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anonymous
 one year ago
Square ABCD was translated using the rule (x, y) → (x – 4, y + 15) to form A'B'C'D'. What are the coordinates of point D in the preimage if the coordinates of point D’ in the image are (9, –8)?
(13, –23)
(5, 7)
(18, 1)
(–6, –4)
anonymous
 one year ago
Square ABCD was translated using the rule (x, y) → (x – 4, y + 15) to form A'B'C'D'. What are the coordinates of point D in the preimage if the coordinates of point D’ in the image are (9, –8)? (13, –23) (5, 7) (18, 1) (–6, –4)

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LynFran
 one year ago
Best ResponseYou've already chosen the best response.2i was once told subtract the original from the spicy... to me it didnt make sense but here how it work\[\left(\begin{matrix}9 \\ 8\end{matrix}\right)\left(\begin{matrix}4 \\ 15\end{matrix}\right)=\left(\begin{matrix}13 \\ 23\end{matrix}\right)\]

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0As an alternative to @lynfran's method, you could also look at it as follows: The transformation is T: (x,y) > (x4, y+15) to go from ABCD to A'B'C'D'. The inverse transformation, \(T^{1}\) is given by: \(T^{1}\) : (x,y) > (x+4, y15) So apply the inverse transformation to D' to get back D.
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