14 points, i need an answer
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     Perimeter= 24 in.

PART-B       Area of inside frame=6×4=24 square inches.

2)  

4)  Dimensions of outside frame= 12.5 ft×17.5 ft.

1)

PART-A

the outside frame has the measures as: 7 in. and 5 in.

i.e. the frame is in shape of a rectangle.

Perimeter of the outside edge= 7+7+5+5=14+10=24 in.

Hence, Perimeter= 24 in.

PART-B

Now we are asked to find the area of the inside of a frame as the frame is in the shape of a rectangle.

We know that the area of rectangle= l×b

where l represents the length of the frame and b represents the breadth of the rectangle.

Here we have l=6 in.

and b=4 in.

Hence,  Area of inside frame=6×4=24 square inches.

2)

Now we are given that the area of the frame was 64 inches².

So the possible dimensions could be:

4)

Now we are using a scale factor as:

1 in.=2.5 ft.

the dimension of the outside frame was: 5 in.×7 in.

As 1 in.=2.5 ft.

5 in.=5×2.5=12.5 ft.

and 7 in.=7×2.5=17.5 ft.

Hence, the new dimension after using the scale factor is:

12.5 ft×17.5 ft.

c. 16 units

step-by-step explanation:

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22.83 repeating

step-by-step explanation:

137 ÷ 6 is 22.8333333 therefore we understand that 3 is repeating.

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