Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

The function [tex]c(n)[/tex] below relates the number of bushels of apples picked at a pick-your-own orchard to the final cost for the apples.

It takes as input the number of bushels of apples picked after paying an entry fee to an orchard, and it returns as output the cost of the apples (in dollars).

[tex]\[ c(n) = 15n + 30 \][/tex]

Which equation below represents the inverse function [tex]n(c)[/tex], which takes the cost of the apples as input and returns the number of bushels picked as output?

A. [tex]\[ n(c) = \frac{c-30}{15} \][/tex]

B. [tex]\[ n(c) = \frac{c-15}{30} \][/tex]

C. [tex]\[ n(c) = \frac{c+15}{30} \][/tex]

D. [tex]\[ n(c) = \frac{c+30}{15} \][/tex]

Sagot :

GinnyAnswer
To find the inverse of the function \( c(n) = 15n + 30 \), we need to:

1. Swap \( c \) and \( n \) in the equation.
2. Solve for \( n \) in terms of \( c \).

Here is the step-by-step process:

1. Start with the original equation:
[tex]\[ c(n) = 15n + 30 \][/tex]

2. Swap \( c \) and \( n \):
[tex]\[ c = 15n + 30 \][/tex]

3. Isolate \( n \) by first subtracting 30 from both sides:
[tex]\[ c - 30 = 15n \][/tex]

4. Divide both sides by 15 to solve for \( n \):
[tex]\[ n = \frac{c - 30}{15} \][/tex]

So, the inverse function \( n(c) \) is:
[tex]\[ n(c) = \frac{c - 30}{15} \][/tex]

Therefore, the correct equation is:

A. [tex]\( n(c) = \frac{c - 30}{15} \)[/tex]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.