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Mia is going out to eat with some friends and has a total of $15 to spend. She got a water to drink and plans to leave a $3 tip. The price of the entrees range from $5.98 to $24…
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Mia is going out to eat with some friends and has a total of $15 to spend. She got a water to drink and plans to leave a $3 tip. The price of the entrees range from $5.98 to $24…
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Feb 15, 2019
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Mia is going out to eat with some friends and has a total of $15 to spend. She got a water to drink and plans to leave a $3 tip. The price of the entrees range from $5.98 to $24.99. She must also remember that there is a 7% sales tax applied to the cost of her meal. The total cost she pays for her meal, including the tip, can be modeled by the function f(c), where c represents the price of the entree she orders.
The domain of the function is__ ≤ c ≤ __.
Mia's friend Clara loans her $5 so that Mia has more entrees to choose from on the menu. With this additional $5, the domain of the function representing the price of an entree Mia can afford is now __≤ c ≤ __.
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Aktau
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1.
First we are going to find the cost function.
We know for our problem that there is a 7% sales tax applied to the cost of her meal and she plans to leave 3$ as tip. Let
represent the cost of the meal:
where
is the cost of the entree
is the total cost she will pay
We know that the minimum cost of an entree is $5.98, so the domain of our functions so far is:
Now, to find the upper limit of the domain, we are going to take advantage of the fact the she only has $15 to spend, so we can replace the total cost with 15 and solve for
:
We can conclude that the domain of the function is
2.
We know that Mia's friend Clara loans her $5, so Mia's total money now is $15+$5=$20. Since the minimum cost for an entree remains the same ($5.98), the lower limit of our domain remains the same:
Now, to find the upper limit of our domain, we are going to replace the total cost with 20 an solve for
:
We can conclude that the domain of the function representing the price of an entree Mia can afford is now
Uttam Panchasara
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Feb 20, 2019
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Uttam Panchasara
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