Math
posted by Coug .
A photo width is 2 in. more than it's length. A borderof 1 in is placed around the photo, and the area covered by the photo and it's border is 120 sq. inches. Find the dimensions of the photo itself.

It helps to draw out what the photo in the frame will look like, so you'll draw a rectangle with length "L" and width of "L + 2" (because it says the width [think w] is 2 more than the length "L").
Now, it says a border of 1 in. is placed around the photo. When you draw it, you see there is two inches added to both the length and width (one inch on either side). Now, the total width of the photo frame with the photo in it is L + 4, and the total length of it L + 2. The 120 square inches applies to the surface area of the photo AND frame, and the formula for surface area is A = LW. Soooo:
A = LW
120 = (L + 4)(L + 2)
Now solve for L to find your length:
L^2 + 6L + 8 = 120
L^2 + 6L  112 = 0
(L + 14)(L  8) = 0
Because you cannot have negative dimensions, discard the 14 you get. You now know that L = 8, so now with this information go find your width of the ORIGINAL photo, not the frame.
width = L + 2 = 8 + 2 = 10
So, the photo is 8 x 10 inches. Hope I helped! 
thank you i got it right

A rectangle picture was reduced in size from 18 inches by 30 to 6 inches by 10 inches. What is the scale factor used to reduce the picture?
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