<span>1.
Photo description: A picture of the Eiffel tower, to be stuck on a mat. Dimensions (including units): 4 in x 6 in 2. Since 2x would be added to each dimension: Length: 6 + 2x (inches)
Width: 4 + 2x (inches)
3. Area: A = LW = (6+2x)(4+2x) square inches 4. F: (6)(4) = 24, O: (6)(2x) = 12x, I: (2x)(4) = 8x, L: (2x)(2x) = 4x^2
Polynomial expression: Adding the FOIL terms up: 4x^2 + 20x + 24 5. The area should be in square inches, since we multiplied length (in inches) by width (in inches).
6. Multiply factors using the distribution method: (6+2x)(4+2x) = 6(4+2x) + 2x(4+2x) = 24 + 12x + 8x + 4x^2 = 24 + 20x + 4x^2 This is identical to the expression in Part 4. 7. x: 24 + 20x + 4x^2 If x = 1.0 in: Area = 24 + 20(1) + 4(1)^2 = 48 in^2 If x = 2.0 in: Area = 24 + 20(2) + 4(2)^2 = 80 in^2
8. If a white mat costs $0.03 per square inch and a black mat costs
$0.05 per square inch, determine the cost of each size of black and
white mat.
x Total area of mat Cost of white mat Cost of black mat
1.0 in, A = 48 in^2, (0.03)(48) = $1.44, (0.05)(48) = $2.40
2.0 in, A = 80 in^2, (0.03)(80) = $2.40, (0.05)(80) = $4.00
9. The cheapest option would be the white mat with 1-in margins on all sides, which would cost $1.44. Without any further criteria on aesthetics or size limitations, this is the most viable option.</span>
