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

- katieb

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- anonymous

##### 1 Attachment

- anonymous

Zed

- anonymous

hey tinkerbell :)

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## More answers

- anonymous

help

- anonymous

sure, just reading the file

- anonymous

ok

- watchmath

hello satellite :)

- anonymous

brb zed im going to get a drink

- anonymous

man that is confusing. pick two numbers, m and n, one even one odd. lets say i pick m = 4, n 1
then take
\[m^2-n^2=4^2-1^2=15\] 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

watchmath!! long time no see. i just mentioned you the other day

- anonymous

saying i missed your interesting questions/puzzles

- watchmath

yes, you are a legend now here :). Good job!

- anonymous

legend smegend. when i get to 10,000 medals i am going to have a whiskey and call it a day

- anonymous

hope to see you more regularly.

- watchmath

:). I will think some interesting problem for you

- anonymous

ok im here sorry had to grab a bottle of water i have been sick but im back now

- anonymous

ok, 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

@tinkerbell, i wrote out a method for finding triples for you above. the worksheet you sent is rather vague (method on line) etc

- anonymous

i have been down ill with the flu really bad

- anonymous

i can walk you through another one if you like

- anonymous

tinkerbell do you need to use different methods each time?

- anonymous

It 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

this is the only one i see , a^2+b^2 = c^2.

- anonymous

maybe 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

thats what i see

- anonymous

did you open the link on the worksheet?

- anonymous

my assignment

- anonymous

this link
http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples

- anonymous

that confused me

- anonymous

i see. it looks like your worksheet is asking for 5 different triples. i wrote one above. would you like to do another one?

- anonymous

can you rewrite it and explain as you go

- anonymous

ok we can try the first one
If a is odd, then b = a2/2 âˆ’ 1/2 and c = b + 1
If a is even, then b = a2/4 âˆ’ 1 and c = b + 2
pick an odd number

- anonymous

7

- anonymous

ok then to find "b" we compute
\[b=\frac{7^2}{2}-\frac{1}{2}\] what do you get?

- anonymous

there any other way we can do this with out fractions

- anonymous

will 7^2 =49 so it would be 49/2-1/2

- anonymous

this is a fraction but don't fret
\[\frac{7^2}{2}-\frac{1}{2}=\frac{49-1}{2}=\frac{48}{2}=24\]

- anonymous

a 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

and you are supposed to check that
\[7^2+24^2=25^2\] which you can do with a calculator

- anonymous

slow down

- anonymous

let me know if you have any questions about what i wrote

- anonymous

no i have no questions so far let me type it into the template real qick

- anonymous

ok that gives me 625

- anonymous

good so we have one. now we can try another one

- anonymous

yes please

- anonymous

Ready when you are

- anonymous

ok now we try a different method

- anonymous

pick two numbers m and n, where one is even and one is odd. make them not too big

- anonymous

88 and 13

- anonymous

whoa nice and small!

- anonymous

you did not say that

- anonymous

we want to make this easy, not hard. pick small numbers

- anonymous

11 and 17

- anonymous

ok i will pick them, one even and the other odd. i pick 5 and 2
now we compute
\[5^2-2^2=25-4=21\]

- anonymous

then 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

ok

- anonymous

now we have two. do you have to use a different method for each one?

- anonymous

yes i would like if it is possible

- anonymous

i will le let you know when im done entering it in the template

- anonymous

well maybe we can do one more, but if i were you i would use the last method again with two different numbers

- anonymous

ok hang on im 3^2-8^2=9-64=55

- anonymous

am i right so far

- anonymous

you are on the right track, but you should make it
\[8^2-3^2=64-9=55\] so you don't get negative numbers

- anonymous

so 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

2x 8x 3=48

- anonymous

8^2 + 3^2 is to check right

- anonymous

no that is the third number, the long side

- anonymous

73

- anonymous

the three numbers are
\[55,48,73\] and the check is
\[55^2+48^2=73^2\]

- anonymous

ok be right with you

- anonymous

unfortunately 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

i just have to rememer to put the bigger number in front

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