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Mathematics 23 Online
OpenStudy (anonymous):

The function shows a linear relationship. Find the slope.

OpenStudy (anonymous):

|dw:1416694665525:dw|

OpenStudy (anonymous):

plot the number on a graph it'll help

OpenStudy (anonymous):

Okay.

OpenStudy (anonymous):

But I mean, how is it linear? It doesn't like, have a pattern.

OpenStudy (anonymous):

im not quite sure

OpenStudy (jdoe0001):

\(\large \begin{array}{ccllll} x&y \\\hline\\ 0&1\\ {\color{red}{ 2}}&{\color{blue}{ 5}}\\ 5&11\\ {\color{red}{ 6}}&{\color{blue}{ 13}}\\ 7&15 \end{array} \qquad \begin{array}{llll} \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ &({\color{red}{ 2}}\quad ,&{\color{blue}{ 5}})\quad &({\color{red}{ 6}}\quad ,&{\color{blue}{ 13}}) \end{array} \\\quad \\ slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}} \end{array}\)

OpenStudy (anonymous):

So jdoe, would I have to do that for every interval?

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