For the function fƒ (t) = (a) Evaluate f (1). f(1) = i 4 (b) Solve f (t) = 4. t - 17/10 2t + 3 5'

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,...
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**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.*
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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