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Answer the ODD numbered items only
Determine whether the tables and graphs below express a direct variation between the variables or not.If they do,find the constant of variation k,and the equation that defines the relation
DIRECT VARIATION
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1.
[tex] \sf \large \tt \green{ DIRECT \: \: VARIATION}[/tex]
[tex] \large\begin{array}{ |c|c|c|c|c| } \hline x & 1 & 2 & 3 &4 \\ \hline y & – 3 & – 6 & – 9 & – 12 \\ \hline \end{array}[/tex]
[tex] \implies \sf \large y = kx[/tex]
[tex]\implies \sf \large – 3 = k(1)[/tex]
[tex]\implies \sf \large \frac{ – 3}{1} = \frac{k( \cancel1)}{ \cancel1} \\ [/tex]
[tex]\implies \sf \large \therefore \orange{ \: k = – 3} \\ [/tex]
[tex]\implies \sf \large – 6 = k(2)[/tex]
[tex]\implies \sf \large \frac{ – 6}{2} = \frac{k( \cancel2)}{ \cancel2} \\ [/tex]
[tex]\implies \sf \large \therefore \orange{ \: k = – 3} \\ [/tex]
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3.
[tex]\sf \large \tt \green{NOT \: \: A \: \: DIRECT \: \: VARIATION}[/tex]
[tex] \large\begin{array}{ |c|c|c|c|c| } \hline x & 2 & 3 & 4 &5 \\ \hline y & 1 & 2 & 3 & 4 \\ \hline \end{array}[/tex]
[tex] \implies \sf \large y = kx[/tex]
[tex]\implies \sf \large 1 = k(2)[/tex]
[tex]\implies \sf \large \frac{ 1}{2} = \frac{k( \cancel2)}{ \cancel2} \\[/tex]
[tex]\implies \sf \large \therefore \orange{ \: k = \frac{1}{2} } \\ [/tex]
[tex]\implies \sf \large 2= k(3)[/tex]
[tex]\implies \sf \large \frac{ 2}{3} = \frac{k( \cancel3)}{ \cancel3} \\[/tex]
[tex]\implies \sf \large \therefore \orange{ \: k = \frac{2}{3} } \\ [/tex]
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5.
[tex]\sf \large \tt \green{ DIRECT \: \: VARIATION}[/tex]
[tex]\implies \sf \large y = kx[/tex]
[tex]\implies \sf \large 5 = k(1)[/tex]
[tex]\implies \sf \large \frac{ 5}{1} = \frac{k( \cancel1)}{ \cancel1} \\ [/tex]
[tex]\implies \sf \large \therefore \orange{ \: k = 5} \\ [/tex]
[tex]\implies \sf \large 10 = k(2)[/tex]
[tex]\implies \sf \large \frac{ 10}{2} = \frac{k( \cancel2)}{ \cancel2} \\ [/tex]
[tex]\implies \sf \large \therefore \orange{ \: k = 5} \\ [/tex]
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