Answer:

INVERSE VARIATION

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1.

\begin{gathered} \implies \sf \large x = \frac{k}{y} \\ \end{gathered}⟹x=yk

\begin{gathered}\implies \sf \large 1 = \frac{k}{2} \\ \end{gathered}⟹1=2k

\begin{gathered}\implies \sf \large 1 \times 2 = \frac{k}{ \cancel2} \times \cancel2 \\ \end{gathered}⟹1×2=2k×2

\implies \sf \large 2 = k⟹2=k

\color{lightblue} \bold{Constant \: of \: the \: Variation:} \: \: \tt \Large \green{2}ConstantoftheVariation:2

\:

2.

\begin{gathered} \implies \sf \large m = \frac{k}{n} \\ \end{gathered}⟹m=nk

\begin{gathered} \implies \sf \large 2 = \frac{k}{36} \\ \end{gathered}⟹2=36k

\begin{gathered} \implies \sf \large 2 \times 36= \frac{k}{ \cancel{36}} \times \cancel{36} \\ \end{gathered}⟹2×36=36k×36

\implies \sf \large 72 = k⟹72=k

\color{lightblue} \bold{Constant \: of \: the \: Variation:} \: \: \tt \Large \green{72}ConstantoftheVariation:72

\:

3.

\begin{gathered} \implies \sf \large a = \frac{k}{b} \\ \end{gathered}⟹a=bk

\begin{gathered} \implies \sf \large 5 = \frac{k}{1} \\ \end{gathered}⟹5=1k

\begin{gathered} \implies \sf \large 5 \times 1 = \frac{k}{ \cancel1} \times \cancel1 \\ \end{gathered}⟹5×1=1k×1

\implies \sf \large 5 = k⟹5=k

\color{lightblue} \bold{Constant \: of \: the \: Variation:} \: \: \tt \Large \green{5}ConstantoftheVariation:5

\:

4.

\begin{gathered} \implies \sf \large l = \frac{k}{w} \\ \end{gathered}⟹l=wk

\begin{gathered} \implies \sf \large 6 = \frac{k}{2} \\ \end{gathered}⟹6=2k

\begin{gathered}\implies \sf \large 6 \times 2 = \frac{k}{ \cancel2} \times \cancel2 \\ \end{gathered}⟹6×2=2k×2

\implies \sf \large 12 = k⟹12=k

\color{lightblue} \bold{Constant \: of \: the \: Variation:} \: \: \tt \Large \green{12}ConstantoftheVariation:12

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