**Do the Math
Here is a nice easy maths problem
Prove that a^n + b^n = c^n has no solution if a, b, c and n are all integers with a, b and c not being 0 and n being greater than 2. A shiny penny to the first to do it. 



Let me guess you had to read Fermat's Enigma in school? I could copy the solution from Wiles but I'm not sure that anyone wants to read about elliptic curves, the Epsilon conjecture, or the TaniyamaShimura Conjecture. That was a great book though.


Where is my pumpkin?!


OK....... im takin up Accounting in college...i passed? I dunno sh1t about math im in trouble arent i XD




Where is my pumpkin?!


I think I'll give an easier problem.
Why is this proof wrong? a = b a2 = ab a2  b2 = abb2 (ab)(a+b) = b(ab) a+b = b b+b = b 2b = b 2 = 1 (Besides the obvious answer that 2 is in fact not equal to 1.) 

Where is my pumpkin?!


should be...:
a = a [ nothing to add / subtract / divide / multiplied ] a2 = a2 [ cuz there were no equations and only numbers and variables are given ] a2  b2 = a2  b2 [ same as vv ] (ab)(a+b) = (ab)(a+b) [ the same as v ] a+b = a+b ( cuz you can't add a and b, and it is understood that only the same variables can be added or subtracted / multiplied / divided ) b+b = 1b raised to the 2nd power?? i guess... 2b = 2b 2 = 2 arghhhhhh im so stupid in math 



you do not account for the possibility that you could have divided by 0 when cancelling out the (ab) parts.
Try this; Solve; f(x) = x^3  7 x^2 + 14x  7 f(x) = 0 and find the area of the portion enclosed by the graph (y = f(x)) and the x axis between the first two zeros 

N/A


why the hell would u want a math thread?




Hopchow wrote: s7ph3r wrote: 6 the owner carries six bananas which he feeds to the camel in the beggining, after eating a thousand bananas the camel arrives with six bananas to spare Let's assume the owner can't carry any bananas, I'm sorry I should have mentioned that. The problem with your answer though is that you left 1994 bananas back at the plantation. I guess I should give a hint the camel cannot do the entire distance in one trip. He can't travel more than 500 kilometers and still manage to return. the problem is that if the camel eats 1 banana per km and travles 500 km eating 500 out of a thousand bananas he still has to eat 500 to get back to fetch more which means that he will eiher starve on the way or carry the rest of the ones he took there back while eating them and leaving him with none. Unless the camel doesnt need to eat any bananas on the way back... 



this thread is just started because someone didn't wanted to make his homework:P
you mix 2 sollutions with eachother 1: 200 ml PH=1.5 2: 75 ml PH= 9.5 what is the final PH when you mix these two? 



s7ph3r wrote: Hopchow wrote: s7ph3r wrote: 6 the owner carries six bananas which he feeds to the camel in the beggining, after eating a thousand bananas the camel arrives with six bananas to spare Let's assume the owner can't carry any bananas, I'm sorry I should have mentioned that. The problem with your answer though is that you left 1994 bananas back at the plantation. I guess I should give a hint the camel cannot do the entire distance in one trip. He can't travel more than 500 kilometers and still manage to return. the problem is that if the camel eats 1 banana per km and travles 500 km eating 500 out of a thousand bananas he still has to eat 500 to get back to fetch more which means that he will eiher starve on the way or carry the rest of the ones he took there back while eating them and leaving him with none. Unless the camel doesnt need to eat any bananas on the way back... Yes that is a problem, but what if he only went say 100 km dropped 800 bananas there and then went back for another load? 

Where is my pumpkin?!


Hopchow wrote: s7ph3r wrote: Hopchow wrote: s7ph3r wrote: 6 the owner carries six bananas which he feeds to the camel in the beggining, after eating a thousand bananas the camel arrives with six bananas to spare Let's assume the owner can't carry any bananas, I'm sorry I should have mentioned that. The problem with your answer though is that you left 1994 bananas back at the plantation. I guess I should give a hint the camel cannot do the entire distance in one trip. He can't travel more than 500 kilometers and still manage to return. the problem is that if the camel eats 1 banana per km and travles 500 km eating 500 out of a thousand bananas he still has to eat 500 to get back to fetch more which means that he will eiher starve on the way or carry the rest of the ones he took there back while eating them and leaving him with none. Unless the camel doesnt need to eat any bananas on the way back... Yes that is a problem, but what if he only went say 100 km dropped 800 bananas there and then went back for another load? that wouldnt work either. eats 200 bananas for 800 moved 100 km. once all the bananas are there you repeat losing 200 bananas per load. same result. *tries to look smart by stating the law of work, energy and power*... cant remember it but I know its important and aplies to this. 



s7ph3r wrote: Hopchow wrote: s7ph3r wrote: Hopchow wrote: s7ph3r wrote: 6 the owner carries six bananas which he feeds to the camel in the beggining, after eating a thousand bananas the camel arrives with six bananas to spare Let's assume the owner can't carry any bananas, I'm sorry I should have mentioned that. The problem with your answer though is that you left 1994 bananas back at the plantation. I guess I should give a hint the camel cannot do the entire distance in one trip. He can't travel more than 500 kilometers and still manage to return. the problem is that if the camel eats 1 banana per km and travles 500 km eating 500 out of a thousand bananas he still has to eat 500 to get back to fetch more which means that he will eiher starve on the way or carry the rest of the ones he took there back while eating them and leaving him with none. Unless the camel doesnt need to eat any bananas on the way back... Yes that is a problem, but what if he only went say 100 km dropped 800 bananas there and then went back for another load? that wouldnt work either. eats 200 bananas for 800 moved 100 km. once all the bananas are there you repeat losing 200 bananas per load. same result. *tries to look smart by stating the law of work, energy and power*... cant remember it but I know its important and aplies to this. By moving 100km at a time you end up with 2500 bananas 900 km away. If you were to continue to do this until you got to the market at 800 km you'd have 2000, at 700 km 1700, at 600 km 1400, at 500 km, 1100 at that point it is most efficient to just take the 1000 load straight into the market and leave the other 100 behind. You end up at the market with 500 bananas. This problem is all about making that process as efficient as possible. 

Where is my pumpkin?!


Hopchow wrote: s7ph3r wrote: Hopchow wrote: s7ph3r wrote: Hopchow wrote: s7ph3r wrote: 6 the owner carries six bananas which he feeds to the camel in the beggining, after eating a thousand bananas the camel arrives with six bananas to spare Let's assume the owner can't carry any bananas, I'm sorry I should have mentioned that. The problem with your answer though is that you left 1994 bananas back at the plantation. I guess I should give a hint the camel cannot do the entire distance in one trip. He can't travel more than 500 kilometers and still manage to return. the problem is that if the camel eats 1 banana per km and travles 500 km eating 500 out of a thousand bananas he still has to eat 500 to get back to fetch more which means that he will eiher starve on the way or carry the rest of the ones he took there back while eating them and leaving him with none. Unless the camel doesnt need to eat any bananas on the way back... Yes that is a problem, but what if he only went say 100 km dropped 800 bananas there and then went back for another load? that wouldnt work either. eats 200 bananas for 800 moved 100 km. once all the bananas are there you repeat losing 200 bananas per load. same result. *tries to look smart by stating the law of work, energy and power*... cant remember it but I know its important and aplies to this. By moving 100km at a time you end up with 2500 bananas 900 km away. If you were to continue to do this until you got to the market at 800 km you'd have 2000, at 700 km 1700, at 600 km 1400, at 500 km, 1100 at that point it is most efficient to just take the 1000 load straight into the market and leave the other 100 behind. You end up at the market with 500 bananas. This problem is all about making that process as efficient as possible. ha, you are too clever for me well done. you win and I am at last saisfied with knowing the answer. but I still think that the banana trader is getting a raw deal... 



why? don't you like this thread? i assume this thread is a very challenging game XD 

