Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
![**Function Evaluation and Solution Example**
For the function \( f(t) = \frac{2t + 3}{5} \):
**(a) Evaluate \( f(1) \).**
\[ f(1) = \frac{2(1) + 3}{5} = \frac{2 + 3}{5} = \frac{5}{5} = 1 \]
*Incorrect entry was \( 4 \).*
**(b) Solve \( f(t) = 4 \).**
To solve for \( t \):
\[ \frac{2t + 3}{5} = 4 \]
Multiply both sides by 5:
\[ 2t + 3 = 20 \]
Subtract 3 from both sides:
\[ 2t = 17 \]
Divide by 2:
\[ t = \frac{17}{2} \]
*Incorrect entry was \(\frac{17}{10}\).*
---
**Hints:**
- **(a)** Substitute the given value for \( t \) in the function and simplify.
- **(b)** Set the given expression equal to 4 and solve for \( t \).
*Note: Ensure to simplify and verify your answers for correctness.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faeeca323-3b43-4b4c-a978-afa81230e85e%2F01413d2e-a311-4a5e-8324-a1132880674b%2Fx53ul1h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Function Evaluation and Solution Example**
For the function \( f(t) = \frac{2t + 3}{5} \):
**(a) Evaluate \( f(1) \).**
\[ f(1) = \frac{2(1) + 3}{5} = \frac{2 + 3}{5} = \frac{5}{5} = 1 \]
*Incorrect entry was \( 4 \).*
**(b) Solve \( f(t) = 4 \).**
To solve for \( t \):
\[ \frac{2t + 3}{5} = 4 \]
Multiply both sides by 5:
\[ 2t + 3 = 20 \]
Subtract 3 from both sides:
\[ 2t = 17 \]
Divide by 2:
\[ t = \frac{17}{2} \]
*Incorrect entry was \(\frac{17}{10}\).*
---
**Hints:**
- **(a)** Substitute the given value for \( t \) in the function and simplify.
- **(b)** Set the given expression equal to 4 and solve for \( t \).
*Note: Ensure to simplify and verify your answers for correctness.*
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