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Mathematics 22 Online
OpenStudy (anonymous):

What is the distance between points (9, 4) and (–3, 4) on a coordinate plane?

OpenStudy (asdasdasd):

Distance Formula : \[\sqrt{(x^2-x^1)(y^2-y^2)}\]

OpenStudy (asdasdasd):

lol failed.

OpenStudy (asdasdasd):

i messed up hold on

OpenStudy (jdoe0001):

\(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ &({\color{red}{ 9}}\quad ,&{\color{blue}{ 4}})\quad &({\color{red}{ -3}}\quad ,&{\color{blue}{ 4}}) \end{array}\qquad d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)

OpenStudy (asdasdasd):

\[\sqrt{(x^2-x^1)^2(y^2-y^1)^2}\]

OpenStudy (asdasdasd):

there you go.

OpenStudy (anonymous):

14?

OpenStudy (asdasdasd):

I don't think so.

OpenStudy (asdasdasd):

-3-9 = 12, square 12 and it gives you 144

OpenStudy (asdasdasd):

4-4= 0, square 0 and it will give you 0

OpenStudy (asdasdasd):

144+0= 144

OpenStudy (asdasdasd):

Now find the square root of 144

OpenStudy (anonymous):

12

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