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Mathematics 19 Online
OpenStudy (anonymous):

How to find perimeter of a hexagon on a graph?

OpenStudy (jdoe0001):

got graph? ** shamelessly taken from 'got milk' **

OpenStudy (anonymous):

yes it's number 3

OpenStudy (jdoe0001):

so you know is simpler to just use a .gif or .jpeg anyhow

OpenStudy (jdoe0001):

can you see the coordinates of the vertices? all 6 corners

OpenStudy (anonymous):

lol I don't know how to do that.

OpenStudy (anonymous):

yes I see the coordinates

OpenStudy (jdoe0001):

you're only asked for XU as far as I can tell though, no the whole perimeter...but anyhow, so long you know the coordinates of each point, you can always find the "distance" between them with \(\bf \textit{distance between 2 points}\\ \quad \\ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)

OpenStudy (jdoe0001):

\(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ U&(-5\quad ,&7)\quad X&(2\quad ,&2) \end{array}\qquad d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)

OpenStudy (anonymous):

(-7, 5)

OpenStudy (anonymous):

thanks!!

OpenStudy (jdoe0001):

hmmm ... actually ... I got the wrong... coordinate for ... ahmm U :S

OpenStudy (jdoe0001):

should be (-2, 7) , but anyhow, you'd use the distance formula

OpenStudy (jdoe0001):

\(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ U&(-2\quad ,&7)\quad X&(2\quad ,&2) \end{array}\qquad d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)

OpenStudy (anonymous):

hey can you help me with #5 @jdoe0001

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