Question

which sequences of transformations map lmn onto lmn? select all that apply. a reflection across the y - axis followed by rotation 90° clockwise around the origin a rotation 90° counterclockwise around the origin followed by a reflection across the y - axis a reflection across the x - axis followed by a rotation 90° counterclockwise around the origin a rotation 90° clockwise around the origin followed by a reflection across the x - axis

Explanation:

Step1: Recall transformation rules

Let's consider the general rules for reflection and rotation. Reflection across the y - axis changes the sign of the x - coordinate ($(x,y)\to(-x,y)$), reflection across the x - axis changes the sign of the y - coordinate ($(x,y)\to(x, - y)$), a 90 - degree clockwise rotation around the origin has the transformation $(x,y)\to(y,-x)$ and a 90 - degree counter - clockwise rotation around the origin has the transformation $(x,y)\to(-y,x)$.

Step2: Analyze the first option

For a reflection across the y - axis: $(x,y)\to(-x,y)$. Then for a 90 - degree clockwise rotation around the origin of $(-x,y)$ gives $(y,x)$.

Step3: Analyze the second option

For a 90 - degree counter - clockwise rotation around the origin: $(x,y)\to(-y,x)$. Then for a reflection across the y - axis of $(-y,x)$ gives $(y,x)$.

Step4: Analyze the third option

For a reflection across the x - axis: $(x,y)\to(x,-y)$. Then for a 90 - degree counter - clockwise rotation around the origin of $(x,-y)$ gives $(y,x)$.

Step5: Analyze the fourth option

For a 90 - degree clockwise rotation around the origin: $(x,y)\to(y,-x)$. Then for a reflection across the x - axis of $(y,-x)$ gives $(y,x)$.

Answer:

All of the above options (a reflection across the y - axis followed by rotation 90° clockwise around the origin; a rotation 90° counterclockwise around the origin followed by a reflection across the y - axis; a reflection across the x - axis followed by a rotation 90° counterclockwise around the origin; a rotation 90° clockwise around the origin followed by a reflection across the x - axis) map $\triangle LMN$ onto $\triangle L'M'N'$.