△DEF has vertices D (2,2), E (-2,-1), and F (-3,5). Complete the following charts indicating the location of △D′E′F′ and △D′′E′′F′′

△DEF
is reflected in the y-axis. Then it is translated along the vector ⟨3,−5⟩.
D (2,2) D`E`F` (?,?), D"E"F" (?,?)
E (-2,1) D`E`F` (?,?) D"E"F" (?,?)
F (-3,5) D`E`F` (?,?) D"E"F" (?,?)

Given:

D (2,2), E (-2,-1), and F (-3,5).

The reflection about the y-axis follows the following rule:
sy: (x,y)->(-x,y)
For example:
sy: D(2,2)->D'(-2,2)

The translation along a vector (p,q) follows the following rule:
T(p,q): (x,y)->(x+p, y+q)
Example:
T(3,-5): D'(-2,2)-> D"(-2+3,2-5)=D"(1,-3)

You can proceed in a similar way for E and F.
Post your results for checking if you wish.

To determine the locations of the reflected and translated triangles, we can follow these steps:

1. Reflecting Triangle △DEF in the y-axis:
To reflect the coordinates across the y-axis, we need to change the signs of the x-coordinates while keeping the y-coordinates the same.

D (2, 2) would become D' (-2, 2)
E (-2, -1) would become E' (2, -1)
F (-3, 5) would become F' (3, 5)

2. Translating Triangle △D'E'F' along the vector ⟨3, -5⟩:
To translate a point along a vector, we add the vector's components to the original coordinates.

D' (-2, 2) + ⟨3, -5⟩ = D" (1, -3)
E' (2, -1) + ⟨3, -5⟩ = E" (5, -6)
F' (3, 5) + ⟨3, -5⟩ = F" (6, 0)

Now we can fill in the chart:

△D'E'F':
D' (-2, 2) E' (2, -1) F' (3, 5)

△D"E"F":
D" (1, -3) E" (5, -6) F" (6, 0)

Please note that this explanation assumes the vector is being added to each individual coordinate, resulting in the translated points.