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To determine whether the functions \( f(x) = \frac{2x+1}{10} \) and \( g(x) = \frac{10}{2x-1} \) are inverses, we can check if \( f(g(x)) = x \) and \( g(f(x)) = x \). Let's compute these one by one.

1. **Check \( f(g(x)) \):**
   - First, substitute \( g(x) \) into \( f(x) \):
   \[
   f(g(x)) = \frac{2 \cdot \frac{10}{2x - 1} + 1}{10}
   \]
   Simplify this expression step by step.

2. **Check \( g(f(x)) \):**
   - Next, substitute \( f(x) \) into \( g(x) \):
   \[
   g(f(x)) = \frac{10}{2 \cdot \frac{2x+1}{10} - 1}
   \]
   Simplify this to see if the result is \( x \).

I'll now compute both expressions to verify whether the functions are inverses.The computations yield the following results:

1. \( f(g(x)) = \frac{2x + 19}{10(2x - 1)} \)
2. \( g(f(x)) = \frac{25}{x - 2} \)

Since neither \( f(g(x)) \) simplifies to \( x \) nor \( g(f(x)) \) simplifies to \( x \), the functions \( f(x) = \frac{2x+1}{10} \) and \( g(x) = \frac{10}{2x-1} \) are **not inverses** of each other.

Therefore, the correct answer is **False**.

Would you like more details on any part of the process? Here are some follow-up questions for further practice:

1. What does it mean for two functions to be inverses in general?
2. How can you algebraically derive the inverse of a given function?
3. Why is verifying \( f(g(x)) = x \) and \( g(f(x)) = x \) necessary to confirm that functions are inverses?
4. Can a function always have an inverse?
5. What would the graph of two inverse functions look like?

**Tip:** When finding the inverse of a function, remember to swap \( x \) and \( y \) and solve for \( y \).

The functions f(x) = (2x + 1) / 10 and g(x) = 10 / (2x - 1) are inverses. True or False?

Learn how to determine if the functions f(x) = (2x + 1) / 10 and g(x) = 10 / (2x - 1) are inverses by verifying if f(g(x)) = x and g(f(x)) = x. This problem covers key algebraic concepts, including inverse functions and function composition, suitable for students in Grades 9-12.

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