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Use the Given Value of K to Complete the Table y=kx2y = \frac { k } { x ^ { 2 } }

Question 4

Multiple Choice

Use the given value of k to complete the table for the inverse variation model​ y=kx2y = \frac { k } { x ^ { 2 } } Plot the points on a rectangular coordinate system. x810121416y=kx2k=5\begin{array}{l}\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\\hline\end{array}\\k = 5\end{array}


A) ​ x810121416y=kx2564120514451965256\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\\hline\end{array}
 Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array}  A) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}   ​   B) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
B) ​ x810121416y=kx2564564564564564\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\\hline\end{array}
 Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array}  A) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}   ​   B) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
C) ​ x810121416y=kx2525651965144120564\begin{array}{|c|c|c|c|c|c|}\hline x & 8 & 10 & 12 & 14 & 16 \\\hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\\hline\end{array}
 Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array}  A) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}   ​   B) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
D) ​ x810121416y=kx25641205144120564\begin{array}{|c|c|c|c|c|c|}\hline x & 8 & 10 & 12 & 14 & 16 \\\hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\\hline\end{array}
 Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array}  A) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}   ​   B) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
E) ​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\& 8 & 10 & 12 & 14 & 16 \\\hline\end{array}
 Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array}  A) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}   ​   B) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D) ​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E) ​  \begin{array} { | c | c | c | c | c | c | }  \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​

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