Solve the equation for the person above!

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Zinrius
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Solve the equation for the person above!

Post by Zinrius » 13 Apr 2011, 21:53

Basically, post an equation and solve it, then make your own equation while MIXING the equation of the previous poster with yours. ( You can only add 2 more digits to the forumla )

Let me start =D

2 + 1 = ?

(p/s, I'm mathematically impaired)
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Re: Solve the equation for the person above!

Post by 697134002 » 13 Apr 2011, 21:58

2 + 1 = 3 ÷ (


(Note - the bracket is part of it)

Edit: Missed the 'Make your own and solve it' part.

3x ÷ 4 = 5
x = 6 1/3
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Re: Solve the equation for the person above!

Post by Hahsime » 14 Apr 2011, 01:58

I have got one for you:
Integral.png
I forgot to include the dx on the end of the integrand.
It is pretty simple.

EDIT: Actually it is not. I just did it, it is kind of a pain and wolfram gets a different answer than me, which is bad considering I have a calc. exam in a few hours....
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Re: Solve the equation for the person above!

Post by vallorn » 14 Apr 2011, 05:16

/me cracks knuckles. i LOVE intergrals. :D
(X^2)(X+2)^-1
now intergrate to get.
(1/3X^3)(1/-2 X+1)^-2

cant remember the area under a curve right now...
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Re: Solve the equation for the person above!

Post by Hahsime » 14 Apr 2011, 07:22

vallorn wrote:/me cracks knuckles. i LOVE intergrals. :D
(X^2)(X+2)^-1
now intergrate to get.
(1/3X^3)(1/-2 X+1)^-2

cant remember the area under a curve right now...
Not quite. Hint: You have to do long division on the integrand to get it into the right form.
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Re: Solve the equation for the person above!

Post by vallorn » 14 Apr 2011, 09:54

Hashime wrote:
vallorn wrote:/me cracks knuckles. i LOVE intergrals. :D
(X^2)(X+2)^-1
now intergrate to get.
(1/3X^3)(1/-2 X+1)^-2

cant remember the area under a curve right now...
Not quite. Hint: You have to do long division on the integrand to get it into the right form.
yeah its kind of difficult to see what im writing on the PC therefore i make mistakes like that... i think simply multiplying it out would work just as well but of course mr derp here forgot to do that :x .

here goes Again:
(X^2)(X+2)^-1

hmm so im multiplying the second bracket by X^2...
lets see...
X^2+2X ??? i think... (kind of rusty with algebraic fractions like this)
then integrate...

1/3 X^3 + X^2

that better?
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Re: Solve the equation for the person above!

Post by Hahsime » 14 Apr 2011, 10:23

vallorn wrote:
Hashime wrote:
vallorn wrote:/me cracks knuckles. i LOVE intergrals. :D
(X^2)(X+2)^-1
now intergrate to get.
(1/3X^3)(1/-2 X+1)^-2

cant remember the area under a curve right now...
Not quite. Hint: You have to do long division on the integrand to get it into the right form.
yeah its kind of difficult to see what im writing on the PC therefore i make mistakes like that... i think simply multiplying it out would work just as well but of course mr derp here forgot to do that :x .

here goes Again:
(X^2)(X+2)^-1

hmm so im multiplying the second bracket by X^2...
lets see...
X^2+2X ??? i think... (kind of rusty with algebraic fractions like this)
then integrate...

1/3 X^3 + X^2

that better?
Not even close, are you sure you are integrating?

I also forgot to add a dx on the end of the integrand
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Re: Solve the equation for the person above!

Post by DuplicateValue » 14 Apr 2011, 11:06

Er, I haven't done one of those in a while, so this will probably be horribly wrong.

X^2/(X+2) dx
X^2 . 1/(X+2)

Let (X+2) = u
du/dx = X^2
du = X^2 dx

New Limits --->
u = (10)+2 = 12
u = (0)+2 = 2

1/u du
[lnu] (lims 12 and 2)

[ln12 - ln2]
[2.48490665 - 0.6931471806]

= 1.791759469
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Re: Solve the equation for the person above!

Post by Hahsime » 14 Apr 2011, 14:32

DuplicateValue wrote:Er, I haven't done one of those in a while, so this will probably be horribly wrong.

X^2/(X+2) dx
X^2 . 1/(X+2)

Let (X+2) = u
du/dx = X^2
du = X^2 dx

New Limits --->
u = (10)+2 = 12
u = (0)+2 = 2

1/u du
[lnu] (lims 12 and 2)

[ln12 - ln2]
[2.48490665 - 0.6931471806]

= 1.791759469
You need to split up the terms in the integrand to be able to integrate this.
This is done using polynomial long division.
Once you get past that step it should look easy.
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Re: Solve the equation for the person above!

Post by DuplicateValue » 14 Apr 2011, 14:46

Hashime wrote:You need to split up the terms in the integrand to be able to integrate this.
This is done using polynomial long division.
Once you get past that step it should look easy.
Really? I was almost certain I was right there.

This is how I was shown how to integrate quotients at any rate.

Edit: Just to be clear, when I said X^2 . 1/X+2, I meant (X^2)(1/X+2).
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Hahsime
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Re: Solve the equation for the person above!

Post by Hahsime » 14 Apr 2011, 18:50

wolframalpha-20110414185342810.gif
Here are the steps as an indefinite integral. You can see where one would put the numbers in.
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Re: Solve the equation for the person above!

Post by vallorn » 15 Apr 2011, 05:36

OH GOD LOGARITHMS! :shock: /me dislikes this solution...
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Re: Solve the equation for the person above!

Post by Cyphafrost » 23 May 2011, 22:27

This is a clever one of mine.

9x-7i > 3(3x-7u)
Spoiler! :
i < 3u
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