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anonymous
 one year ago
WILL FAN AND MEDAL!
Triangle XYZ has vertices X(1, 3), Y(0, 0), and Z(–1, 2). The image of triangle XYZ after a rotation has vertices
X'(–3, 1), Y'(0, 0), and Z'(–2, –1). Which rule describes the transformation?
R0, 90°
R0, 180°
R0, 270°
R0, 360°
anonymous
 one year ago
WILL FAN AND MEDAL! Triangle XYZ has vertices X(1, 3), Y(0, 0), and Z(–1, 2). The image of triangle XYZ after a rotation has vertices X'(–3, 1), Y'(0, 0), and Z'(–2, –1). Which rule describes the transformation? R0, 90° R0, 180° R0, 270° R0, 360°

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Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0make a sketch of triangles XYZ and X'Y'Z'

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0the 90 degree rotation matrix \[\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]\]now multiplying by each point we get \[x\left(\begin{matrix}1 \\ 3\end{matrix}\right)*\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]=x'\left(\begin{matrix}1*0+3*1 \\ 1*1+3*0\end{matrix}\right)=\left(\begin{matrix}3 \\ 1\end{matrix}\right)\]\[y'\left(\begin{matrix}0 \\ 0\end{matrix}\right)\] \[z\left(\begin{matrix}1 \\ 2\end{matrix}\right)*\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]=z'\left(\begin{matrix}1*0+2*1 \\ 1*1+2*0\end{matrix}\right)=\left(\begin{matrix}2 \\ 1\end{matrix}\right)\]
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