anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
Zed
anonymous
  • anonymous
hey tinkerbell :)

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anonymous
  • anonymous
help
anonymous
  • anonymous
sure, just reading the file
anonymous
  • anonymous
ok
watchmath
  • watchmath
hello satellite :)
anonymous
  • anonymous
brb zed im going to get a drink
anonymous
  • 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
  • anonymous
watchmath!! long time no see. i just mentioned you the other day
anonymous
  • anonymous
saying i missed your interesting questions/puzzles
watchmath
  • watchmath
yes, you are a legend now here :). Good job!
anonymous
  • anonymous
legend smegend. when i get to 10,000 medals i am going to have a whiskey and call it a day
anonymous
  • anonymous
hope to see you more regularly.
watchmath
  • watchmath
:). I will think some interesting problem for you
anonymous
  • anonymous
ok im here sorry had to grab a bottle of water i have been sick but im back now
anonymous
  • 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
  • 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
  • anonymous
i have been down ill with the flu really bad
anonymous
  • anonymous
i can walk you through another one if you like
anonymous
  • anonymous
tinkerbell do you need to use different methods each time?
anonymous
  • 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
  • anonymous
this is the only one i see , a^2+b^2 = c^2.
anonymous
  • 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
  • anonymous
thats what i see
anonymous
  • anonymous
did you open the link on the worksheet?
anonymous
  • anonymous
my assignment
anonymous
  • anonymous
this link http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples
anonymous
  • anonymous
that confused me
anonymous
  • 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
  • anonymous
can you rewrite it and explain as you go
anonymous
  • anonymous
ok 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
  • anonymous
7
anonymous
  • anonymous
ok then to find "b" we compute \[b=\frac{7^2}{2}-\frac{1}{2}\] what do you get?
anonymous
  • anonymous
there any other way we can do this with out fractions
anonymous
  • anonymous
will 7^2 =49 so it would be 49/2-1/2
anonymous
  • 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
  • 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
  • anonymous
and you are supposed to check that \[7^2+24^2=25^2\] which you can do with a calculator
anonymous
  • anonymous
slow down
anonymous
  • anonymous
let me know if you have any questions about what i wrote
anonymous
  • anonymous
no i have no questions so far let me type it into the template real qick
anonymous
  • anonymous
ok that gives me 625
anonymous
  • anonymous
good so we have one. now we can try another one
anonymous
  • anonymous
yes please
anonymous
  • anonymous
Ready when you are
anonymous
  • anonymous
ok now we try a different method
anonymous
  • anonymous
pick two numbers m and n, where one is even and one is odd. make them not too big
anonymous
  • anonymous
88 and 13
anonymous
  • anonymous
whoa nice and small!
anonymous
  • anonymous
you did not say that
anonymous
  • anonymous
we want to make this easy, not hard. pick small numbers
anonymous
  • anonymous
11 and 17
anonymous
  • 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
  • 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
  • anonymous
ok
anonymous
  • anonymous
now we have two. do you have to use a different method for each one?
anonymous
  • anonymous
yes i would like if it is possible
anonymous
  • anonymous
i will le let you know when im done entering it in the template
anonymous
  • 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
  • anonymous
ok hang on im 3^2-8^2=9-64=55
anonymous
  • anonymous
am i right so far
anonymous
  • 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
  • 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
  • anonymous
2x 8x 3=48
anonymous
  • anonymous
8^2 + 3^2 is to check right
anonymous
  • anonymous
no that is the third number, the long side
anonymous
  • anonymous
73
anonymous
  • anonymous
the three numbers are \[55,48,73\] and the check is \[55^2+48^2=73^2\]
anonymous
  • anonymous
ok be right with you
anonymous
  • 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
  • anonymous
i just have to rememer to put the bigger number in front

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