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Comments:
<0> wait
<0> xerxes1358: you are doing GoF now?
<1> Yes
<2> How can (-1s^11)/7 not be in its simplest form ??
<0> oh
<0> okay, carry on
<2> Hawkes Learning says it isn't simple enough.....
<0> Defender: maybe just write it as -(1/7) s^11
<3> -(s^11)/7
<1> so that is 3t+4t^(1/2)+2
<1> ?
<2> ok
<0> Defender: yeah or like Meldon says
<3> Defender they might want it in a specific form
<3> like i showed
<3> mymathlab for me requires that too
<0> xerxes1358: right
<1> is it possible to further add 3t+4t^(1/2)
<3> JohnFlux I forget how 3 Sqrt[t]^2 reduces to 3t
<3> xerxes1358 yea
<1> sqrt == ^(1/2)
<1> Meldon, how
<3> 4t^(1/2) = 4 Sqrt[t]
<0> xerxes1358: you can't reduce that further
<0> xerxes1358: you basically have a polynomial
<3> but the latter is considered "simpler"
<0> right
<0> i prefer to write 4t
<0> :P
<0> but then the windows luddites can't read it
<3> JohnFlux nvm i remember
<3> JohnFlux i can
<3> JohnFlux how do you type math expressions
<1> ah if I cant further simplkify I must have typed something wrong to mbot
<4> My brain doesn't feel like working so I'm having a little trouble with this problem I've come up with. Lets say you can make an investment in tomatoes whatever. It doesn't really matter. It costs n + 6/5x where n is the number of other investments currently and x is the number of tomatoes you invest in. You can only invest in multiples of 5. Every even 5 produces 12 dollars and every odd...
<4> ...produces 13. So if you x is 10 then you will recieve 25 dollars back. Also if there are say 6 other investments and x is 10 then you will have to spend 18 on the investment. For a given x when will equilbrium be reached. When would making an investment not be worth the money.
<3> 2 + 4 Sqrt[t] + 3t is simplest
<3> xerxes1358
<4> That's the best way I've figured out how to put it so far.
<3> unless you want to solve for t
<1> Meldon, may I ask why you dont get out the sqrt ?
<3> why would you want to
<2> Well I have another very similar error. It says that (-1z^4)/3 is not in its simplest form. What am I doing wrong this time?
<3> you could write 2 + 4*t^(1/2) + 3t
<5> Defender: drop the 1
<2> ok
<3> but doing anything else changes what that expression represents given any t
<5> Defender: 1x is the same as x
<3> yea so z^4/3
<6> except with a negative sign
<3> yea sorry
<3> -z^4/3
<3> they should have made there website with JLink mathematica
<1> whats is JLink ?
<3> allows Java to interface with a mathematica kernal on a computer/server
<3> kernel
<1> I see
<3> you can program a java app to tell mathematica to solve something and return it
<3> liek what im doing
<3> but im not using JLink
<3> im using MathLink (for apps)
<7> anyone knows about graph drawing software? dot/graphviz is very good but i need to create an animated sequences of graphs which allows dinamical edge insertion/deletion
<5> charlls: flash? :)
<8> Ok - got another problem I'm stuck on
<8> I'm supposed to calc the kinectic energy on this one
<9> Hmm...
<8> Ok - so an object weighing 10^9 smashed into some other object at 50 km/s
<9> There's a set {{},{{}},{{{}}},{{{{}}}}...}, right?
<10> iratsu: yep
<10> ihope_: you just wrote it down :)
<8> Ok - so the object is traveling at 50,000 m/sec
<10> err
<8> so - the energy released is .5 * m*** * Velocity^2
<8> So - ahh 1/2 * 10^9 * 50,0000
<8> Right?
<8> the result is in joules
<9> JabberWalkie: eh, I'll take that as a "yes".
<10> Cpudan80: sounds ok....but i fail to see how this is math
<1> % inverse g(x)=2x+1
<11> xerxes1358: 1 + 2*x
<9> And it's clearly infinite.
<9> I'm thinking the whatchamie is omega.
<8> I could toss it over to the physics channel I guess
<10> ihope_: i dont know about that....mabey it ends and some point ;)
<8> JabberWalkie: So - you think that's right (the answer is 2.5E13 joules)?
<8> This damn computer
<8> It gives you no clue as to wether or not your answer is right
<8> or - ehh - close to right
<10> Cpudan80: 1.25E18
<8> JabberWalkie: where did I screw up?
<10> Cpudan80: you forgot to square the veloctity
<8> I did?
<10> yes
<8> hmm
<7> JohnFlux2, thats actually a good idea i wouldn't have thought... i'll search to see if there is some graph drawing stuff for flash
<8> AH HA!
<8> Thanks Jabber
<10> np ;)
<1> Is inverse g(x)=2x+1 correct syntax ?
<10> looks ok
<10> errr
<10> wait
<1> I get (manually) 1/2x-1/2
<10> you want g^-1(x) for the inverse of g(x)
<1> I get (manually) 1/2(x)-(1/2)
<1> Yes
<1> g^-1
<1> JabberWalkie, I have to type g^-1(x) ?
<1> like g^-1(x)=2x+1 ?
<1> night all
<8> Ok everybody - one more question
<8> If a star collapses inward until it is 4 times smaller than it was originally (while losing no m***), how many times faster will its surface be moving?
<8> Conservation of momentum says that it can't be lost
<8> so - I say the speed has to up by a factor of 4
<12> hmm, cute question
<8> But that isn't right
<8> Thomas2_: lol
<12> 4 times smaller by total volume?
<8> That's the exact question
<8> I ***ume so - conservation of m*** takes presedence here
<3> -(Pi/6000) Sqrt[64 h]
<12> right. if it's 4 times smaller by volume, I think the answer should be the cube root of 4
<3> How do I simplify that?
<3> I forgot how to take 64 out.
<12> 64 = 8^2
<3> Thomas2_ so how do you write it in terms of Sqrt[h]
<12> Cpudan80: think about any given particle, and figure out how the radius of its orbit changes
<12> Meldon: sqrt (a x b) = sqrt (a) x sqrt (b)
<8> Well I mean
<8> it's 4 times smaller - but it could be by like radius
<8> so then it would just get heavier
<12> no, I think the point is that its m*** doesn't change, but its volume does, no?
<3> ok so -(Pi/6000) Sqrt[8] Sqrt[8 h]
<12> erm, no, just move the 64
<3> well that is correct
<8> Well this question has a follow up
<12> you want sqrt [h] in your answer, no?
<3> yes
<8> If it was spinning around once every 100 hours initially, how long will it take to spin around after it collapses?
<8> I say that it would take 25 hours
<8> I mean think about it
<8> If the sun did something nuts and became 1/4 of its size
<8> and its m*** didnt change
<8> I donno
<8> this is a strange question
<3> So -(Pi/6000) Sqrt[64] Sqrt[h]
<3> How do I simplify -(Pi/6000) Sqrt[64]?
<12> Cpudan80: sure. I think you need to distinguish between volume and diameter though
<12> Meldon: do you know what sqrt[64] is?
<3> 64^(1/2)
<8> Well volume = 4/3 pi r^3 right?
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