Respuesta :
part A)
[tex]\bf \begin{array}{ccllll} hours(x)&velocity(y)\\ -----&-----\\ 2&50\\ 6&54 \end{array}\\\\ -------------------------------\\\\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 2}}\quad ,&{{ 50}})\quad % (c,d) &({{ 6}}\quad ,&{{ 54}}) \end{array}[/tex]
[tex]\bf slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{54-50}{6-2}\implies \cfrac{4}{4}\implies 1 \\\\\\ % point-slope intercept y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-50=1(x-2)\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form} \\\\\\ y-50=x-2\implies \boxed{y-x=48}\impliedby \begin{array}{llll} standard\\ form \end{array}[/tex]
part B)
well, to graph a LINEar equation, since it's just a LINE, you simply need two points to graph a line, and since we know that y - x = 48.... if you want to know what's "y" when x = 7? or 7 hours, well
y - (7) = 48
y = 48 + 7
y = 55
so.. that's the point for the 7th hour, 7, 55, and you can pick any other point above, and graph it away.
[tex]\bf \begin{array}{ccllll} hours(x)&velocity(y)\\ -----&-----\\ 2&50\\ 6&54 \end{array}\\\\ -------------------------------\\\\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 2}}\quad ,&{{ 50}})\quad % (c,d) &({{ 6}}\quad ,&{{ 54}}) \end{array}[/tex]
[tex]\bf slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{54-50}{6-2}\implies \cfrac{4}{4}\implies 1 \\\\\\ % point-slope intercept y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-50=1(x-2)\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form} \\\\\\ y-50=x-2\implies \boxed{y-x=48}\impliedby \begin{array}{llll} standard\\ form \end{array}[/tex]
part B)
well, to graph a LINEar equation, since it's just a LINE, you simply need two points to graph a line, and since we know that y - x = 48.... if you want to know what's "y" when x = 7? or 7 hours, well
y - (7) = 48
y = 48 + 7
y = 55
so.. that's the point for the 7th hour, 7, 55, and you can pick any other point above, and graph it away.
