Question

I’m doing homework and this may be easy for others but I’m not that good at learning quickly. Take ur time it’s fine. Please help

Answer 1

the statement that "you walk at a constant rate" tells us that the relationship is indeed proportional, namely, there's a value in the input that's consistent or common yielding the output.

Check the picture below.

to get the equation of any straight line, we simply need two points off of it, for the distance "from school" table, let's use those two points in that table

`(\stackrel{x_1}{10}~,~\stackrel{y_1}{0.5})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{1.5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{1.5}-\stackrel{y1}{0.5}}}{\underset{\textit{\large run}} {\underset{x_2}{30}-\underset{x_1}{10}}}\implies \cfrac{1}{20}`

`\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}{0.5}=\stackrel{m}{\cfrac{1}{20}}(x-\stackrel{x_1}{10})\implies y-0.5=\cfrac{1}{20}x-\cfrac{1}{2} \\\\\\ y=\cfrac{1}{20}x-\cfrac{1}{2}+0.5\implies y=\cfrac{1}{20}x-\cfrac{1}{2}+\cfrac{1}{2}\implies \boxed{y=\cfrac{1}{20}x}`