Given that th<span>e coordinates of the vertices of △DEF are D(2, −1) , E(7, −1) , and F(2, −3) and the coordinates of the vertices of △D′E′F′ are D′(0, −1) , E′(−5, −1) , and F′(0, −3) .
Notice that the y-coordinates of the pre-image and that of the image are the same, which means that there is a reflection across the y-axis.
A refrection across the y-axis results in the change in sign of the x-coordinates of the pre-image and the image while the y-coordinate of the image remains the same as that of the pre-image.
A refrection across the y-axis of </span>△DEF with vertices D(2, −1) , E(7, −1) , and F(2, −3)
will result in and image with vertices (-2, -1), (-7, -1) and (-2, -3) respectively.
Notice that the x-coordinate of the final image △D′E′F′ with vertices <span>D′(0, −1) , E′(−5, −1) , and F′(0, −3) is 2 units greater than the vertices of the result of recting the pre-image across the y-axis.
This means that the result of refrecting the pre-image was shifted two places to the right.
Therefore, </span>the sequence of transformations that maps △DEF to △D′E′F′ are reflection across the y-axis and translation 2 units right.
Answer:Do you have a pic of the map? It would make it easier to solve if I could see the map. Its basic area. A=base times height. If the park is 50 m long and is a square, you would do 50 times 50=2500
