September 10, 200619 yr X^n+Y^n=Z^n now if we do what i like to call substitution X^n+Y^n-Z^n=0 devide all 3 by 1^n x+y-z=0 so x+y=z or x-z=-y, or y-z=-x x and y both = 2, Z equals four, n=1 I don't think that x and y can be equal and n has to be >2 since the original written problem had no limitations written, it would work for just that Sigs made by Runemetsa, Nuzza, Dark_Shadow, Kuroi and Me.liquid blobs attacking eachother American Anime... YAY!
September 10, 200619 yr X^n+Y^n=Z^n now if we do what i like to call substitution X^n+Y^n-Z^n=0 devide all 3 by 1^n x+y-z=0 so x+y=z or x-z=-y, or y-z=-x x and y both = 2, Z equals four, n=1 I don't think that x and y can be equal and n has to be >2 since the original written problem had no limitations written, it would work for just that What? You didn't read the topic? It's All the numbers have to be intergers and n has to be greater than 2 the x and y thing was just speculation
September 10, 200619 yr X^n+Y^n=Z^n Ha! There, I found something that satisfies all the requirements of your equation. X = 2 Y = 1 Z = 3 N = 3 Which comes to: 8 + 1 = 9 If there's anything wrong with that answer, please tell me, because I think I may be a genius yes? My greatest ambition is to kill every member of the human race.However I am a realist and therefore know that I probably wont be able to.
September 10, 200619 yr X^n+Y^n=Z^n Umm ... well the equation balances if you value; X = 1 Y = 1 Z = 2 N = 1 There, did I just solve the equation that has baffled mathematicians for hundreds of years? If you had taken a few minutes to read the topic you would have found that it baffled them for N>2, not N<=2 I just posted something! ^_^ to the terrorist...er... kirbybeam.
September 10, 200619 yr If you had taken a few minutes to read the topic you would have found that it baffled them for N>2, not N<=2 Heeheehee, read my above post, I just solved your un-solvable equation. My greatest ambition is to kill every member of the human race.However I am a realist and therefore know that I probably wont be able to.
September 10, 200619 yr If you had taken a few minutes to read the topic you would have found that it baffled them for N>2, not N<=2 Heeheehee, read my above post, I just solved your un-solvable equation. EDIT: Ah sheesh, you go and edit it, anyways it'd be X = 2 Y = 1 Z = 3 N = 3 Which comes to: 8 + 1 = 9 2^3+1^3=3^3 which would be 8+1=27, which it doesn't :P I just posted something! ^_^ to the terrorist...er... kirbybeam.
September 10, 200619 yr Oh yeah .. I was thinking 3*3, not 3^3. Damn, got all excited over nothing .. My greatest ambition is to kill every member of the human race.However I am a realist and therefore know that I probably wont be able to.
September 10, 200619 yr X^n+Y^n=Z^n Ha! There, I found something that satisfies all the requirements of your equation. X = 2 Y = 1 Z = 3 N = 3 Which comes to: 8 + 1 = 9 If there's anything wrong with that answer, please tell me, because I think I may be a genius yes? oooh kashi gj gj its probably wrong but i dunt see what's wrong with it. EDIT: yeah wrong Trix.--quit WoW as of 12/07Thank you 4be2jue for the wonderful sig and avatar!
September 10, 200619 yr Oh yeah .. I was thinking 3*3, not 3^3. Damn, got all excited over nothing .. Lol! Everytime I though about the equation I came up with the same thing, then noticed the same problem that 3^3 isn't 9 :x
September 10, 200619 yr HAHAHA!! This was bugging me, so I looked it up .. If N>2, then there are no whole number solutions for X Y Z! You were right about what you said in the title, given requirements of X Y Z being integers and N>2, there is no solution to this problem. Tell your calculus teacher he's an [wagon]. If N>2 you can solve it if X Y Z aren't whole numbers, but otherwise it really is impossible. Jerk, you had everyone thinking hard about it for no reason. My greatest ambition is to kill every member of the human race.However I am a realist and therefore know that I probably wont be able to.
September 10, 200619 yr looks like your calculus teacher was having a bit of a laugh with you there. Shame that you spent a long time trying to prove it. To answer the question above: how about x=2 y=3 z=5 n=1 this is a triangle with zero area! Here's one to get your revenge on your calculus teacher with: Prove that 1=2!! (there is a a flaw in this, see if you can figure it out) A=B AB=BB AB-AA=BB-AA A(B-A)=(B-A)(B+A) A=B+A as A=B, then A=2A, therefore 1=2! :D No, silly rabbit, you're dividing by zero! (At the bolded part) No, silly rabbit, you're not dividing! (You're simplifying) A=B AB=BB AB-AA=BB-AA A(B-A)=(B-A)(B+A) 1(1-1)=(1-1)(1+1), when simplified, is 1=2
September 10, 200619 yr Integrate: x*tan(x)dx :lol: 1. We are not here to do your homework for you! 2. Integration and differentiation have nothing to do with this! While they may not...lemme just do this problem! Integration by parts! u= x dV=tan(x)dx du=1 dx v= -ln(cos(x)) -xln(cos(x))-(inetragl of){-xln(cos(x)} u = x dv=-ln(cos(x)) du = 1 dx v= tan(x) -xln(cos(x))-tan(x)+(integral of){tan(x)] -xln(cos(x)) - tan(x) - ln(cox(x)) + C I love calculus.
September 10, 200619 yr looks like your calculus teacher was having a bit of a laugh with you there. Shame that you spent a long time trying to prove it. To answer the question above: how about x=2 y=3 z=5 n=1 this is a triangle with zero area! Here's one to get your revenge on your calculus teacher with: Prove that 1=2!! (there is a a flaw in this, see if you can figure it out) A=B AB=BB AB-AA=BB-AA A(B-A)=(B-A)(B+A) A=B+A as A=B, then A=2A, therefore 1=2! :D No, silly rabbit, you're dividing by zero! (At the bolded part) No, silly rabbit, you're not dividing! (You're simplifying) A=B AB=BB AB-AA=BB-AA A(B-A)=(B-A)(B+A) 1(1-1)=(1-1)(1+1), when simplified, is 1=2 What are you [developmentally delayed]ed? 1-1=0 so 1*0=0 and 2*0=0 so what you proved is that 0=0 not 1=2
September 10, 200619 yr Here's one to get your revenge on your calculus teacher with: Prove that 1=2!! (there is a a flaw in this, see if you can figure it out) A=B AB=BB AB-AA=BB-AA A(B-A)=(B-A)(B+A) A=B+A as A=B, then A=2A, therefore 1=2! :D Take it back to the very beginning: A=B Everything on from here is void, as you could substitute in B for every A. However, something else slightly odd happens when you do this :P : BB=BB BB-BB=BB-BB B(B-B)=(B-B)(B-B) B=(B-B) If B=2, this equation now reads 2=2-2=0 2 quite clearly does not equal 0! See if you can spot the bit where it goes twisted. 99/99 Fletching, 99/99 Cooking, 96/99 Strength
September 10, 200619 yr Take it back to the very beginning: A=B Everything on from here is void, as you could substitute in B for every A. However, something else slightly odd happens when you do this :P : BB=BB BB-BB=BB-BB B(B-B)=(B-B)(B-B) B=(B-B) If B=2, this equation now reads 2=2-2=0 2 quite clearly does not equal 0! See if you can spot the bit where it goes twisted. 3rd line you have BB-BB = 2BB-2BB is 0=0 you then multiply both sides by 0 (B-B) so the next line should also read 0=0 :-) babelfish - level 180 60th placestrongguy - level 173 69th place
September 10, 200619 yr Here's one to get your revenge on your calculus teacher with: Prove that 1=2!! (there is a a flaw in this, see if you can figure it out) A=B AB=BB AB-AA=BB-AA A(B-A)=(B-A)(B+A) A=B+A as A=B, then A=2A, therefore 1=2! :D Take it back to the very beginning: A=B Everything on from here is void, as you could substitute in B for every A. However, something else slightly odd happens when you do this :P : BB=BB BB-BB=BB-BB B(B-B)=(B-B)(B-B) B=(B-B) If B=2, this equation now reads 2=2-2=0 2 quite clearly does not equal 0! See if you can spot the bit where it goes twisted. I can tell it's wrongbut Idk how to explain it Integrate: x*tan(x)dx :lol: 1. We are not here to do your homework for you! 2. Integration and differentiation have nothing to do with this! While they may not...lemme just do this problem! Integration by parts! u= x dV=tan(x)dx du=1 dx v= -ln(cos(x)) -xln(cos(x))-(inetragl of){-xln(cos(x)} u = x dv=-ln(cos(x)) du = 1 dx v= tan(x) -xln(cos(x))-tan(x)+(integral of){tan(x)] -xln(cos(x)) - tan(x) - ln(cox(x)) + C I love calculus. Dude, that's not math. IT has like 3 digits in it. That's more like 1st grade English. :roll: Trix.--quit WoW as of 12/07Thank you 4be2jue for the wonderful sig and avatar!
September 10, 200619 yr Take it back to the very beginning: A=B Everything on from here is void, as you could substitute in B for every A. However, something else slightly odd happens when you do this :P : BB=BB BB-BB=BB-BB B(B-B)=(B-B)(B-B) B=(B-B) If B=2, this equation now reads 2=2-2=0 2 quite clearly does not equal 0! See if you can spot the bit where it goes twisted. 3rd line you have BB-BB = 2BB-2BB is 0=0 you then multiply both sides by 0 (B-B) so the next line should also read 0=0 :-) Lol, correct. Its actually the second line, BB-BB=BB-BB. If you convert that to numbers, so B = 2, you have (4-4)=(4-4)=0. =D> 99/99 Fletching, 99/99 Cooking, 96/99 Strength
September 11, 200619 yr xln(cos(x))-(inetragl of){-xln(cos(x)} this line is wrong: xln(cos(x))-(inetragl of){-xln(cos(x)} it should be: -xln(cos(x))-(integral of){-ln(cos(x))} and as you don't know the integral of -ln(cos(x)) you can't continue. The integral of x*tan(x) doesn't have a real solution thats why I posted it in this topic. It's impossible to do with the normal methods of integration you are taught.
September 11, 200619 yr I don't get it. There are no common variables and it is completely simplified. I can't see what else you can do with it. And I don't see what were trying to prove. Which Final Fantasy Character Are You?Final Fantasy
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