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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 1:59:54 PM

MissAsylum

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Omg...so much math speak.....

Doctor,my head.

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 2:00:36 PM

mnottertail

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It pays dividends.....or is it divisors?

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 2:34:30 PM

VaguelyCurious

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quote:

ORIGINAL: MasterG2kTR

Can't be that either because you have no idea what the value of x or y is. You can only end up with an exponential expression without a definitive solution until you know the actual values of x & y.

No no no, you've misunderstood; we're not talking about the whole expression, just the integer that x^5 and y^8 are multiplied by.

And I'm glad that the sixes were fictional, because I REALLY DIDN'T UNDERSTAND. But now we have an expression that we're all agreed on, right?

(3888^(1/4))(x^(5/4))(y^2)

Right?

Speak now, or forever hold your peace.

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 3:19:48 PM

Hippiekinkster

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42.

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 3:51:17 PM

MasterG2kTR

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quote:

ORIGINAL: VaguelyCurious

quote:

ORIGINAL: MasterG2kTR

Can't be that either because you have no idea what the value of x or y is. You can only end up with an exponential expression without a definitive solution until you know the actual values of x & y.

No no no, you've misunderstood; we're not talking about the whole expression, just the integer that x^5 and y^8 are multiplied by.

And I'm glad that the sixes were fictional, because I REALLY DIDN'T UNDERSTAND. But now we have an expression that we're all agreed on, right?

(3888^(1/4))(x^(5/4))(y^2)

Right?

Speak now, or forever hold your peace.

gaaaaaaahhhhhhhhhhhhh!!! not even close.....(sigh).......

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 3:58:27 PM

IrishMist

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LMAO

I showed this to my daughter...who is the math whiz in this family lol...she has been trying to work it...she finally told me that if she ever comes across a problem like this, she's gonna burn her book

She's still trying to figure it out though lol...so fucking cute

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 4:01:03 PM

VaguelyCurious

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quote:

ORIGINAL: MasterG2kTR

gaaaaaaahhhhhhhhhhhhh!!! not even close.....(sigh).......
.

But that's...exactly the same as what you said. Look.

You said it was this: 6xy^2 4^√3x
I said it was this: (3888^(1/4))(x^(5/4))(y^2)

So I'm going to write them as an identity. (I have no idea what the code for an identity symbol is. I'm just gonna use an equals, so sue me):

(3888^(1/4))(x^(5/4))(y^2) = 6xy^2 4^√3x

now 3888^(1/4) = 6(3^(1/4)), so we can divide both sides by that, leaving:

(x^(5/4))(y^2) = x(y^2)(x^1/4))

divide both sides by y^2 and we have:

x^(5/4) = x(x^(1/4))

Which is obviously true.

So that thing which is 'not even close' is exactly the same.

Maybe you oughta stop hitting your head against that wall, it looks painful.

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 4:03:12 PM

VaguelyCurious

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quote:

ORIGINAL: IrishMist

LMAO

I showed this to my daughter...who is the math whiz in this family lol...she has been trying to work it...she finally told me that if she ever comes across a problem like this, she's gonna burn her book

She's still trying to figure it out though lol...so fucking cute

I promise you it's not as bad as it looks. It looks weird because we're writing it in flat lines, which means you need shitloads of brackets and stupid hat things ^^. If we wrote it out by hand or with proper equation code it would be a lot clearer and simpler.

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 4:19:48 PM

MasterG2kTR

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Ok....I'm gonna take a final stab at this one to explain it and break it down

the original problem is

in the first segment the expression simplifies to xy^2(^4√3x) or 1xy^2(^4√3x)
the second segment simplifies to                       3xy^2(^4√3x)
the third segment simplifies to                           2xy^2(^4√3x)

giving you a total of                                          6xy^2(^4√3x)
there is one other way you can re-write this      6xy^2(3x^-4) that's 3x to the negative fourth power
these are the only two possible answers to the original problem

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 4:22:47 PM

DomKen

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quote:

ORIGINAL: VaguelyCurious

quote:

ORIGINAL: IrishMist

LMAO

I showed this to my daughter...who is the math whiz in this family lol...she has been trying to work it...she finally told me that if she ever comes across a problem like this, she's gonna burn her book

She's still trying to figure it out though lol...so fucking cute

I promise you it's not as bad as it looks. It looks weird because we're writing it in flat lines, which means you need shitloads of brackets and stupid hat things ^^. If we wrote it out by hand or with proper equation code it would be a lot clearer and simpler.

written out in a reasonably clear one line format it is:
(x*(3x*y^8)^1/4) + (y*(243x^5*y^4)^1/4) + (y^2*(48x^5)^1/4)

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 4:29:45 PM

VaguelyCurious

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quote:

ORIGINAL: MasterG2kTR

these are the only two possible answers to the original problem

Bullshit. They aren't the 'only possible answers'. You've just failed to simplify the x terms, that's all.

You can take this: 6xy^2(^4√3x)

and take the x out of that fourth root, so instead of (3x)^1/4, you have (3^(1/4))(x^(1/4))

and then you take that x^(1/4) and you multiply it by the first x, to give you x^(5/4)

giving you 6(x^(5/4))(y^2)(3^(1/4))

which is what I said in the first place.

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 4:34:54 PM

VaguelyCurious

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Also, this:

quote:

ORIGINAL: MasterG2kTR

there is one other way you can re-write this 6xy^2(3x^-4) that's 3x to the negative fourth power

is wrong. A negative power isn't a root, it's a reciprocal: a^-x=1/(a^x)

Is that what this whole argument's been about? Have you just misunderstood what a^(1/4) means?

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 4:37:08 PM

NakisisaX

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what is 6 and the square miles of missoula, montana, alex.

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 4:43:23 PM

dovie

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FR, This thread caused a bloomers change. Just saying.

dovie *whew!

< Message edited by dovie -- 10/12/2011 4:44:55 PM >

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 7:03:59 PM

LanceHughes

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quote:

ORIGINAL: MasterG2kTR

*looks for ruler*

Hold out those knuckles young man

(you know what's coming next I assume)

Yup....trip to the principle's office

GOOD! I hear he's quite the closet cock-sucker! LOL!

ETA: THANKS for the compliment given by calling me a "young man."

< Message edited by LanceHughes -- 10/12/2011 7:05:55 PM >

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 7:15:56 PM

LanceHughes

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quote:

ORIGINAL: VaguelyCurious

Also, this:

quote:

ORIGINAL: MasterG2kTR

there is one other way you can re-write this 6xy^2(3x^-4) that's 3x to the negative fourth power

is wrong. A negative power isn't a root, it's a reciprocal: a^-x=1/(a^x)

Is that what this whole argument's been about? Have you just misunderstood what a^(1/4) means?

QFT !

And the scary part is that he's a math teacher. At least all I did was have an extra factor of 2. But as quoted above: A negative power isn't a root, it's a reciprocal.

I saw that same "big" error and it just went HUH? in my brain. Kept reading to find yours quoted above.

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 9:03:33 PM

LanceHughes

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quote:

ORIGINAL: VaguelyCurious

Lance, I'm still getting a value that's half yours for that integer, and I don't know why.

The integer (if we're leaving everything inside that fourth root) is: (3^1/4 + 243^1/4 + 48^1/4)^4

243^1/4 = 3(3^(1/4))

48^1/4 = 2(3^(1/4))

so the part inside the 4th power is:

1(3^(1/4)) + 3(3^(1/4)) + 2(3^(1/4)) = 6(3^(1/4))

And when you ^4 that you get 3(6^4) = 3888

So I don't know why your figure is double mine.

'Cause yours is correct. I had a 6 where I shoulda, coulda, woulda had a 3. Hence, double yours.

< Message edited by LanceHughes -- 10/12/2011 9:14:39 PM >

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 9:13:38 PM

LanceHughes

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quote:

ORIGINAL: NakisisaX
what is 6 and the square miles of missoula, montana, alex.

I'll take "Weird Math" for $2,000, Alex.
Ans: Fourth root of 3
What is 6 plus the square miles of missoula and montana?

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 9:43:21 PM

HannahLynHeather

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sorry, but heather is asleep, and i'm not fucking waking her.

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RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 10:01:37 PM

HannahLynHeather

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hey lance, new mail.

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