Problem 58
Question
In Exercises \(57-76,\) solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe? \(D=R T\) for \(R\)
Step-by-Step Solution
Verified Answer
The formula when solved for 'R' is \(R = \frac{D}{T}\)
1Step 1: Identify the Formula
The provided formula is \(D = RT\). Given that 'D' represents distance, 'R' is rate or speed, and 'T' is time, this formula describes how distance travelled is a product of speed and time.
2Step 2: Isolate the Variable R
To solve the equation for 'R', you need to isolate 'R'. That can be achieved by dividing both sides of the equation by 'T'. Doing so, you get \(R = \frac{D}{T}\)
3Step 3: Interpret the Result
The rearranged equation, \(R = \frac{D}{T}\), means that the rate (R) can be determined by dividing the distance (D) by the time (T).
Key Concepts
Distance-Rate-Time EquationIsolating VariablesAlgebraic Manipulation
Distance-Rate-Time Equation
Understanding the relationship between distance, rate, and time is essential in various fields, from physics to everyday life logistics. The distance-rate-time equation is a simple yet fundamental concept that connects these three variables. In its basic form, the equation is written as (D = RT), where 'D' stands for the distance traveled, 'R' is the rate or speed at which the travel occurs, and 'T' represents the time taken to cover that distance.
To use this equation effectively, one must grasp that the distance is directly proportional to both the rate and the time. This means that if you increase the speed while keeping the time constant, you'll cover a greater distance, and likewise, traveling for a longer time at the same speed will also result in a longer distance traveled. Conversely, if you need to calculate the speed or time, you can rearrange the equation accordingly. The ease of this calculation makes it an excellent tool for solving real-world problems, such as estimating travel times or planning trips.
To use this equation effectively, one must grasp that the distance is directly proportional to both the rate and the time. This means that if you increase the speed while keeping the time constant, you'll cover a greater distance, and likewise, traveling for a longer time at the same speed will also result in a longer distance traveled. Conversely, if you need to calculate the speed or time, you can rearrange the equation accordingly. The ease of this calculation makes it an excellent tool for solving real-world problems, such as estimating travel times or planning trips.
Isolating Variables
Isolating a variable is a critical step in solving equations and understanding the relationships between different mathematical elements. It involves manipulating an equation so that the variable of interest stands alone on one side of the equation, making its value clearly identifiable. When confronting an equation like (D = RT), isolating a variable, such as 'R', allows us to find how the rate relates to distance and time when the other values are known.
Practical Use
- For planning: Determine how fast you must travel to reach a destination in a given time.
- For analysis: Calculate the speed of a vehicle based on the distance covered and the time taken.
- For academic problems: Find unknown values in physics, chemistry, or even finance equations.
Algebraic Manipulation
Algebraic manipulation is the mathematical art of rearranging equations and expressions to solve for specific variables. It involves a variety of operations such as adding, subtracting, multiplying, dividing, and factoring, all of which help in solving algebraic equations. When we isolate 'R' in the formula (D = RT), we perform algebraic manipulation by dividing both sides by 'T'. This process, which yields (R = D/T), demonstrates a direct application of algebraic manipulation.
Benefits of Mastering Algebraic Manipulation
- It allows the simplification of complex equations.
- It enhances problem-solving efficiency.
- It creates flexibility to solve for different variables in an equation based on the situation's requirements.
Other exercises in this chapter
Problem 58
Solve each equation in by making an appropriate substitution. $$ \left(y-\frac{10}{y}\right)^{2}+6\left(y-\frac{10}{y}\right)-27=0 $$
View solution Problem 58
Solve each equation in Exercises \(55-64\) using the quadratic formula. $$ x^{2}+5 x+2=0 $$
View solution Problem 58
Solve each inequality in Exercises 57-84 by first rewriting each one as an equivalent inequality without absolute value bars. Graph the solution set on a number
View solution Problem 58
In Exercises \(51-58,\) determine whether each equation is an identity, a conditional equation, or an inconsistent equation. $$ \frac{3}{x-3}=\frac{x}{x-3}+3 $$
View solution