1ba1fe…: “Opposite a fine eyes”

$\newenvironment {prompt}{}{} \newcommand {\accordion }[0]{} \newcommand {\ungraded }[0]{} \newcommand {\xmDefaultPreamble }[0]{xmPreamble.tex} \newcommand {\xmDefaultPrintstyle }[0]{xmPrintstyle.sty} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{}$

On the $xy$-plane above a line contains the points $(1,8)$ and $(5,2)$. This line is the graph of a linear function. What is the formula for this linear function?

\[ y(x) = \answer {-\frac {3}{2}} (x - \answer {5}) + 2 \]

On the $rt$-plane above, a line contains the points $(-5,-2)$ and $(3,9)$. This line is the graph of a linear function. What is the formula for this linear function?

\[ t(r) = \answer {\frac {11}{8}} (r - \answer {3}) +9 \]

On the $pn$-plane above, a line contains the points $(-5,-1)$ and $(4,8)$. This line is the graph of a linear function. What is the formula for this linear function?

\[ n(p)= \answer {1} (p + 5) + \answer {-1} \]

2025-05-17 23:58:11