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Help. Thanks. http://fotos.fotoflexer.com/e35202bc5d3c01a65b83e6b92ecb48e4.jpg You should be able to see it. Thanks:)

Mathematics
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which one?
both. its 11 problems
help with whatever

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Other answers:

any help is appreciated
they all use the distance formula, which is pythagoras
ok.
can u do the first as demonstration ill try the others?
page B number 1 i will do if i can read it properly
tnx
you have two points labelled A and B and no numbers to go with them so lets count
mm,ok
ok
to get from A to B you go 7 units right, and 2 units down at least that is what i see
uh huh
let me verify
|dw:1351133689882:dw|
brb,counting
how did you get seven again?
we want \(d\) which is the hypotenuse of the right triangle with one side 7 and one side 2 by pythagoras, we get \[d^2=7^2+2^2\] \[d=\sqrt{7^2+2^2}=\sqrt{49+4}=\sqrt{53}\]
i counted how many steps i had to go the the right
did you count a or b?
i just counted them, i did not compute anything at all just went "one, two, three, four, five, six, seven"
ok.
so we aare solving for the hypotnuese
ok, so sq rt 53 is about 7.2
yes in each case that is the basis for the distance formula
like for number 3 on the same page, from A to B is 4 steps right, 4 steps up distance is \[d-\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{32}\]

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