## Algebra 1

a) d=1.8w b)No, you can't project a picture 7 ft wide. c) The maximum width of image is 6$\frac{2}{3}$ ft
a) The function for that relation is d=1.8w because the distance of the projector from wall you have to multiply the images' width with 1.8 which makes 1.8 as the slope and w as the independent variable because it is affecting the dependent variable which is distance. b)No, you can't fit an image with width of 7 ft because the max distance of projector from Wall is 12 ft which means d=12 and to find width we divide 12 by 1.8. 12=1.8w w=$\frac{12}{1.8}$ = 6.7 ft c) Using the work from part b, we can conclude that 6.7 or 6$\frac{2}{3}$ ft is the maximum width of the image with 12 ft as the distance between the projector and wall.