Problem 65
Question
Solve for the indicated variable. Solve for \(r: t=D r\)
Step-by-Step Solution
Verified Answer
\(r = \frac{t}{D}\)
1Step 1: Identify the Equation
Start with the given equation: \( t = Dr \). This equation relates the variables \(t\), \(D\), and \(r\). Our goal is to solve for \(r\).
2Step 2: Isolate the Variable
To solve for \(r\), we need to isolate \(r\) on one side of the equation. The current equation is \( t = Dr \). To isolate \(r\), divide both sides of the equation by \(D\).
3Step 3: Simplify the Equation
After dividing both sides by \(D\), the equation becomes: \( \frac{t}{D} = r \). This equation shows that \(r\) is equal to \(\frac{t}{D}\).
Key Concepts
Solving for VariablesEquation ManipulationVariables and Constants
Solving for Variables
Solving for variables is a fundamental skill in algebra. It involves finding the value of a variable that makes an equation true. In the exercise given, we're asked to solve for the variable \( r \) in the equation \( t = Dr \).
When you solve for a variable, you essentially isolate it on one side of the equation, so it's alone. This gives you a clearer view of what the variable equals or represents.
This solving technique is a building block for tackling more complicated equations in the future.
When you solve for a variable, you essentially isolate it on one side of the equation, so it's alone. This gives you a clearer view of what the variable equals or represents.
- For instance, starting with the equation \( t = Dr \), our goal is to have \( r \) on one side and everything else on the other.
- Think of it as organizing a room: all the elements related to \( r \) (like ''D'') need to be moved away to make space and clearly see what \( r \) is equal to.
This solving technique is a building block for tackling more complicated equations in the future.
Equation Manipulation
Equation manipulation involves performing operations to both sides of an equation to keep it balanced. This is like adjusting both sides of a scale to maintain equilibrium while achieving the desired arrangement.
In our equation \( t = Dr \), equation manipulation is used to isolate \( r \). By dividing both sides of the equation by \( D \), we effectively "cancel out" the \( D \) attached to \( r \), leading us to \( \frac{t}{D} = r \).
This careful balancing act ensures the equality is still valid while allowing us to manipulate the equation to our needs.
In our equation \( t = Dr \), equation manipulation is used to isolate \( r \). By dividing both sides of the equation by \( D \), we effectively "cancel out" the \( D \) attached to \( r \), leading us to \( \frac{t}{D} = r \).
- Remember, whatever operation you do to one side, you must do the same to the other side.
- This can be adding, subtracting, multiplying, or dividing, as long as the end goal of isolating the variable is achieved.
This careful balancing act ensures the equality is still valid while allowing us to manipulate the equation to our needs.
Variables and Constants
In every algebraic equation, understanding the roles of variables and constants is crucial. Variables represent unknown values we aim to solve, while constants are fixed values that provide context and constraints.
Taking the equation \( t = Dr \) from our exercise, \( r \) is our variable because it's what we're solving for—its value isn't initially known.
Understanding this distinction helps in shaping the solution approach. Recognizing what's a variable and what's a constant guides how we manipulate the equation to find the desired value.
Taking the equation \( t = Dr \) from our exercise, \( r \) is our variable because it's what we're solving for—its value isn't initially known.
- Constants, like \( D \) in this context, are typically known quantities or coefficients that give structure to the equation.
- \( t \) could either be a constant or another variable depending on the problem context.
Understanding this distinction helps in shaping the solution approach. Recognizing what's a variable and what's a constant guides how we manipulate the equation to find the desired value.
Other exercises in this chapter
Problem 64
The breaking distance of an automobile is directly proportional to the square of its speed. The time, \(t\), it takes an object to fall is directly proportional
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$$ -2 x+14 x 3-49 x $$
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Simplify. $$ 1 x+3-x x 2-6 x+9+3 x 2-9 $$
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If two objects with masses 50 kilograms and 100 kilograms are \(1 / 2\) meter apart, then they produce approximately \(1.34 \times 10-6\) newtons (N) of force.
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