--- ### Algebraic Problem: Speed and Distance #### Problem Statement A car travels 210 miles in the same time that a truck travels 150 miles. If the car's speed is 24 mph faster than the truck's, find the car's speed and the truck's speed. #### Instructions **a) Set up an algebraic equation (or equations) to solve the problem.** **b) Solve the equation in part a.** #### Solution Steps **Step A: Setting Up the Equation** Let's define the speeds of the car and the truck as follows: - Let \( t \) be the speed of the truck in mph. - The speed of the car is \( t + 24 \) mph. Since both vehicles travel for the same amount of time, we can use the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \] For the car: \[ 210 = (t + 24) \times \text{Time} \] For the truck: \[ 150 = t \times \text{Time} \] Since the times are equal, we can set up the following equations: \[ \text{Time for car} = \frac{210}{t + 24} \] \[ \text{Time for truck} = \frac{150}{t} \] Therefore, \[ \frac{210}{t + 24} = \frac{150}{t} \] **Step B: Solving the Equation** Let's solve the equation \(\frac{210}{t + 24} = \frac{150}{t}\): 1. Cross-multiply to get rid of the fractions: \[ 210t = 150(t + 24) \] 2. Expand and simplify the equation: \[ 210t = 150t + 3600 \] 3. Subtract \(150t\) from both sides: \[ 60t = 3600 \] 4. Divide both sides by 60: \[ t = 60 \] Thus, the speed of the truck (\( t \)) is 60 mph. The speed of the car is: \[ t + 24 = 60 + 24 = 84 \text{ mph} \] Therefore, the speed of the car is 84 mph, and the speed of the truck is 60 mph. **Evaluating Functions for Indicated Values** The task is to evaluate the function for the given value: **4) \( f(x) = x^2 - 5x + 7 \); find \( f(-3) \)** In the next steps, we will substitute \( x = -3 \) into the function and solve it to find \( f(-3) \). 1. Start by plugging in \( x = -3 \) into the function: \[ f(-3) = (-3)^2 - 5(-3) + 7 \] 2. Evaluate the expressions inside the function: \[ (-3)^2 = 9 \] \[ -5(-3) = 15 \] 3. Substitute these values back into the function: \[ f(-3) = 9 + 15 + 7 \] 4. Sum the results to get the final value: \[ f(-3) = 31 \] So, \[ f(-3) = 31 \] This method can be used to evaluate any function for an indicated value by substituting the given value into the function and simplifying.

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,...
icon
Related questions
Question
---

### Algebraic Problem: Speed and Distance

#### Problem Statement

A car travels 210 miles in the same time that a truck travels 150 miles. If the car's speed is 24 mph faster than the truck's, find the car's speed and the truck's speed.

#### Instructions

**a) Set up an algebraic equation (or equations) to solve the problem.**

**b) Solve the equation in part a.**

#### Solution Steps

**Step A: Setting Up the Equation**

Let's define the speeds of the car and the truck as follows:

- Let \( t \) be the speed of the truck in mph.
- The speed of the car is \( t + 24 \) mph.

Since both vehicles travel for the same amount of time, we can use the formula:

\[ \text{Distance} = \text{Speed} \times \text{Time} \]

For the car:

\[ 210 = (t + 24) \times \text{Time} \]

For the truck:

\[ 150 = t \times \text{Time} \]

Since the times are equal, we can set up the following equations:

\[ \text{Time for car} = \frac{210}{t + 24} \]

\[ \text{Time for truck} = \frac{150}{t} \]

Therefore,

\[ \frac{210}{t + 24} = \frac{150}{t} \]

**Step B: Solving the Equation**

Let's solve the equation \(\frac{210}{t + 24} = \frac{150}{t}\):

1. Cross-multiply to get rid of the fractions:
   \[ 210t = 150(t + 24) \]
   
2. Expand and simplify the equation:
   \[ 210t = 150t + 3600 \]

3. Subtract \(150t\) from both sides:
   \[ 60t = 3600 \]

4. Divide both sides by 60:
   \[ t = 60 \]

Thus, the speed of the truck (\( t \)) is 60 mph.

The speed of the car is:
\[ t + 24 = 60 + 24 = 84 \text{ mph} \]

Therefore, the speed of the car is 84 mph, and the speed of the truck is 60 mph.
Transcribed Image Text:--- ### Algebraic Problem: Speed and Distance #### Problem Statement A car travels 210 miles in the same time that a truck travels 150 miles. If the car's speed is 24 mph faster than the truck's, find the car's speed and the truck's speed. #### Instructions **a) Set up an algebraic equation (or equations) to solve the problem.** **b) Solve the equation in part a.** #### Solution Steps **Step A: Setting Up the Equation** Let's define the speeds of the car and the truck as follows: - Let \( t \) be the speed of the truck in mph. - The speed of the car is \( t + 24 \) mph. Since both vehicles travel for the same amount of time, we can use the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \] For the car: \[ 210 = (t + 24) \times \text{Time} \] For the truck: \[ 150 = t \times \text{Time} \] Since the times are equal, we can set up the following equations: \[ \text{Time for car} = \frac{210}{t + 24} \] \[ \text{Time for truck} = \frac{150}{t} \] Therefore, \[ \frac{210}{t + 24} = \frac{150}{t} \] **Step B: Solving the Equation** Let's solve the equation \(\frac{210}{t + 24} = \frac{150}{t}\): 1. Cross-multiply to get rid of the fractions: \[ 210t = 150(t + 24) \] 2. Expand and simplify the equation: \[ 210t = 150t + 3600 \] 3. Subtract \(150t\) from both sides: \[ 60t = 3600 \] 4. Divide both sides by 60: \[ t = 60 \] Thus, the speed of the truck (\( t \)) is 60 mph. The speed of the car is: \[ t + 24 = 60 + 24 = 84 \text{ mph} \] Therefore, the speed of the car is 84 mph, and the speed of the truck is 60 mph.
**Evaluating Functions for Indicated Values**

The task is to evaluate the function for the given value:

**4) \( f(x) = x^2 - 5x + 7 \); find \( f(-3) \)**

In the next steps, we will substitute \( x = -3 \) into the function and solve it to find \( f(-3) \).

1. Start by plugging in \( x = -3 \) into the function:
   \[ f(-3) = (-3)^2 - 5(-3) + 7 \]

2. Evaluate the expressions inside the function:
   \[ (-3)^2 = 9 \]
   \[ -5(-3) = 15 \]

3. Substitute these values back into the function:
   \[ f(-3) = 9 + 15 + 7 \]

4. Sum the results to get the final value:
   \[ f(-3) = 31 \]

So, 
\[ f(-3) = 31 \]

This method can be used to evaluate any function for an indicated value by substituting the given value into the function and simplifying.
Transcribed Image Text:**Evaluating Functions for Indicated Values** The task is to evaluate the function for the given value: **4) \( f(x) = x^2 - 5x + 7 \); find \( f(-3) \)** In the next steps, we will substitute \( x = -3 \) into the function and solve it to find \( f(-3) \). 1. Start by plugging in \( x = -3 \) into the function: \[ f(-3) = (-3)^2 - 5(-3) + 7 \] 2. Evaluate the expressions inside the function: \[ (-3)^2 = 9 \] \[ -5(-3) = 15 \] 3. Substitute these values back into the function: \[ f(-3) = 9 + 15 + 7 \] 4. Sum the results to get the final value: \[ f(-3) = 31 \] So, \[ f(-3) = 31 \] This method can be used to evaluate any function for an indicated value by substituting the given value into the function and simplifying.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 4 steps with 3 images

Blurred answer
Recommended textbooks for you
Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:
9780134463216
Author:
Robert F. Blitzer
Publisher:
PEARSON
Contemporary Abstract Algebra
Contemporary Abstract Algebra
Algebra
ISBN:
9781305657960
Author:
Joseph Gallian
Publisher:
Cengage Learning
Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra And Trigonometry (11th Edition)
Algebra And Trigonometry (11th Edition)
Algebra
ISBN:
9780135163078
Author:
Michael Sullivan
Publisher:
PEARSON
Introduction to Linear Algebra, Fifth Edition
Introduction to Linear Algebra, Fifth Edition
Algebra
ISBN:
9780980232776
Author:
Gilbert Strang
Publisher:
Wellesley-Cambridge Press
College Algebra (Collegiate Math)
College Algebra (Collegiate Math)
Algebra
ISBN:
9780077836344
Author:
Julie Miller, Donna Gerken
Publisher:
McGraw-Hill Education