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Need Math Help?

Merkuri

Smash Lord
Joined
Apr 1, 2010
Messages
1,861
Thanks drugged fox.

Can someone please tell me what the variable K in summation notation(or general algebra?) represents? This variable is repeatedly used in equation where K has not be defined. I would really appreciate it if someone could help me out with this. Please and thank you.
 

Merkuri

Smash Lord
Joined
Apr 1, 2010
Messages
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it's used in direct variation equations i think.
Despite the fact that I am doing summation notation I am not actually good at math. What does the K variable actually mean? How am I supposed to read Xk?

Figured it out. I appreciate anyone who was trying to help.
 

forward

Smash Champion
Joined
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Tucson Arizona
No shiv I am subscribed to this topic. And I can't name search because any thread (all of them) with reference to a forward tilt/smash and looking forward to the next tournament come up if I name search.
 

Merkuri

Smash Lord
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Can someone explain to me what the ^ symbol means in math? I have no idea.

It means that whatever number that comes before the symbol should be raised to the power of the symbol that comes after it.
 

Winnar

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oh my goodness how did I not know this thread existed

Linear algebra isn't my strong suit so I'll let this one slide to someone else
 

terr13

Smash Journeyman
Joined
Sep 13, 2006
Messages
268
So demanding, and some confusing notation as well. Is that e^(3tsin8t) or e^(3t) * sin(6t)?
Anyways, assuming it goes A B C D read from left to right, top down, the inverse would be the matrix
(D -B -C A)/Det, where the determinant is AD-BC.
 

ruhtraeel

Smash Ace
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Messages
707
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Calgary, Alberta, Canada
So demanding, and some confusing notation as well. Is that e^(3tsin8t) or e^(3t) * sin(6t)?
Anyways, assuming it goes A B C D read from left to right, top down, the inverse would be the matrix
(D -B -C A)/Det, where the determinant is AD-BC.
It's
e^(3t) * sin(8t)

I tried the determinant formula, but it comes out as something ridiculously complex and wrong LOL

My answer was:

a = (-6e^(3t)sin(8t))/(1/((-12e^(5t)(sin(8t))^2)+12e^(5t)(cos(8t))^2))

b = (2e^(3t)cos(8t))/(1/((-12e^(5t)(sin(8t))^2)+12e^(5t)(cos(8t))^2))

c = (6e^(2t)cos(8t))/(1/((-12e^(5t)(sin(8t))^2)+12e^(5t)(cos(8t))^2))

d = (2e^(2t)sin(8t))/(1/((-12e^(5t)(sin(8t))^2)+12e^(5t)(cos(8t))^2))


I have unlim tries
 

terr13

Smash Journeyman
Joined
Sep 13, 2006
Messages
268
The determinant should come out nicely. The power of e for the two terms should be the same, so you can factor it out. Then you have sin^2 + cos^2 = 1, and the coefficients should match up. That's just my guess, they always make the answers come out nicely.
 

ruhtraeel

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The determinant should come out nicely. The power of e for the two terms should be the same, so you can factor it out. Then you have sin^2 + cos^2 = 1, and the coefficients should match up. That's just my guess, they always make the answers come out nicely.
Perhaps I am going about this the wrong way. How about using the matrix inverse algorithm to solve it?

[A I] = [I A^(-1)]


EDIT: Geez, now I'm getting cot(8t) and a bunch of other wierd things. I reached

[1 -e^t*cot(8t)]
[0 -6e^(3t)*sin(8t)+6e^(2t)cos(8t)]

multiplied by

[1/(2e^(2t)*sin(8t) 0]
[0 6e^(2t)*cos(8t)]



Now I have to make the first matrix = to 1 0 0 1(identity) by doing row operations. But I ask you, what can you multiply to Row 2 to make it the same as -e^t*cot(8t) to get 0???
 

terr13

Smash Journeyman
Joined
Sep 13, 2006
Messages
268
Ah, that works as well, and is better in general, but when they give you a nice 2x2 matrix I just get too tempted to shortcut it. It will get messy extremely fast if you do it that way. Let me try to get the determinant.

If A = [2e^(2t)sin(8t)....... −2e^(3t)cos(8t)]
.........[−6e^(2t)cos(8t)..... −6e^(3t)sin(8t)]

Then determinant = -12e^(5t) (sin(8t))^2 -12e^(5t) (cos(8t))^2
By sin^2+cos^2 = 1, we have -12e^(5t).
 

ruhtraeel

Smash Ace
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Wolfram got it right.

Determinant was: -12e^(-5t)

a = ((e^(-2t))/6)*(3sin(8t))

b = ((e^(-2t))/6)*(-cos(8t))

c = ((e^(-2t))/6)*((-3e^(-t)*cos(8t)))

d = ((e^(-2t))/6)*(-e^(-t)*sin(8t))



All in all, I'm too mindgamey. Math requires more tech skill.
 

terr13

Smash Journeyman
Joined
Sep 13, 2006
Messages
268
Can you clean that up a little bit, I really can't tell what you're trying to write.

EDIT: Math is all mindgamey, just need basic tech skills.
 

Velox

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Feb 14, 2007
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Somebody talk about tensors. They're molesting me, but I think I remember what they look like for the sketch artist. Just give me four profound sentences about them in general or something, I have a test on Wednesday over E&M and relativity tensor stuff. Like just something to make them seem less esoteric (I once saw this word used in a text book "bessel functions are no more esoteric than your familiar childhood acquaintances sine and cosine" and I lol'ed)
 

Sasha

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Oct 24, 2007
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Berkerey, CA
O_O How would you factor this?

6X^3(1-x^2)^-3(X+5)^3-3x^2(1-x^2)^-4(X+5)^5
I can't tell if this is one term with 6 powers or two terms with 3 powers each.

I also can't tell if you're raising the exponents to a power, or if you're raising the term to a power.

More parentheses and spaces would be nice. Then I can help ya. :)

@Velox:

I only know about tensors in theory. I don't have any mathematical practice with 'em. I know they're a way to show a relation between two sets of vectors. More detail below:

http://en.wikipedia.org/wiki/Tensor
 

Velox

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Well, I know generally what they are, they're pretty ways to denote transformations from what I can tell, but ugh, descriptions of them are always so unclear and needlessly technical..
 

Grandeza

Smash Master
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Nov 11, 2007
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Brooklyn,New York
Can someone explain to me how I can find the value of the six trigonometric functions for the quadrantal angles using a unit circle. I really don't understand it and would rather understand it then have to memorize them all =\
 

MintyFlesh

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I can't tell if this is one term with 6 powers or two terms with 3 powers each.

I also can't tell if you're raising the exponents to a power, or if you're raising the term to a power.
It's two terms with three powers, and I'm raising the term in parenthesis to a power.

6X^3(1-x^2)^-3(X+5)^3 - 3x^2(1-x^2)^-4(X+5)^5
 

Kinzer

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Kinzer
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2251-6533-0581
Could somebody give me the basic run-down of vectors and finding magnitude/direction and sum and difference of? Then some vectors with coefficients like 2->U + (-1/2->P). I'd also like to know how to work with some that involve the order of operations like "find ->R + ->S - ->V."

The textbook did not really give me any practical examples for my exercises nor did it (clearly) go over how to find angles of (/the difference of) two vectors, let alone just angles in general. It'd say that the answer was something along of the lines of "the direction of ->O + ->I," nothing like "135°."

Thank you so much in advance if you can help me with some Pre-Calc.

:093:
 

terr13

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Grandeza: For the quadrantal angles, just remember that Cosine represents the x-value, and sine represents the y value. When the angle is 0, then we are look at the point on the x access, to the right of 0. The value there would be (1,0), so cosine(0) = 1, and sine(0) = 0.

Kinzer: Vectors only have magnitude and direction, they are not fixed upon any point. You can move vectors around, and it doesn't change their magnitude and direction. Coefficients change the magnitude of the vector, but not the direction. Multiplying by -1 means turning the vector in the opposite direction, aka 180 degree. Adding Vectors is just putting the tail of one of the vectors at the head of the other, and then completing the triangle. IE A+B means put the tail of B at the head of A, and draw the vector from the tail of A to the head of B. To find the angle, you just use trigonometry.
EDIT: A 2 dimensional vector can be given by either magnitude and direction, or X and Y coordinates. You can switch between the two by using the formula x = r cos(theta) and y = r sin(theta), r^2 = x^2+y^2, where r is magnitude and theta is direction in angle. You can add vectors by adding their x and y components and then calculating r and theta.

Minty Flesh: I don't think that's factorable. I could be wrong, since I haven't taken Algebra in awhile, but I don't think you can get anything worthwhile. Can anybody else do this one?
 

terr13

Smash Journeyman
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I don't think that's legal. Do you mean something like this is valid?
2^(3^4) + 2^(6^2) = 2^(3^2) [2^(3^2)+2^(2)]

EDIT: Wolfram Alpha found nothing.
 

terr13

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Sep 13, 2006
Messages
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I guess it is kind of factorable, but probably not in the way that you're thinking. Say we have 16-2, which is the same as 2^(2^2) -2^(1^1). Then we can factor out a 2, but we have 2 (2^[(2^2)-1] -1), or 2(2^3-1). Doesn't really clean up anything.

EDIT: I don't mean to be rude, but to be honest I trust Wolfram Alpha over your teacher, but here's the link, in case I'm typing something wrong.
http://www.wolframalpha.com/input/?i=factor+x^(y^z)+-+x^(2y^(2z))
 

rsr2

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I have a feeling, I am going to visit this thread a lot in the next couple of months. I just started Algebra and that is my worst subject.
 

Grandeza

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For my math class, I have to write a page on a 17th century application of trigonometry. I googled it and have tried but have not found anything. Any help? =\ Thanks!
 

Dxt XXII

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Miami, Florida


I need help...... :(
The simplest way to look at it is that a revolution would be made up of an infinite amount of circles (one at every x value) with the radius being the y value at that point.

The area of each circle would then be (pi)r^2, and the volume of the revolution would be the addition of every single circle.

For this, you would plug the equation, x^2, as the radius into the equation for the area of a circle. You do this because the radius is the y value, which is dependent on x.

Since you need to find the addition of an infinite amount of areas, you would integrate the equation for the area, which is (pi)(x^2)^2, with x^2 being the y value, and therefore the radius.

Integrating that would give you (pi)(x^5)/5, to be evaluated over 1 and 3.

After you integrate, you would plug in the values 1 and 3, which would be F(3) - F(1) = (3.14)(3^5)/5 - (3.14)(1^5)/5, which would give you your answer.

Hope this helps :)
 

terr13

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Sep 13, 2006
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Not super confident on this but here's my take.

All the corners should move the same distance. If we look at the lower left corner, than with a single 90 degree turn of the square, it should end up where the upper right corner was. We can also see that the path that it takes is 1/4 of a circle, since it has turned 90 degrees. Then we can say the total path taken by that corner is equal to the circumference of that circle. We can also tell, that the radius of that circle is equal to S, since two clockwise turns of the square puts the lower left at the lower right, and it is 2S away from where it was before. So the circumference of the circle would be 2s pi.
 

Corpsecreate

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Ah I dont think that is quite correct, one 90 degree rotation will end up where the top left corner is, then the top right, then the bottom right, then the bottom left. The part thats tricky is that when a corner is on the bottom right, it doesnt move at all because it becomes the pivot point.
 

terr13

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It becomes the top left of the square, but the actual position will be where the top right currently is, since the square has moved forward by S. With 4 rotations, the square will have moved 4s forward, and all the corners will be in the same position relative to all others. I did realize about the pivot point, so I think the answer is actually slightly different. The bottom left and top right corners should both move the same, or S(pi)/2 in a single spin. The upper left corner actually moves a little bit more, since it moves 1/4 the circumference of the circle that inscribes the 4 squares, ie the radius of that circle is s(sqrt2) and not s. Therefore the answer would be s(pi)[2+sqrt(2)]/2.
 

Corpsecreate

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Your answer is the same as mine just expressed differently. If your interested I can offer another very similar question that is quite harder. I've done it myself but it would be nice to see if someone else gets the same answer :p
 

MintyFlesh

Smash Ace
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Jul 30, 2010
Messages
577
I ask here too much :p

How can you convert this to a rectangular equation?

r^2=cos(theta)

I replaced R^2 with x^2+ y^2, and then multipled both sides with r, to get

r(x^2+y^2)=rcos(theta)

Which became

r(x^2+y^2)=y

But after then I'm stuck D=
 
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