Maths- Gym for Brainiacs
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Re: Maths- Gym for Brainiacs
Arghh!!!!
Ok I did this shit again and it got me k=-1/2. If I'm wrong I'm not doing it again
Ok I did this shit again and it got me k=-1/2. If I'm wrong I'm not doing it again
urbaNRoots- First of his name
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Re: Maths- Gym for Brainiacs
Why do you guess k? Equal what you calculated to 2/3. Let me see:
(4+4k)/ ( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/ ( 3*(1+k)*sqrt(1+k)) = 2/3 and so on... You almost solved the shit
(4+4k)/ ( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/ ( 3*(1+k)*sqrt(1+k)) = 2/3 and so on... You almost solved the shit
Babun- Fan Favorite
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Re: Maths- Gym for Brainiacs
k=+-sqrt(2)-1
Finally did the integral but struggled to solve k
Finally did the integral but struggled to solve k
urbaNRoots- First of his name
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Re: Maths- Gym for Brainiacs
kiranr wrote:Yeah, i did that and integrated over the interval.
The interval is -1/(k+1)^(1/2) to +1/(k+1)^(1/2)
But the integral of g(x) minus integral of f(x) over the interval gets very complicated to solve for k.
I only tried it once and then decided to assume it is ellipse. I may be making some mistake. I will try again when i get the time, but if you solve it before, tell us the answer.
That's exactly what i did, However It's integral over integral in the end and i don't think it will end pretty. Although it might be a case of f'(x)/g(x)
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Re: Maths- Gym for Brainiacs
German highschool
(4+4k)/ ( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/ ( 3*(1+k)*sqrt(1+k)) = 2/3 ,eradicate (1+k) from the fraction on the left
=> 4/(3*sqrt(1+k)) = 2/3 | divide both sides by 2/3
=> 2/ sqrt(1+k) = 1 | ( )²
=> 4/ (1+k) = 1² <=> 1+k= 4 => k=3
I don't know whether the answer is right or wrong but k=3 according to urban's prior calculations.
Let's continue at home. According to urbanroot's prior calculation:Babun wrote:Why do you guess k? Equal what you calculated to 2/3. Let me see:
(4+4k)/ ( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/ ( 3*(1+k)*sqrt(1+k)) = 2/3 and so on... You almost solved the shit
(4+4k)/ ( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/( 3* sqrt( 1+k)^3) = 2/3
4*(1+k)/ ( 3*(1+k)*sqrt(1+k)) = 2/3 ,eradicate (1+k) from the fraction on the left
=> 4/(3*sqrt(1+k)) = 2/3 | divide both sides by 2/3
=> 2/ sqrt(1+k) = 1 | ( )²
=> 4/ (1+k) = 1² <=> 1+k= 4 => k=3
I don't know whether the answer is right or wrong but k=3 according to urban's prior calculations.
Are you a highschooler?urbaNRoots wrote:k=+-sqrt(2)-1
Finally did the integral but struggled to solve k
Babun- Fan Favorite
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Re: Maths- Gym for Brainiacs
Wtf is that School teaching their kids lmao
RealGunner- Admin
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Re: Maths- Gym for Brainiacs
She said it's from their exercise book I asked her to post a photo of that bookRealGunner wrote:Wtf is that School teaching their kids lmao
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Re: Maths- Gym for Brainiacs
i can see that u guys need the spanksters help
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Re: Maths- Gym for Brainiacs
And i was planning to do my masters in Germany sack that now lol
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Re: Maths- Gym for Brainiacs
Masters in what?RealGunner wrote:And i was planning to do my masters in Germany sack that now lol
Spanky Verify please whether k=3 is the correct answerspanky489 wrote:i can see that u guys need the spanksters help
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Re: Maths- Gym for Brainiacs
Software Engineering with Games Concept
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Re: Maths- Gym for Brainiacs
Babun wrote:Are you a highschooler?
I study computer engineering, tried to solve this as I remembered the formula but got caught
Last edited by urbaNRoots on Mon Sep 10, 2012 5:32 pm; edited 1 time in total
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Re: Maths- Gym for Brainiacs
You could give it a try Ask Urban, he studies something smiliar to software engineering in Germany I guessRealGunner wrote:Software Engineering with Games Concept
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Re: Maths- Gym for Brainiacs
RealGunner wrote:
That's exactly what i did, However It's integral over integral in the end and i don't think it will end pretty. Although it might be a case of f'(x)/g(x)
Integral over integral RG?
I just got k = 8 now
After intergration, it would be x - (k*x^3)/3 - (x^3)/3. Then you sub 1/root(k+1) and -1/root(k+1) into this and equate to 2/3.
so it will be
x ( 1 - ((x^2)/3)(k+1)) over which you sub +1/root(k+1) to -1/root(k+1).
When you sub this, you get 2/root(k+1) = 2/3. Hence k = 8.
Please check it guys.
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Re: Maths- Gym for Brainiacs
@kiranr, very strange but according to Urban's calculations and my continuation we had 4/(3*root(k+1)) = 2/3 at that point.kiranr wrote:RealGunner wrote:
That's exactly what i did, However It's integral over integral in the end and i don't think it will end pretty. Although it might be a case of f'(x)/g(x)
Integral over integral RG?
I just got k = 8 now
After intergration, it would be x - (k*x^3)/3 - (x^3)/3. Then you sub 1/root(k+1) and -1/root(k+1) into this and equate to 2/3.
so it will be
x ( 1 - ((x^2)/3)(k+1)) over which you sub +1/root(k+1) to -1/root(k+1).
When you sub this, you get 2/root(k+1) = 2/3. Hence k = 8.
Please check it guys.
@Spanky, test for k=3 or k=8 whether the calculations were right You're our specialist in applied maths
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Re: Maths- Gym for Brainiacs
For all the interested parites, that's what German high schoolers need to know in maths in order to be able to study at an uni:
http://www.sendspace.com/file/pt4y8f
http://www.sendspace.com/file/pt4y8f
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Re: Maths- Gym for Brainiacs
urbaNRoots wrote:Babun wrote:Are you a highschooler?
I study computer engineering, tried to solve this as I remembered the formula but got caught
In Germany ?
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Re: Maths- Gym for Brainiacs
k=3 was right kiranr:
Just tested Urban
Just tested Urban
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Re: Maths- Gym for Brainiacs
just got home
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Re: Maths- Gym for Brainiacs
spanky489 wrote:just got home
Fun Question
Requirements: the ability to count till 9
Difficulty: unknown
Nine 3s are ordered in a way that their horizontal, diagonal and vertical sum is 9. Order the nine 3s in a way that 10 series with the sum of 9 are created The solution is graphical of course
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Re: Maths- Gym for Brainiacs
i spent half an hour on this one and got nowhere
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Re: Maths- Gym for Brainiacs
The solution is easy.fatman123 wrote:i spent half an hour on this one and got nowhere
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Re: Maths- Gym for Brainiacs
do you make those lines into a 9 ?
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Re: Maths- Gym for Brainiacs
Yeah, horizontal, vertical and diagonal lines with three 3s each which result in 9. YOu need 10 of those linesRealGunner wrote:do you make those lines into a 9 ?
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Re: Maths- Gym for Brainiacs
3-------3
----3----
3---3---3
----3----
3------3
----3----
3---3---3
----3----
3------3
kiranr- First Team
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Re: Maths- Gym for Brainiacs
kiranr wrote:3-------3
----3----
3---3---3
----3----
3------3
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