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PostPosted: March 9th, 2007, 12:46 pm 
Teh Ladybird Fwee Queen
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:hide: x Lots.

I'll, erm, let someone else help you on that.

:erm:

*Gulps*

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PostPosted: March 9th, 2007, 6:22 pm 
Vala
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Long devision or synthetic?

Long devision is harder
Divide 2x^4+3X^3+5X-1 by X^2-2X+2
Turn the dividend to 2X^4+3^3+0X^2+5X-1 because when dividing polynomials all the powers have to be present.

Since there are no dividing signs on the computer, the things in brackets are being divided.
X^2-2X+2 [2X^4+3^3+0X^2+5X-1]

Take 2X^4 and divide it by X^2 to get 2X^2 and put the prduct over the third part of the divisor (the third part because there are tree parts in the dividend, X^2,-2X, and +2)


SYNTHETIC (easier and much,, much shorter.)
..................................2X^2
X^2-2X+2[2X^4+3^3+0X^2+5X-1]

Then multiply the first three parts of the divisor by 2X^2 and subtract.

..................................2X^2
X^2-2X+2[2X^4+3^3+0X^2+5X-1]
.............. -2X^4-4X^3+4X^2
_____________________________(that's a subtration line)
.........................7X^3-4x^2+5x (the 5X drops down like in normal division)

Then divide 7X^3 by X^2 and repeat. Keep doing this until you reach the end!


SYNTHETIC. Shorter and much, much easier

This only works if the divisor is a polynomial to the first power.

http://www.purplemath.com/modules/synthdiv.htm
I can't write it out on here. Take a lookat this site. It explains it well.


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PostPosted: March 10th, 2007, 11:16 am 
Balrog
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Thanks a lot Nauriel. I finally got it!!!! *yay* :wave:

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PostPosted: March 11th, 2007, 1:46 pm 
Maia
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Oh yeah Este....can you add History(Esc. the Crusades)

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PostPosted: March 11th, 2007, 4:12 pm 
Teh Ladybird Fwee Queen
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Aye, all done. ;)

Well done Nauriel for helping Fíriel_18190. :D

Wahey for success! :bounce:

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PostPosted: March 11th, 2007, 6:19 pm 
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Fíriel_18190 wrote:
Thanks a lot Nauriel. I finally got it!!!! *yay* :wave:

:blink: You understood that? I didn't even understand what I was writing! Your powers of perception amaze me! I'm glad I could help!


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PostPosted: March 12th, 2007, 10:02 am 
Futon-Revolutionist
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^I didn't! YAY! :P

*Patiently waits for a question he can answer*

Oh, you don't have my strengths up there. Uh...
World History
Latin (If I can't answer it myself I have my textbook. :P)
Science

Tah-dah!


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PostPosted: March 12th, 2007, 10:05 am 
Istari
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Hello I would like to join....I have problems * coughprealgebracough* sometimes,but my strong suits are Literature, Science, and Texas History

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PostPosted: March 12th, 2007, 2:36 pm 
Balrog
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Nauriel Rochnur wrote:
Fíriel_18190 wrote:
Thanks a lot Nauriel. I finally got it!!!! *yay* :wave:

:blink: You understood that? I didn't even understand what I was writing! Your powers of perception amaze me! I'm glad I could help!

Well the website also helped a bit. It's a really good site for Maths.

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PostPosted: March 12th, 2007, 3:48 pm 
Teh Ladybird Fwee Queen
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Astinos wrote:
^I didn't! YAY! :P

*Patiently waits for a question he can answer*

Oh, you don't have my strengths up there. Uh...
World History
Latin (If I can't answer it myself I have my textbook. :P)
Science

Tah-dah!

Thanks Finny. ;)

Mimsy wrote:
Hello I would like to join....I have problems * coughprealgebracough* sometimes,but my strong suits are Literature, Science, and Texas History

Welcome to the club. :happy:

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PostPosted: March 20th, 2007, 8:37 pm 
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So, if anyone wishes to help clarify physics for me, specifically torque and rotational motion, I'd be eternally grateful. My teacher can't teach, it's pretty much a learn by yourself class.

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PostPosted: March 20th, 2007, 10:50 pm 
Vala
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Same here. My teacher is nice and all, but he's not very good at teaching.
So...Physics....physics...physics! (That joke is wasted on all of you who don't watch Doctor Who)

Torque is rotational force. Say you have a .5 meter wrench. Using that wrench to tighten a bolt will have a set amount of torque. Now, if you had a 1 meter wrench and used the same amount of force you did with the .5 meter wrench, the torque would be twice as much. Or, if you used only half the force you could still get the same amount of work done.

There is an equation: torque=(distance to center)x(force) where distance is measured in meters and force in newtons.
So, if the wrench was .5 meters long and you used 1 neton of force to turn it, the torque would be .5 newton meters (Newton meters is the unit)
Therefore it stands to reason that if 1 newton of force was applied to a 1 meter wrench the torque would be 1 newton meter.

This equation isn't too hard, so I don't think I need to explain it any more.

On a similar note there is the equation (total force on lever arm A)x(distance to center)=(total force on lever arm B)x (distance from center)

This equation can be used to solve "seesaw" problems. Imagine a seesaw. One child weights 30 Kg and is sitting at the edge of the seesaw (This seesaw is 5 meters on both ends, making it 10 meters in total) Another child is 60kg and is also on the seesaw. The seesaw is balanced. So therefore the equation is 30 (5)= 60X 150=60X 2.5=X So the child is 2.5 meters away from the center of the seesaw.

I’m not quite sure what you need to know about rotational motion…its basically something traveling in a circle. I’m a bit foggy on that. I remember a little…tangential motion means that if the force holding the object in rotational motion is taken away, the object will move in a strait line (if there’s no gravity of air resistance)

Oh! And rotational inertia! (also called moment of inertia) The equation is I=M r² where I is rotaitional inertia, M is the mass, and r is the radius. So say there is a 10Kg mass at the end of a 5 meter string. The rotational inertia would be (10)x(5²), or 250. I’ve forgotten the units…could it be kg m² ? I think it is.

There was some other things…I wish I remembered…I’ll have to look it up tomorrow. I hope I’ve helped so far.

EDIT! I have more (oh joy oh joy)
Angular momentum = (rotational inertia)x (rotational velocity) and the units are kg m²/s

So if you wanted to find the angular momentum of the 10kg mass at the end of the 5m string that is traveling at 3 m/s
(angular momentum)=(rotational inertia)x(rotational velocity)
angular momentum= (250)x(3)
angular momentum=750 kg m²/s
Eh...for some reason that doesn't seem right. I don't know why...I'm pretty sure i did it right.

And now comes the last part...
conservation of angular momentum: If nothing external acts on a system that is already rotating, the angular mementum remains the same.


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PostPosted: March 22nd, 2007, 9:51 pm 
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Ah, the joys of Physics. Thanks Nauriel, that helped quite a bit!

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PostPosted: March 23rd, 2007, 10:28 am 
Balrog
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A Physics. I'm sooo glad I was able to drop it. I hate Physics.

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PostPosted: March 29th, 2007, 6:52 am 
Teh Ladybird Fwee Queen
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I'm in Physics now actually, I'm terrible at it, it scare me. :hide: :lol:

But for some reason unknown, the teacher has let us do 'whatever we want' on the laptops. :blink: Very odd.

Well done for helping out peeps, don't forget you can join in with explanations or add your own, don't feel you can't also explain if someone has already done so. :)

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PostPosted: March 29th, 2007, 11:55 am 
Maia
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Um... Some latin help would be hugely appreciated, I've been teaching myself from this book since saturday, but it's not too clear in places. When asking questions, does the verb go after the accusitive as usual, or before it?

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So...Physics....physics...physics! (That joke is wasted on all of you who don't watch Doctor Who)

Exactamundo :P (again, wasted on people who don't watch the show)

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