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anonymous
 5 years ago
i need someone to walk me step by step on a Pythagorean theorem i am having trouble with it and just a little bit of extra help
anonymous
 5 years ago
i need someone to walk me step by step on a Pythagorean theorem i am having trouble with it and just a little bit of extra help

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure, just reading the file

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0brb zed im going to get a drink

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0man that is confusing. pick two numbers, m and n, one even one odd. lets say i pick m = 4, n 1 then take \[m^2n^2=4^21^2=15\] also \[2mn=2\times 4\times 1=8\] and finally \[m^2+n^2=4^2+1^2=17\] those three number, \[15,8,17\] will be a pythagorean triple. that is \[15^2+8^2=17^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath!! long time no see. i just mentioned you the other day

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0saying i missed your interesting questions/puzzles

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0yes, you are a legend now here :). Good job!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0legend smegend. when i get to 10,000 medals i am going to have a whiskey and call it a day

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hope to see you more regularly.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0:). I will think some interesting problem for you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok im here sorry had to grab a bottle of water i have been sick but im back now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, but right now i am puzzling over how to make truth tables interesting because they are boring me to death. if you have any good puzzles or exercises involving them, let me know.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@tinkerbell, i wrote out a method for finding triples for you above. the worksheet you sent is rather vague (method on line) etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have been down ill with the flu really bad

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can walk you through another one if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tinkerbell do you need to use different methods each time?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It can be proved as the 2 dimensional case of the Parseval equality I think! Functional Analysis is goood. If not, then just remember this: The square on the hypotenuse is equal to the sum of the squares on the other two sides!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is the only one i see , a^2+b^2 = c^2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe it is not clear what you are looking for. not any three numbers a, b and c, but three whole numbers a, b and c with \[a^2+b^2=c^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you open the link on the worksheet?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this link http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i see. it looks like your worksheet is asking for 5 different triples. i wrote one above. would you like to do another one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you rewrite it and explain as you go

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok we can try the first one If a is odd, then b = a2/2 − 1/2 and c = b + 1 If a is even, then b = a2/4 − 1 and c = b + 2 pick an odd number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok then to find "b" we compute \[b=\frac{7^2}{2}\frac{1}{2}\] what do you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there any other way we can do this with out fractions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0will 7^2 =49 so it would be 49/21/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is a fraction but don't fret \[\frac{7^2}{2}\frac{1}{2}=\frac{491}{2}=\frac{48}{2}=24\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a nice whole number so \[a=7,b=24\] and now \[c=b+1=24+1=25\] and that is your "triple" \[7,24,25\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and you are supposed to check that \[7^2+24^2=25^2\] which you can do with a calculator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me know if you have any questions about what i wrote

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no i have no questions so far let me type it into the template real qick

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good so we have one. now we can try another one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok now we try a different method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0pick two numbers m and n, where one is even and one is odd. make them not too big

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we want to make this easy, not hard. pick small numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i will pick them, one even and the other odd. i pick 5 and 2 now we compute \[5^22^2=254=21\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then i take \[2\times 5\times 2=20\] and finally \[5^2+2^2=25+4=29\] and the three numbers \[21,20,29\] are also a triple and you can check that \[21^2+20^2=29^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we have two. do you have to use a different method for each one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i would like if it is possible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i will le let you know when im done entering it in the template

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well maybe we can do one more, but if i were you i would use the last method again with two different numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok hang on im 3^28^2=964=55

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you are on the right track, but you should make it \[8^23^2=649=55\] so you don't get negative numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you picked 8 and 3,and now you have 55 next two numbers will be \[2\times 8\times 3\] and \[8^2+3^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.08^2 + 3^2 is to check right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no that is the third number, the long side

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the three numbers are \[55,48,73\] and the check is \[55^2+48^2=73^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0unfortunately i have to run. but you have 3 so far, and it looks like you know what you are doing, so i think you should be in good shape

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i just have to rememer to put the bigger number in front
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