1) part a-find the perimeter of the outside edges of the frame. ( explain and show work)
part b-find the area of the inside of the frame where the picture would belong and the area of the entire frame using the outside values. ( explain and show work)
2) if the area of the frame was 64 inches², what would be 2 possible sets of dimensions that could be used to make a frame with that area? ( explain and show work)
4) the frame is being enlarged to hang on a gallery wall. using the scale factor of 1in=2.5ft., give the new dimensions of the outside edges of the frame. ( explain and show work)

1) Part A-Find the PERIMETER of the outside edges of the frame. (Please explain and show work)
 The perimeter of a rectangle in is given by:
 P = 2L + 2w
 Where,
 L = long
 W = width
 We have then that the perimeter of the outer edges is
 P = 2 * (5) + 2 * (7)
 P = 24 in
 Answer
 the PERIMETER of the outside edges of the frame is 24 in.

 Part B-Find the AREA of the inside of the frame where the picture would belong and the area of the entire frame using the outside values. (Please explain and show work)
 The area of a rectangle is given by
 A = w * l
 where
 l = long
 w = width
 we have then that for
 the inside of the frame:
 A = (4) * (6) = 24in ^ 2
 the area of the entire frame: 
 A = (5) * (7) = 35in ^ 2

 2) IF the area of the frame was 64 inches², what would be 2 possible sets of dimensions that could be used to make a frame with that area? (Please explain and show work)
 For this case we can assume for example that the frame is a square.
The area of a square is given by
 A = L ^ 2
 Where L is the side of the square
 Equaling the area to 64 inches² we have:
 A = L ^ 2 = 64
 Clearing L
 L = (64) ^ (1/2)
 L = 8 in
 answer
 2 possible sets of dimensions that could be used to make a frame with that area would be (8 in) * (8 in)

 4) The frame is being enlarged to hang on a gallery wall. Using the scale factor of 1in = 2.5ft., Give the new dimensions of the outside edges of the frame. (Please explain and show work)

 The first thing we should do is move the feet to inches
 (2.5) * (12) = 30in
 Then, the new dimensions of the outer edges of the frame are
 L = (7) * (30) = 210 in
 w = (5) * (30) = 150 in
 answer
 the new dimensions of the outside edges of the frame are
 L = 210 in
 w = 150 in

Gocohcpuccphjvpjvvjiv