![\bf \begin{array}{|cc|ll} \cline{1-2} \stackrel{Days}{d}&\stackrel{Price}{P}\\ \cline{1-2} 1&\$115\\ 2&\$150&\leftarrow 115+\stackrel{35(1)}{35}\\ 3&\$185&\leftarrow 115+\stackrel{35(2)}{35+35}\\ 4&\$220&\leftarrow 115+\stackrel{35(3)}{35+35+35}\\ d&&\leftarrow 115+35d\\ \cline{1-2} \end{array}~\hfill \stackrel{part~A}{P(d)=35d+115} \\\\\\ ~\hspace{34em}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Ccc%7Cll%7D%20%5Ccline%7B1-2%7D%20%5Cstackrel%7BDays%7D%7Bd%7D%26%5Cstackrel%7BPrice%7D%7BP%7D%5C%5C%20%5Ccline%7B1-2%7D%201%26%5C%24115%5C%5C%202%26%5C%24150%26%5Cleftarrow%20115%2B%5Cstackrel%7B35%281%29%7D%7B35%7D%5C%5C%203%26%5C%24185%26%5Cleftarrow%20115%2B%5Cstackrel%7B35%282%29%7D%7B35%2B35%7D%5C%5C%204%26%5C%24220%26%5Cleftarrow%20115%2B%5Cstackrel%7B35%283%29%7D%7B35%2B35%2B35%7D%5C%5C%20d%26%26%5Cleftarrow%20115%2B35d%5C%5C%20%5Ccline%7B1-2%7D%20%5Cend%7Barray%7D~%5Chfill%20%5Cstackrel%7Bpart~A%7D%7BP%28d%29%3D35d%2B115%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chspace%7B34em%7D)
part B)
the slope is always in a linear equation, the coefficient of the independent variable, in this case namely "35".
35 or 35/1 dollars/day means
for every passing day the car is rented out, the charge is $35, so if you rent the car "d" days, you'll be charged 35d.
Answer:
0ft and 60ft
Step-by-step explanation:
Given
The attached function
Required
Determine the valid values of the domain of the function
To do this, we simply consider the starting point and the end point of the trajectory on the x-axis (i.e. the horizontal distance).
From the attached graph, the horizontal distance starts from 0 and ends at 180.
This implies that the domain is: ![0 \le x \le 180](https://tex.z-dn.net/?f=0%20%5Cle%20x%20%5Cle%20180)
From the options, the values that fall in this bracket are 0ft and 60ft
Answer:
186.73 x (-0.0175) = −3.267775
Five hundredths minus two hundredths