Graphing out those points, you'll find that (-2,5) & (3,-2) are opposite each other, and the same goes for (5,6) & (-4,-3). So, all you have to do is find the midpoints for those two pairs of coordinates, and they should be equal. Not much of a formula here...for example:

Let's take (-2,5) & (3,-2) first. Now, you have to say in your head what you had to do to the first number to get the second number. So, to get from -2 to 3, you have to add 5. Cut that number in half, and you get 2.5. Add this number to -2, and you get .5. That's your x-midpoint for the first pair. Same thing for the y-midpoint. You subtracted 7 from 5 to get -2, so cut 7 in half and you get 3.5. Subtract 3.5 from 5 and you get 1.5. Thus, your midpoint for the first pair of coordinates is (.5,1.5). Since the problem is asking for an intersection, and given your math level, the next midpoint should be exactly the same. Let's go see.

Aha, now (5,6) and (-4,-3). Same thing. To get the x-midpoint, look for the pattern. You had to subtract 9 to get from 5 to -4, so cut 9 in half and you get 4.5. Subtract 4.5 from 5, and that gives you .5. Look familiar?

Ok, now look at the two y-coordinates. You had to subtract 9 to get from 6 to -3, so cut 9 in half and you get 4.5. Subtract 4.5 from 6 and you get none other than 1.5.

Thus, the point of intersection between the diagonals is (.5,1.5). The most important part is graphing the coordinates and knowing which verticies (sp?) you have to look at. Hope that makes sense...it takes ALOT less time to actually do it in your head than it took to explain it. [img]graemlins/lol.gif[/img]

***CaptHayfever*** · 02-27-2005, 01:25 PM

^^Interestingly enough, there was just an episode of Numb3rs about that a couple weeks ago.
Anyway, yeah, it hasn't been proven yet, and I think The Lurker is well aware of this fact.

And remember, "I'm-a Luigi, number one!"

The Missing Link · 02-27-2005, 01:51 PM

Alternate methodology (works for non-parallelogram quadrilaterals as well!):

As SML stated, you must find the two sets of points that actually create the two diagonals. This time, it's pretty easy since each of the points is in its own quadrant of the graph. So pick one diagonal as the one with points in Quadrant II and IV and the second with Quadrants I and III.

Create two parametric equations to describe the two diagonals as follows:

Diagonal from (-2, 5) to (3, -2):
x(t) = x1 + t(x2 - x1) = -2 + 5t,
y(t) = y1 + t(y2 - y1) = 5 - 7t.

Diagonal from (5, 6) to (-4, -3)
x'(u) = x1' + u(x2' - x1') = 5 - 9u
y'(u) = y1' + u(y2' - y1') = 6 - 9u.

Whenever 1 >= t >= 0 and 1 >= u >= 0, both equations will describe a point on the diagonal. [This is because, in both equations, (x, y)(0) = (x1, y1) and (x, y)(1) = (x2, y2), as well as (x', y')(0) = (x1', y1') and (x', y')(1) = (x2', y2').] The goal is to find the t and u value that will solve the system.

Take the x-equations:

x(t) = -2 + 5t
x'(u) = 5 - 9u.

Since at the intersection point the x's must be equal, so at that point, x(t) = x'(u). Thus,

-2 + 5t = 5 - 9u
-7 + 5t = -9u
u = 7/9 - 5t/9.

Since the y's must also be equal, we set y(t) = y'(u), substitute u = 7/9 - 5t/9, and solve for t.

5 - 7t = 6 - 9u
5 - 7t = 6 - 9(7/9 - 5t/9)
5 - 7t = 6 - 7 + 5t
5 - 7t = -1 + 5t
6 = 12t
t = 1/2

[Thus also means that u = 7/9 - 5/9(1/2) = (14 - 5)/18 = 9/18 = 1/2.]

Thus, the intersection occurs when t = 1/2 (and when u = 1/2). Thus, we find (x, y)(1/2).

x(1/2) = -2 + 5(1/2) = -2 + 5/2 = 1/2
y(1/2) = 5 - 7(1/2) = 5 - 7/2 = 3/2

[As a check, we compute (x', y')(1/2).

x'(1/2) = 5 - 9(1/2) = 5 - 9/2 = 1/2
y'(1/2) = 6 - 9(1/2) = 6 - 9/2 = 3/2.]

Thus, the intersection occurs at (1/2, 3/2).

"There are some who call me... Link?"

"Carpe Gaium Domesticum!" (Seize the Cucco!)
Zelda: The Grand Adventures | Triforce MUCK
ザ行方不明リンク
悪いユウモアの賢人

Kukurio · 02-27-2005, 04:52 PM

Quote:

Originally posted by The Minish Link:
You have to prove the Reimann Hypothesis for homework?? According to our friends at MathWorld, no proof of that has yet been constructed.

I was just trying to see if anyone noticed.

FYI, it's Riemann, not Reimann.

[ February 27, 2005, 04:54 PM: Message edited by: The Lurker ]

imported_Andrew · 02-27-2005, 07:28 PM

You guys (and that MathWorld site, good god) make my head hurt. Math will be the death of me. But thanks

ocarinagirl · 02-27-2005, 09:04 PM

^Now you know why I'm in no position to help with math. Because I think my head just exploded. =O

Damn you, TML.

Swordmaster Link · 02-27-2005, 09:21 PM

Yeah, TML's way looks confusing.

The Missing Link · 02-28-2005, 01:54 AM

That's only because you shortcut all of the "special-case" steps by it being a parallelogram.

Not to mention that TML's way is used in computer graphics as a means of rendering a 2D display.

"There are some who call me... Link?"

"Carpe Gaium Domesticum!" (Seize the Cucco!)
Zelda: The Grand Adventures | Triforce MUCK
ザ行方不明リンク
悪いユウモアの賢人

02-24-2005, 04:57 PM	#1
ocarinagirl Senior Member Join Date: Jan 2001 Location: Ehhh? What? Where am I?! Gender: Posts: 3,010 Thanks: 0 Thanked 0 Times in 0 Posts	Now, this subject is the reason that I wanted to elect a different member to run. I don't like math. It is the bane of my very existance. I can't stand it. So it wouldn't make much sense for me to head a math topic, would it? I'm open to suggestions as to who should be TEH MATHZ GURU!!!111 although in my mind, I'm thinking either TML or Jay would be excellent picks (you have to post sometime, Jay...mwahaha..) Although I might be able to help with...Geometry. Or trig. A little. [img]tongue.gif[/img]

02-24-2005, 11:25 PM	#2
[sage] Veteran Member Join Date: Dec 2000 Location: Florida Gender: Posts: 18,304 Thanks: 1 Thanked 13 Times in 10 Posts	Why do you need someone to be the guru, why not just let anyone who can answer a question answer.

02-24-2005, 11:37 PM	#3
ocarinagirl Senior Member Join Date: Jan 2001 Location: Ehhh? What? Where am I?! Gender: Posts: 3,010 Thanks: 0 Thanked 0 Times in 0 Posts	I...said in my other topics that anyone can answer a question in a topic. The "guru" would just be the main person who helps. Other people can help if they want. [img]tongue.gif[/img] But yeah, I guess I don't really NEED someone running a topic. I just thought it would be a good idea, but eh...=/

02-25-2005, 02:45 AM	#4
The Missing Link Professional Lurker Join Date: Dec 2000 Location: New Hyrule, Washington, US Gender: Posts: 17,128 Thanks: 405 Thanked 606 Times in 368 Posts	I'll be happy to be math guru if you need one. "There are some who call me... Link?" "Carpe Gaium Domesticum!" (Seize the Cucco!) Zelda: The Grand Adventures \| Triforce MUCK ザ行方不明リンク悪いユウモアの賢人

02-25-2005, 09:53 PM	#5
Eagle Eye Diddy bops Join Date: Feb 2004 Location: Birmingham, AL Gender: Posts: 15,014 Thanks: 1,616 Thanked 1,084 Times in 597 Posts	Same

02-26-2005, 12:25 AM	#6
SuperLuigi64 Senior Member Join Date: Jun 2000 Location: Louisiana Gender: Posts: 7,699 Thanks: 0 Thanked 0 Times in 0 Posts	I don't know bout Andre, but I can vouch for Team L being a math god ^_^

02-26-2005, 01:06 PM	#8
Kukurio Senior Member Join Date: Apr 2001 Posts: 3,645 Thanks: 0 Thanked 0 Times in 0 Posts	Ok, I have a question. For homework, I'm supposed to prove that the zeros of the zeta function (which are not negative even integers) all have a real part equal to 1/2. Any tips?

02-26-2005, 02:35 PM	#10
Swordmaster Link Super Bodyguard & King of the Arcade Join Date: Apr 2000 Location: Wherever you want me to be Gender: Posts: 32,132 Thanks: 253 Thanked 951 Times in 640 Posts Blog Entries: 2	I was just saying it to bust your chops, Andi. I saw your post. [img]tongue.gif[/img]

02-27-2005, 12:26 PM	#12
imported_Andrew Banned Join Date: Dec 1969 Posts: 0 Thanks: 0 Thanked 0 Times in 0 Posts	I got a easy ass 10th grade problem I need help on, because I'm a dumbass (that and I can't remember the formula): Given the parallelogram below, find the point of intersection of the diagonals. Point locactions are, (-2, 5) (5,6) (-4,-3) (3,-2) ready...go!

02-27-2005, 01:25 PM	#14
*CaptHayfever* Smash SuperMod Join Date: Jul 2002 Location: (n) - the place where I am Gender: Posts: 27,661 Thanks: 1,991 Thanked 2,486 Times in 1,513 Posts	^^Interestingly enough, there was just an episode of Numb3rs about that a couple weeks ago. Anyway, yeah, it hasn't been proven yet, and I think The Lurker is well aware of this fact. And remember, "I'm-a Luigi, number one!"

02-27-2005, 01:51 PM	#15
The Missing Link Professional Lurker Join Date: Dec 2000 Location: New Hyrule, Washington, US Gender: Posts: 17,128 Thanks: 405 Thanked 606 Times in 368 Posts	Alternate methodology (works for non-parallelogram quadrilaterals as well!): As SML stated, you must find the two sets of points that actually create the two diagonals. This time, it's pretty easy since each of the points is in its own quadrant of the graph. So pick one diagonal as the one with points in Quadrant II and IV and the second with Quadrants I and III. Create two parametric equations to describe the two diagonals as follows: Diagonal from (-2, 5) to (3, -2): x(t) = x1 + t(x2 - x1) = -2 + 5t, y(t) = y1 + t(y2 - y1) = 5 - 7t. Diagonal from (5, 6) to (-4, -3) x'(u) = x1' + u(x2' - x1') = 5 - 9u y'(u) = y1' + u(y2' - y1') = 6 - 9u. Whenever 1 >= t >= 0 and 1 >= u >= 0, both equations will describe a point on the diagonal. [This is because, in both equations, (x, y)(0) = (x1, y1) and (x, y)(1) = (x2, y2), as well as (x', y')(0) = (x1', y1') and (x', y')(1) = (x2', y2').] The goal is to find the t and u value that will solve the system. Take the x-equations: x(t) = -2 + 5t x'(u) = 5 - 9u. Since at the intersection point the x's must be equal, so at that point, x(t) = x'(u). Thus, -2 + 5t = 5 - 9u -7 + 5t = -9u u = 7/9 - 5t/9. Since the y's must also be equal, we set y(t) = y'(u), substitute u = 7/9 - 5t/9, and solve for t. 5 - 7t = 6 - 9u 5 - 7t = 6 - 9(7/9 - 5t/9) 5 - 7t = 6 - 7 + 5t 5 - 7t = -1 + 5t 6 = 12t t = 1/2 [Thus also means that u = 7/9 - 5/9(1/2) = (14 - 5)/18 = 9/18 = 1/2.] Thus, the intersection occurs when t = 1/2 (and when u = 1/2). Thus, we find (x, y)(1/2). x(1/2) = -2 + 5(1/2) = -2 + 5/2 = 1/2 y(1/2) = 5 - 7(1/2) = 5 - 7/2 = 3/2 [As a check, we compute (x', y')(1/2). x'(1/2) = 5 - 9(1/2) = 5 - 9/2 = 1/2 y'(1/2) = 6 - 9(1/2) = 6 - 9/2 = 3/2.] Thus, the intersection occurs at (1/2, 3/2). "There are some who call me... Link?" "Carpe Gaium Domesticum!" (Seize the Cucco!) Zelda: The Grand Adventures \| Triforce MUCK ザ行方不明リンク悪いユウモアの賢人

02-27-2005, 07:28 PM	#17
imported_Andrew Banned Join Date: Dec 1969 Posts: 0 Thanks: 0 Thanked 0 Times in 0 Posts	You guys (and that MathWorld site, good god) make my head hurt. Math will be the death of me. But thanks

02-27-2005, 09:04 PM	#18
ocarinagirl Senior Member Join Date: Jan 2001 Location: Ehhh? What? Where am I?! Gender: Posts: 3,010 Thanks: 0 Thanked 0 Times in 0 Posts	^Now you know why I'm in no position to help with math. Because I think my head just exploded. =O Damn you, TML.

02-27-2005, 09:21 PM	#19
Swordmaster Link Super Bodyguard & King of the Arcade Join Date: Apr 2000 Location: Wherever you want me to be Gender: Posts: 32,132 Thanks: 253 Thanked 951 Times in 640 Posts Blog Entries: 2	Yeah, TML's way looks confusing.

02-28-2005, 01:54 AM	#20
The Missing Link Professional Lurker Join Date: Dec 2000 Location: New Hyrule, Washington, US Gender: Posts: 17,128 Thanks: 405 Thanked 606 Times in 368 Posts	That's only because you shortcut all of the "special-case" steps by it being a parallelogram. Not to mention that TML's way is used in computer graphics as a means of rendering a 2D display. "There are some who call me... Link?" "Carpe Gaium Domesticum!" (Seize the Cucco!) Zelda: The Grand Adventures \| Triforce MUCK ザ行方不明リンク悪いユウモアの賢人