West of a city, a certain eastbound route is straight and makes a steep descent toward the city. The highway has a 11% grade, which means that its slope is [tex]$-\frac{11}{100}$[/tex].
Sure, let's work through this step-by-step to fully understand the problem and the solution.
1. Understand the Given Information: - The highway has a grade value of 1190. - The grade indicates the steepness of the slope in percentage. Here the grade value corresponds to a slope calculation provided as [tex]\(-\frac{11}{100}\)[/tex].
2. Grade and Slope Relationship: - The given grade of [tex]\(-\frac{11}{100}\)[/tex] translates to the slope of the highway in fractional form. This means for every 100 units of horizontal distance, the highway drops 11 units of vertical distance (since it's a descent, the slope is negative).
3. Convert Slope to Percentage: - The next step is to convert the slope [tex]\(-\frac{11}{100}\)[/tex] into a percentage to understand the steepness as commonly expressed in road grades. - The slope [tex]\(-\frac{11}{100}\)[/tex] is equivalent to multiplying by 100 to obtain the percentage. - Therefore, [tex]\(-\frac{11}{100} \times 100 = -11\%\)[/tex].
4. Discuss the Interpretation: - The highway's grade of 1190 corresponds to a slope of [tex]\(-11\%\)[/tex]. - This suggests that the highway descends 11 meters vertically for every 100 meters horizontally. The negative sign is indicative of a downward slope.
Summary: - The highway with a 1190 grade has a slope of [tex]\(-0.11\)[/tex]. - This slope converts to a grade percentage of [tex]\(-11\%\)[/tex].
I hope this detailed explanation clarifies the relationship between the given grade and the corresponding slope and grade percentage. Please feel free to ask if you have any further questions!
