Question

Can you solve this for me I’m stuck?

Answer 1

to get the equation of any straight line, we simply need two points off of it, let's use those two in the picture below.

`(\stackrel{x_1}{65}~,~\stackrel{y_1}{149})\qquad (\stackrel{x_2}{105}~,~\stackrel{y_2}{221}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{221}-\stackrel{y1}{149}}}{\underset{\textit{\large run}} {\underset{x_2}{105}-\underset{x_1}{65}}} \implies \cfrac{ 72 }{ 40 } \implies \cfrac{ 9 }{ 5 }`

`\begin{array}c \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{149}=\stackrel{m}{ \cfrac{ 9 }{ 5 }}(x-\stackrel{x_1}{65})\implies y-149=\cfrac{ 9 }{ 5 }x-117 \\\\\\ y=\cfrac{ 9 }{ 5 }x-117+149\implies \boxed{y=\cfrac{ 9 }{ 5 }x+32} \\\\\\ \stackrel{\textit{Celsius is 70, so x = 70}}{y=\cfrac{ 9 }{ 5 }(70)+32}\implies \boxed{y = 158}`