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$2 A solid has volume 7 cubic units. The equation k=root(3)((V)/(7)) represents the scale factor of k by which the solid must be dilated to obtain an image with volume V cubic units. Select all points which are on the graph representing this equation. A (0,0) B (1,1) C (1,7) D (7,1) E (14,2) F \$(49,2)\\ G \\((56,2)\$ H \$(27,3)\$$

$2$ A solid has volume $7$ cubic units. The equation $k=\sqrt[3]{\frac{V}{7}}$ represents the scale factor of $k$ by which the solid must be dilated to obtain an image with volume $V$ cubic units. Select all points which are on the graph representing this equation. $\newline$ A $(0,0)$ $\newline$ B $(1,1)$ $\newline$ C $(1,7)$ $\newline$ D $(7,1)$ $\newline$ E $(14,2)$ $\newline$ F $\backslash((49,2) \backslash$ $\newline$ G $\backslash((56,2) \backslash)$ $\newline$ H $k$ $0$

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2

A solid has volume

7

cubic units. The equation

k=\sqrt[3]{\frac{V}{7}}

represents the scale factor of

k

by which the solid must be dilated to obtain an image with volume

V

cubic units. Select all points which are on the graph representing this equation.

\newline

(0,0)

\newline

(1,1)

\newline

(1,7)

\newline

(7,1)

\newline

(14,2)

\newline

\backslash((49,2) \backslash

\newline

\backslash((56,2) \backslash)

\newline

k

0

Check Point A: Plug in the value from point A $(0,0)$ into the equation to check if it satisfies the equation: $k=\sqrt[3]{\frac{0}{7}}$ . Calculate $k$ for point A: $k=\sqrt[3]{\frac{0}{7}} = \sqrt[3]{0} = 0$ . Since $k=0$ when $V=0$ , point A $(0,0)$ is on the graph.
Calculate $k$ for Point A: Plug in the value from point B $(1,1)$ into the equation to check if it satisfies the equation: $k=\sqrt[3]{\frac{1}{7}}$ . Calculate $k$ for point B: $k=\sqrt[3]{\frac{1}{7}} = \sqrt[3]{\frac{1}{7}} \approx 0.192$ . Since $k$ does not equal $1$ when $V=1$ , point B $(1,1)$ is not on the graph.
Check Point B: Plug in the value from point C $(1,7)$ into the equation to check if it satisfies the equation: $k=\sqrt[3]{\left(\frac{7}{7}\right)}$ . Calculate $k$ for point C: $k=\sqrt[3]{\left(\frac{7}{7}\right)} = \sqrt[3]{1} = 1$ . Since $k=1$ when $V=7$ , point C $(1,7)$ is on the graph.
Calculate $k$ for Point B: Plug in the value from point D $(7,1)$ into the equation to check if it satisfies the equation: $k=\sqrt[3]{\left(\frac{1}{7}\right)}$ . Calculate $k$ for point D: $k=\sqrt[3]{\left(\frac{1}{7}\right)} = \sqrt[3]{\frac{1}{7}} \approx 0.192$ . Since $k$ does not equal $7$ when $V=1$ , point D $(7,1)$ is not on the graph.
Check Point C: Plug in the value from point E $(14,2)$ into the equation to check if it satisfies the equation: $k=\sqrt[3]{\left(\frac{14}{7}\right)}$ . $\newline$ Calculate $k$ for point E: $k=\sqrt[3]{\left(\frac{14}{7}\right)} = \sqrt[3]{2} \approx 1.260$ . $\newline$ Since $k$ does not equal $2$ when $V=14$ , point E $(14,2)$ is not on the graph.
Calculate $k$ for Point C: Plug in the value from point F $(49,2)$ into the equation to check if it satisfies the equation: $k=\sqrt[3]{\frac{49}{7}}$ . Calculate $k$ for point F: $k=\sqrt[3]{\frac{49}{7}} = \sqrt[3]{7} \approx 1.913$ . Since $k$ does not equal $2$ when $V=49$ , point F $(49,2)$ is not on the graph.
Check Point D: Plug in the value from point G $(56,2)$ into the equation to check if it satisfies the equation: $k=\sqrt[3]{\frac{56}{7}}$ . Calculate $k$ for point G: $k=\sqrt[3]{\frac{56}{7}} = \sqrt[3]{8} = 2$ . Since $k=2$ when $V=56$ , point G $(56,2)$ is on the graph.
Calculate $k$ for Point D: Plug in the value from point H $(27,3)$ into the equation to check if it satisfies the equation: $k=\sqrt[3]{\frac{27}{7}}$ . Calculate $k$ for point H: $k=\sqrt[3]{\frac{27}{7}} = \sqrt[3]{\frac{27}{7}} \approx 1.817$ . Since $k$ does not equal $3$ when $V=27$ , point H $(27,3)$ is not on the graph.