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Author Topic: Mathematics Help Thread  (Read 228740 times)

Tsuchigumo550

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Re: Mathematics Help Thread
« Reply #1335 on: November 03, 2013, 05:31:34 pm »

Hmmm... I think I may also be doing the wrong thing, or at least trying to brute-force my way through this a bit too hard.

Oop, forgot a whole term.

-.0000000000003322(x-45)(x-55)(2x-160)(x-190)(x-200)((x^(4/3))-200)(x-250)
The one area with all the parenthesis reads "x to the power of four thirds minus 200."

f(250)=0, or it should anyway.

I've never dealt with the "replace with constants" thing, so I'm a little lost. Doing this for a pre-algebra class.
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Jim Groovester

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Re: Mathematics Help Thread
« Reply #1336 on: November 03, 2013, 05:42:47 pm »

f(250)=0, or it should anyway.

What exactly are you trying to evaluate?

If you're trying to evaluate that expression at x=250, I have good news for you, it's zero, since one of the factors in your expression evaluates to zero.
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Tsuchigumo550

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Re: Mathematics Help Thread
« Reply #1337 on: November 03, 2013, 05:47:24 pm »

I need those numbers to be multiplied out, but I don't really know how to handle that fractional exponent.

Where x=0, y =250. Or at least, it should; that's what the uncombined numbers give.

Long story short- I need all those numbers put together into a polynomial.
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1338 on: November 03, 2013, 05:59:50 pm »

WolframAlpha is your friend. Look under "expanded form".
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Jim Groovester

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Re: Mathematics Help Thread
« Reply #1339 on: November 03, 2013, 06:09:28 pm »

I need those numbers to be multiplied out, but I don't really know how to handle that fractional exponent.

No, what's the exact problem statement given to you?

I'm pretty sure you're going about this in entirely the wrong manner.
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Tsuchigumo550

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Re: Mathematics Help Thread
« Reply #1340 on: November 03, 2013, 06:38:26 pm »

I didn't know that existed and thank you thank you thank you.

The problem (or at least, goal) is to create a graph that starts at 250,0, goes under the X-axis for ten units, simulating a "rollercoaster". I probably shouldn't be using a fractional exponent, but I did. Like an idiot.

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Neonivek

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Re: Mathematics Help Thread
« Reply #1341 on: November 07, 2013, 07:37:32 am »

Well I am pretty much doomed, but I need help finding ways to extract variables from within an equation.

For example... in here I have

2/4 = r / (4-h)
r = 2h/4

How did that happen?

Not to mention...

2 = (Pi/12)*3*3^2X
X = (2*12^4)/(Pi*3*3^2) = 8/9pi (I have a feeling that 12^4 isn't written correctly)

I cannot remember how to extract variables from equations.
« Last Edit: November 07, 2013, 07:53:16 am by Neonivek »
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Another

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Re: Mathematics Help Thread
« Reply #1342 on: November 07, 2013, 08:04:33 am »

One of the basic principles is mathematics is that for every operation there is a kind of "undo" operation. Subtraction for addition, division for multiplication and so on. Also if 2 things are equal as indicated by equality sign between them - then if you simultaneously make the same operation on both of them - they will still be equal.

That being said - your first example is wrong because 2/4*(4-h)=r/(4-h)*(4-h) -> 2/4*(4-h)=r -> r=2-2/4*h=2-h/2
The second one is written a bit ambiguously and I assume is 2 = (Pi/12)*3*3^2*X instead of 2 = (Pi/12)*3*3^(2*X) because in the later case you would have logarithms. Confusingly the answer uses different convention of order of operations and should be X= 8/(9*Pi) with 12^4 part of intermediate just not making any sense.

In short - just get what part of the equation you need (X or r) to what part of the equation you want by applying reverse operations to both sides.
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Neonivek

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Re: Mathematics Help Thread
« Reply #1343 on: November 07, 2013, 08:17:09 am »

It sucks that my notes are not clear.

I can sort of understand where it came from since the ratio of the triangle doesn't change.

And the radius is half of the height.

Now if it was 2/4 = (2-r)/(4-h) then it would add up.

Since that would be... 2(4-h) = 4 (2-r)

(2(4-h))/4 = 2-r
(4-h)/2 = 2 - r
(4-4-h)/2 = -r
-h/2 = -r
r = h/2
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Furtuka

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Re: Mathematics Help Thread
« Reply #1344 on: November 11, 2013, 02:31:17 pm »

Could someone explain to me how the Gauss-Green formula for double integrals work? Particularly what m and n represent when its written in the form



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Pnx

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Re: Mathematics Help Thread
« Reply #1345 on: November 16, 2013, 10:07:08 pm »

So I'm doing these whole "applies maximum and minimum" problems, and it was going fine until I ran into ones where they abstract it... Like for example:

Find the height and radius of the cone of slant height L whose volume is as large as possible.

I know I should be able to find some values in terms of L, like if I make x equal to the radius, then solve for the height using x, I should then be able to find what x is in terms of L, but it just doesn't seem to work out how it should.
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Mego

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Re: Mathematics Help Thread
« Reply #1346 on: November 17, 2013, 02:44:29 am »

So I'm doing these whole "applies maximum and minimum" problems, and it was going fine until I ran into ones where they abstract it... Like for example:

Find the height and radius of the cone of slant height L whose volume is as large as possible.

I know I should be able to find some values in terms of L, like if I make x equal to the radius, then solve for the height using x, I should then be able to find what x is in terms of L, but it just doesn't seem to work out how it should.

First, we're going to need an equation of one variable for the volume.

Code: [Select]
V = 1/3*pi*h*r^2
h = sqrt(L^2 - r^2)

Substitute the formula for h in:
Code: [Select]
V = 1/3*pi*r^2*sqrt(L^2-r^2)

L is a constant, so V has only one variable: r. So now, you need to find dV/dr:

Code: [Select]
dV/dr = 2/3*pi*r*sqrt(L^2-r^2) - 1/3*pi*r^3/sqrt(L^2-r^2) = (pi*r*(2*L^2-3*r^2))/(3*sqrt(L^2-r^2))

So now, find the critical values. A few sanity bounds need to be established:

Code: [Select]
r^2 < L^2
r =/= 0

The first bound allows us to ignore the denominator. The second bound is due to the fact that the radius of a cone must be positive. So now we have the following:

Code: [Select]
(pi*r*(2*L^2-3*r^2)) = 0

The pi disappears due to being a constant, and we can break up the remaining factors:

Code: [Select]
r = 0
2*L^2 - 3*r^2 = 0

The first is illegal by our bounds. The second can be rearranged into something more meaningful:

Code: [Select]
3*r^2 = 2*L^2
r^2 = 2/3*L^2
r = sqrt(2/3*L^2) = L*sqrt(2/3)

And there you have it, a value for r at an extreme value of the volume function. Because the minimum volume would be 0 (which cannot happen), you can safely conclude that this is a maximum.

Descan

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Re: Mathematics Help Thread
« Reply #1347 on: November 18, 2013, 06:38:54 pm »

fuckthisbooksofuckinghardlikeseriouslyihateyou
« Last Edit: November 18, 2013, 06:47:06 pm by Descan »
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Dutchling

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Re: Mathematics Help Thread
« Reply #1348 on: November 21, 2013, 03:42:05 am »

I am sure I've learned how to do this in highschool, but evidently I forgot it.
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Jim Groovester

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Re: Mathematics Help Thread
« Reply #1349 on: November 21, 2013, 04:07:36 am »

Let Km = a
[S] = b
Vmax = c

Because typing all that out in an expression is annoying. Anyways,

(a + b)/bc = a/bc + b/bc = a/c * 1/b + 1/c

You follow?
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