Question

A car traveling at a rate \(R\) for a time \(T\) will go a distance equal to \(R T .\) If the rate is decreased by 8.5 mi/h and the time is increased by 2.4 h, write an expression for the new distance traveled, and multiply out.

Short answer

The expression for the new distance traveled is D' = R*T + 2.4R - 8.5T - 20.4.

Step-by-step solution

- Define the original rate and time

The original rate of the car is given as R (in mi/h) and the original time the car travels is given as T (in hours). The original distance is found using the formula distance = rate × time, which is given as D = RT.

- Adjust the rate and the time

The rate is decreased by 8.5 mi/h and the time is increased by 2.4 hours. Therefore the new rate is R - 8.5 mi/h and the new time is T + 2.4 hours.

- Write the expression for the new distance

The new distance traveled, D', can be expressed as the product of the new rate and the new time. Hence, the new distance is D' = (R - 8.5) * (T + 2.4).

- Expand the expression

To find the expanded form of the new distance, use the distributive property to multiply each term in the first parenthesis by each term in the second parenthesis. Thus, the expanded form of the new distance is D' = R*T + 2.4R - 8.5T - 20.4.

Key concepts

Technical Mathematics

Technical mathematics is an area of applied mathematics that deals with mathematical techniques commonly used in the fields of engineering, physics, computer science, and other technical and scientific fields. It encompasses various topics including algebra, trigonometry, calculus, and statistics.

In the context of our exercise regarding a car traveling a certain distance, technical mathematics helps us to model and solve problems related to distance-rate-time relationships using algebraic methods. This form of application is crucial in technical fields where such relations are used for project planning, design, and analysis.

Understanding technical mathematics allows us to accurately calculate changes in situations such as the speed of a car or the time taken for a trip when certain variables are altered. It also involves using appropriate units of measurement and understanding the significance of constants and coefficients in formulas.

Algebraic Expression

An algebraic expression is a mathematical phrase that can contain numbers, variables (like the rate R and time T in our problem), and operation symbols. It can be a simple expression or a more complex equation that represents a variety of situations.

In our scenario, the algebraic expression for the new distance traveled by the car \( D' = (R - 8.5) \ast (T + 2.4) \) is an example of a more complex algebraic expression. This expression encapsulates the relationship between the distance traveled, rate of speed, and time of travel in a condensed form. Breaking down such expressions into simpler parts can help students grasp the underlying relationships between these variables.

To effectively study algebraic expressions, one should understand the role of variables and how operations can alter their values. The ability to manipulate algebraic expressions is foundational in solving real-world problems algebraically.

Distributive Property

The distributive property is a cornerstone of algebra that allows us to multiply a single term by each term within a parenthesis. It is expressed as \( a(b + c) = ab + ac \) where a, b, and c are any terms that could be numbers, variables, or more complex expressions.

In our car travel problem, we applied the distributive property to expand the new distance expression \( D' = (R - 8.5) \ast (T + 2.4) \) into \( D' = RT + 2.4R - 8.5T - 20.4 \). To do this, we multiplied the term R by each term inside the parenthesis (T + 2.4) and then multiplied -8.5 by each term in the parenthesis.

This property is particularly useful for simplifying algebraic expressions and is used in everything from basic algebra to more advanced calculus. By mastering the distributive property, students can greatly enhance their problem-solving abilities in algebra and beyond.