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?

To solve this problem, we need to set up an equation based on the given information and then solve for the dimensions of the photo. Let's start by defining the dimensions of the photo itself.

Let's assume that the length of the photo is 'L' inches. According to the problem statement, the width of the photo is 2 inches more than its length, so the width would be 'L + 2' inches.

Now, let's consider the border placed around the photo. The border has a consistent width of 1 inch on all sides. Therefore, the total length and width of the photo with the border would be 'L + 2 + 2' and 'L + 2 + 2', respectively.

The area covered by the photo and its border is given as 120 square inches. To find the dimensions of the photo, we need to subtract the area of the border from the total area.

Total area = (length + 2) * (width + 2)
Photo area = length * width

By subtracting the photo area from the total area, we get the area of the border:

Border area = Total area - Photo area

Now, let's set up the equation and solve for the dimensions of the photo.

Total area - Photo area = Border area
(L + 2 + 2) * (L + 2 + 2) - L * (L + 2) = 120

Simplifying the equation:

(L + 4) * (L + 4) - L * (L + 2) = 120
L^2 + 8L + 16 - L^2 - 2L = 120
6L + 16 = 120
6L = 120 - 16
6L = 104
L = 104 / 6
L = 17.33 (rounded to two decimal places)

Therefore, the length of the photo itself is approximately 17.33 inches.
The width of the photo can be found by adding 2 inches to the length:

Width = L + 2 = 17.33 + 2 = 19.33 (rounded to two decimal places)

Hence, the dimensions of the photo itself are approximately 17.33 inches by 19.33 inches.