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1 Attachment
okay well the theorem is \[z^2=x^2+y^2\] and you know everything has to be an integer (whole number)
you can just try different values. For instance for the first one i might try z=51 and x=24. so 51^2=24^2+y^2. Solving i get y=45. So this works

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Other answers:

can you try 19?
@cherio12 can you try #19 plz
you just have to play with the numbers. I first tried \[z^2=25^2+20^\] but i got z= 32.0156..so this didn't work next i tried \[25^2=20^2+y^2\] y=15 (this is an integer) so it checks out
does this make sense?
thats the answer in the book so i should be able to learn off your 2 answers
wait so on 18# i got 2601=576+y2 then do i get y by itself? @cherio12
yes. you should go 2601-576=y^2. then take the squareroot of both sides to just get y
ok how do you know were to put the numbers on the equation?
what do you mean?
wait oh i see you said move them around and it has to be a whole number
yup, its always (leg)^2+(leg)^2=(hypotenuse)^2
thanks man! helped alot
its commonly written as x^2+y^2=z^2
no problem =D
k
wait so on # 20 i got 91.82592222ect
okay so that means you picked the wrong combo of hypotenuse and legs. try another combo (by the way the hypotenuse will always be the largest of the 3 numbers)
so set it up as x2= 96 + 28?
i would try \[96^2=x^2+28^2\]
do get x by its self?
yes
|dw:1359577137858:dw|
but i still will get that 91.etc right?
i was just showing you the triangle to help show what i meant by x, y and z
i know im saying when i get x by its self and take 9216 and square it i get that 91.00202 number
o you mean for the 96, 28 problem?
yes
\[96^2=x^2+28^2\] \[9216=x^2+784\] \[9216-784=x^2\] \[x=\sqrt{8432}\] x=91.8259 so this way didn't work.
i would then try \[z^2=96^2+28^2\]
but you would still get 91.8259 right?
no i got 100
you add 96^2 and 28^2 in this example instead of subtracting them (like we tried before)
ohhh
now i see thakns agian lol
hey can you help @cherio12 with # 23
sure
okay so i first tried z^2=72^2+75^2 But i got z=103.966 so that's not right then i tried 75^2=x^2+72^2
ok i did the first but didnt try that one what did you get?
21
can you show how you did that?
75^2=x^2+72^2 solve for x
ok thanks i got 21
no problem
hey @cherio12 for #24 you you do the same?
1 Attachment
yes, your two legs are 6 and 3
so would it be 6(2)=3(2)+x2
by(2) do you mean you are squaring it?
no 6 to the power 2 like 6*6
that means squaring, but yes that is correct
wait that doesnt work
i tried both ways and got not a whole number
I don't believe you are supossed to get an integer for this problem
so would 5.19 work?
i got 7.937
6^2+3^2=x^2
ya that was my second way
so would you set up # 25 and # 26 the same way?
1 Attachment
yes for 25, but for 26 you will use a ratio
i gotta go eat..be back later
ok when you get back can you explain ratio?
@cherio12 you back?
yes
okay well for starters find the other leg on the 5,3 triangle
wait on the # 25 its just 11 and x so is it 121 then sqare root answer is 11?
no, there are two 11 sides
oh i see cuase of that line thingy
so the answer is 0
no for 25, it is 11^1+11^2=x^2
so its 15.49
yes
does that make sense why?
yes i just didnt try that way and remember there not whole numbers
so how do you do #26 is it 2 problems in 1
find the other leg in the 5,3 triangle first
5.83
wait no its 4
yes 4
now do the other triangel?
now we can use like triangle ratios 4/5=7/x
so 4 sqare + 5 square = 7sqare?
no, no square
just ratio, just division
so 0.8=7
4/5=7/x .8=7/x .8x=7
oh ok wow being sick for one day of notes can rweally heart you
haha, no problem
wait # 28 its 2x and 2x+4 and the length of the hypot is 4x-4
yes, just apply the Pythagorean theorem

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