[SOLVED] $20 in BTC for an algebra problem

TriggerX

hero member

Activity: 826

Merit: 1000

Quote from: Baldassare on April 24, 2015, 11:55:31 AM

Yes, TriggerX got an answer. Thanks everyone!
Contest closed.

TriggerX, money sent.
Would you mind posting the steps to the solution?

It took 30 minutes to figure it out and I don't want to type it all out again haha. If I have time I may post it here again.

As for the BTC, I have received it, thanks.

Baldassare

member

Activity: 67

Merit: 10

Yes, TriggerX got an answer. Thanks everyone!
Contest closed.

TriggerX, money sent.
Would you mind posting the steps to the solution?

monsterdoge

full member

Activity: 266

Merit: 100

so triggerx already won this contest?

Baldassare

member

Activity: 67

Merit: 10

TriggerX did not request an escrow. I'm checking his solution at this moment. If it's good, I'll send the money.

volatilebtc

member

Activity: 70

Merit: 10

where is the fund being escrowed ? Show us the proof ?

bitspill

legendary

Activity: 2087

Merit: 1015

Quote from: ShetKid on April 24, 2015, 12:58:26 AM

wouldn't be unsolvable

May want to rephrase that a bit

monsterdoge

full member

Activity: 266

Merit: 100

math make me crazy insane lol , goodluck to everyone .

ShetKid

sr. member

Activity: 392

Merit: 250

I am not Math genius, but in high school we were told that if we had 5 variables and 4 equations for it, then it wouldn't be unsolvable . Wouldn't that be true for this case ?

Baldassare

member

Activity: 67

Merit: 10

TriggerX, this looks extremely close! Still need to carefully check

TriggerX

hero member

Activity: 826

Merit: 1000

xi = 2L/mu
alpha = 1/(1+xi)(mu+L)
c = (s_k)/(2*xi*mu*m*alpha*alpha)
k=(1/[(s_k)/m-(1+1/xi)*c*alpha*alpha*2*L)] + (mu+L)/[2*alpha*mu*L]

Is this correct?

Baldassare

member

Activity: 67

Merit: 10

TriggerX,

Let u=1/100, L=2/100, s=1, m=2

Then xi = 6, c = 147/2000000, a = 100/21.

In that case the first three inequalities hold, but the last one doesn't.

BOUNTY STILL OPEN!!!!!

EDIT: I will pay $20 for EITHER a) or b) !!!! Will extend the deadline 24 hours from now.

arrowguys

member

Activity: 101

Merit: 10

the answer is n ^ n + b ^ = c n ^

Agestorzrxx

sr. member

Activity: 462

Merit: 250

you can use Mathematica to solve this problem, it's not hard.

erikalui

legendary

Activity: 2632

Merit: 1094

Quote from: Baldassare on April 22, 2015, 05:20:39 PM

Checking the solution!

Am I correct that you are suggesting the following?

k=1

alpha=(u + L)/2uL

e =4uL - 1

c=(sk/m - 1)/(1 + 1/(4uL - 1))*2u²L/(u + L)²

EDIT: I still need to check it, but I don't think it will work, since your first three lines contain a mistake...
1/k - 2alpha*u*L/(u + L) <=0;
1/k <= 2alpha*uL/(u + L);
k<=(u + L)/2alpha*uL

EDIT2: This solution does not work, since for u = 1, L =2, the second (original) equation does not hold.
(1+4*1*2-1)*(3)/(2*1*2) - 2/(1+2) = 5.3333... which is not <=0

I just checked and noticed I substituted a wrong value for alpha and so the result is incorrect. I can see that another memebr has tried solving the equation and if by chance his solution too is incorrect, I can try again.

oxiyusuf

sr. member

Activity: 462

Merit: 251

Im still confuse with math question, for this one, very difficult for me! Im usually can answer for aljabar question!

TriggerX

hero member

Activity: 826

Merit: 1000

I think I have the answer, can you escrow the funds? Also do you think you can extend the date a few more days?

Edit: Nevermind, I'll just post the answers. Only for part a)

alpha = (mu)/[(mu+L)(mu+3L)]
xi = 3L/u
c = (s_k)(u+L)(u+L)(u+3L)(u+3L)/3L
k = (1/[(s_k)/m-(1+1/xi)*c*alpha*alpha*2*L)] + (mu+L)/[2*alpha*mu*L]

If it is correct, please send the BTC to this address: 13MLwx49jT4a2SSwD17qu1uaG1sbGMdrzi

josef2000

sr. member

Activity: 280

Merit: 250

Bro, you need to try http://dadice.com

Is the calculation for some purpose? Are you trying to let someone else donyou homework Wink

? Or are just trying to have fun when people burn their brain juice away?

bitspill

legendary

Activity: 2087

Merit: 1015

Don't know if you posted it yourself or not but these guys can't solve it. https://2ch.hk/cc/res/98420.html

Baldassare

member

Activity: 67

Merit: 10

Bounty still open!!

Baldassare

member

Activity: 67

Merit: 10

Checking the solution!

Am I correct that you are suggesting the following?

k=1

alpha=(u + L)/2uL

e =4uL - 1

c=(sk/m - 1)/(1 + 1/(4uL - 1))*2u²L/(u + L)²

EDIT: I still need to check it, but I don't think it will work, since your first three lines contain a mistake...
1/k - 2alpha*u*L/(u + L) <=0;
1/k <= 2alpha*uL/(u + L);
k<=(u + L)/2alpha*uL

EDIT2: This solution does not work, since for u = 1, L =2, the second (original) equation does not hold.
(1+4*1*2-1)*(3)/(2*1*2) - 2/(1+2) = 5.3333... which is not <=0

Topic: [SOLVED] $20 in BTC for an algebra problem (Read 2448 times)