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Author Topic: Mathematics Help Thread  (Read 228805 times)

Pnx

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Re: Mathematics Help Thread
« Reply #1215 on: September 15, 2013, 09:37:55 pm »

sqrt(4x^2+5x)+2x becomes sqrt(4+5/x)+2.
This part, right here. What are the rules going into this? I feel like I'm not knowing something I should really know.
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Skyrunner

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Re: Mathematics Help Thread
« Reply #1216 on: September 15, 2013, 10:56:48 pm »

(sqrt(4x2+5x)+2x)/x = sqrt(4x2+5x)/x + 2x/x
 
While normally you can't divide by x without risking an incorrect answer, when talking about x going to infinity, you're allowed to divide both denominator and numerator by x, since we already know x!=0.

Now look at 2sqrt(5). If you want to move the 2 inside the sqrt, you square it and multiply the inside to make sqrt(4*5)=sqrt(20). Likewise, sqrt(4x2+5x)/x becomes sqrt{(4x2+5x)/x2} = sqrt(4 + 5/x), and 2x/x=2.

So the equation in the first line becomes

sqrt(4+5/x)+2
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Errol

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Re: Mathematics Help Thread
« Reply #1217 on: September 16, 2013, 04:22:35 am »

sqrt(4x^2+5x)+2x becomes sqrt(4+5/x)+2.
This part, right here. What are the rules going into this? I feel like I'm not knowing something I should really know.

In terms of roots, sqrt(x*y) equals sqrt(x) * sqrt(y). What Sky's done here is writing 4x^2+5x as x^2*(4+5/x), then individually taking the square roots. This leads to the term becoming x*(sqrt(4+5/x)+2x = x*(sqrt(4+5/x) + 2). Cue cancellation of the x (why we're able to do that is something Sky explained)
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Descan

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Re: Mathematics Help Thread
« Reply #1218 on: September 17, 2013, 03:58:43 pm »

3(x – 6)2 + 4x(x + 3y + 2)
3(x212x + 36) + 4x2 + 12xy + 8x

Where the fuck does that 12x come from? :I
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da_nang

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Re: Mathematics Help Thread
« Reply #1219 on: September 17, 2013, 04:21:07 pm »

3(x – 6)2 + 4x(x + 3y + 2)
3(x212x + 36) + 4x2 + 12xy + 8x

Where the fuck does that 12x come from? :I
(a+b)2 = a2 + 2ab + b2
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Pnx

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Re: Mathematics Help Thread
« Reply #1220 on: September 17, 2013, 04:34:53 pm »

(x-6)2
(x-6) * (x-6)
When multiplying something like this, you multiply all the terms together, typically the easiest way to do this is to split one side apart and multiply it's little bits individually.
((x-6) * (x)) + ((x-6) * (-6))
((x)(x) + (-6)(x)) + ((x)(-6) + (-6)(-6))
(x2 - 6x) + (-6x + 36)
x2 - 12x + 36
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Descan

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Re: Mathematics Help Thread
« Reply #1221 on: September 17, 2013, 04:39:26 pm »

I think I understand, lemme see if I can get another equation to behave the same way.

edit: Yeah, I got it now. Hm. That wasn't explained very well in the book. :I
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Pnx

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Re: Mathematics Help Thread
« Reply #1222 on: September 17, 2013, 04:54:10 pm »

Pretty much every mathematics textbook I've ever been assigned has been horrible at explaining how to do the maths...
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Tarqiup Inua

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Re: Mathematics Help Thread
« Reply #1223 on: September 18, 2013, 09:52:29 am »

Imagine it like adding up the lengths of a line and then squaring the result...

A+B could be imagined like aaaaaabbb (the number of same characters would represent length of the line in this case)

so (A+B)^2 would be

x  aaaaaa bbb

a ++++++***
a ++++++***
a ++++++***
a ++++++***
a ++++++***
a ++++++***
b ****** ---
b ****** ---
b ****** ---

The area covered by + represents area of square with side of length of a. (that's your a^2)
The two areas covered by stasrs both represent area of rectangle with one side of length a and other of length b. (that's two times ab)
The area covered by - represents area of square with side of length of b. (that's your b^2)

I hope this makes it bit more intuitive for you
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1224 on: September 18, 2013, 10:26:21 am »

x aaaaaabbb

a ++++++***
a ++++++***
a ++++++***
a ++++++***
a ++++++***
a ++++++***
b ******---
b ******---
b ******---

Fix'd
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miauw62

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Re: Mathematics Help Thread
« Reply #1225 on: September 18, 2013, 10:43:56 am »

Uh, I have to note an interval so that 1/x < 1.
I got ]1; +infinity[ , but isn't it technically that AND ]-infinity ; -1[ ?
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Pnx

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Re: Mathematics Help Thread
« Reply #1226 on: September 18, 2013, 10:52:08 am »

Actually it would be -infinity to 0 (since all negative numbers are less than 1), then 1 to infinity. But you might not be bothering with that depending on how it's all worded.
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Tarqiup Inua

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Re: Mathematics Help Thread
« Reply #1227 on: September 18, 2013, 11:19:51 am »

x aaaaaabbb

a ++++++***
a ++++++***
a ++++++***
a ++++++***
a ++++++***
a ++++++***
b ******---
b ******---
b ******---

Fix'd
Yeah, looks more neat. Have a cookie! *passes*
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Kirbypowered

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Re: Mathematics Help Thread
« Reply #1228 on: September 19, 2013, 06:04:36 pm »

Blargh, this question is going to kill me.

Basically, I have one function, N(T), and another T(t). I need the composite function N(T(t)). I have the domain of N(T).

I have a feeling that since since N is a function of T, and T is a function of t, this works differently than usual, but I simply cannot find out how it works.
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1229 on: September 19, 2013, 06:16:39 pm »

Just take N and substitute all Ts with T(t)s, then substitute T(t).

Example: N = (T+1)², T = t²+t, then N(T(t)) = (T(t)+1)² = (t²+t+1)².
« Last Edit: September 19, 2013, 06:19:45 pm by MagmaMcFry »
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