The functions \( f(x) = 0.0925x^2 - 0.5x + 66.9 \) and \( g(x) = 0.0925x^2 + 2.7x + 12.6 \) model a car's stopping distance, \( f(x) \) or \( g(x) \), in feet, when traveling at \( x \) miles per hour. Function \( f \) models stopping distance on dry pavement, and function \( g \) models stopping distance on wet pavement. The graphs of these functions are shown for \(\{x \mid x \geq 30\}\). Use this information to complete parts (a) through (d) below.

a. Use the given functions to find the stopping distance on dry pavement and the stopping distance on wet pavement for a car traveling 55 miles per hour.

- The stopping distance on dry pavement is _ feet. (Round to the nearest whole number as needed.)
- The stopping distance on wet pavement is _ feet. (Round to the nearest whole number as needed.)

b. Based on your answers to part (a), which rectangular coordinate graph shows stopping distances on dry pavement and which shows stopping distances on wet pavement?

We substitute the value 55 into the given equations to find the stopping distance on dry pavement and wet pavement at this speed. The graph representing stopping distances should reflect larger distances for wet pavement than dry at the same speed.

Explanation:

To find the stopping distance on dry and wet pavement respectively for a car traveling at 55 miles per hour, you need to substitute 'x' with '55' in the given functions. The stopping distance on dry pavement function f(x)= 0.0925x² - 0.5x + 66.9 becomes f(55)= 0.0925*(55)² - 0.5*55 + 66.9. The stopping distance on wet pavement function g(x)= 0.0925x² +2.7x+ 12.6 becomes g(55)= 0.0925*(55)² +2.7*55+ 12.6. Then calculate these values and round it to the nearest whole number, these are the stopping distances. The graphs that show stopping distances on dry and wet pavements respectively should mirror the patterns of the calculated results: larger values for wet pavement and smaller values for dry pavement given a certain speed.

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