Problem 64
Question
For exercises 61-64, the completed problem has one mistake. (a) Describe the mistake in words or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: In the formula \(A=\frac{10}{B}\), is the relationship between \(A\) and \(B\) a direct variation or an inverse variation? Incorrect Answer: Since as \(B\) increases, \(A\) also increases, this is a direct variation.
Step-by-Step Solution
Verified Answer
The mistake was not recognizing that \(A\) decreases as \(B\) increases. This is an inverse variation.
1Step 1: Identify the given formula
Consider the given formula: \(A = \frac{10}{B}\)
2Step 2: Recall definitions
Direct variation means that as one variable increases, the other variable also increases. Inverse variation means that as one variable increases, the other variable decreases.
3Step 3: Analyze the given relationship
In the given formula \(A = \frac{10}{B}\), analyze how \(A\) changes as \(B\) increases.
4Step 4: Identify and describe the mistake
The incorrect answer states that as \(B\) increases, \(A\) also increases, leading to the conclusion of direct variation. This is incorrect because in the formula \(A = \frac{10}{B}\), as \(B\) increases, \(A\) actually decreases.
5Step 5: Correctly categorize the relationship
Since as \(B\) increases, \(A\) decreases, the relationship is an inverse variation.
Key Concepts
Direct VariationInverse RelationshipAlgebraic FormulasProportional Relationships
Direct Variation
In direct variation, two variables change in the same direction. If one increases, the other also increases. Likewise, if one decreases, the other also decreases. This relationship is described by the formula: \[\begin{equation} y = kx \end{equation}\] Here, \(y\) varies directly as \(x\), and \(k\) is the constant of variation. To understand direct variation better, think of scenarios where this applies:
- The more hours you work, the more money you earn (assuming a constant pay rate).
- The longer you travel at a constant speed, the further you go.
Inverse Relationship
An inverse relationship operates differently from a direct variation. Here, as one variable increases, the other decreases. This can be shown by the formula: \[\begin{equation} y = \frac{k}{x} \end{equation}\] In this formula, \(y\) varies inversely as \(x\), and \(k\) is the constant of variation. To explore this concept further, consider examples of inverse relationships:
- The time it takes to travel a fixed distance varies inversely with speed. If you drive faster, the travel time decreases.
- If you divide a fixed amount of money among more people, the amount each person gets decreases.
Algebraic Formulas
Algebraic formulas help us express relationships between different quantities. They can describe direct variations, inverse variations, and other relationships. The key to working with these formulas is understanding what each variable and constant represents. For example, in the formula \[\begin{equation} A = \frac{10}{B} \end{equation}\] \(A\) is inversely related to \(B\). This means, as \(B\) grows, \(A\) shrinks.To correctly use algebraic formulas, follow these steps:
- Identify the variables and constants in the formula.
- Understand the meaning of each component and how they relate.
- Analyze how changes in one variable affect the other variable(s).
Proportional Relationships
Proportional relationships are foundational in understanding direct and inverse variations. In a proportional relationship, two quantities maintain a constant ratio to each other. For direct variations, this ratio is \(k\) in the formula \[\begin{equation} y = kx \end{equation}\] For inverse variations, the ratio is seen in the product, \[\begin{equation} y = \frac{k}{x}. \end{equation}\]. Consider some real-world proportional relationships:
- The ratio of distance to time at a constant speed remains constant.
- Selling items at a constant price per unit keeps the ratio of total cost to number of items proportional.
Other exercises in this chapter
Problem 62
For exercises 1-66, simplify. $$ \frac{2 c^{3}-2 c^{2}-4 c}{4 c^{3}-8 c^{2}-12 c} $$
View solution Problem 63
For exercises 39-82, simplify. $$ \frac{2 z+6}{z^{2}+3 z+2} \div \frac{z+3}{z+2} $$
View solution Problem 64
For exercises 39-82, simplify. $$ \frac{3 a+6}{a^{2}+4 a+4} \div \frac{a+4}{a+2} $$
View solution Problem 65
For exercises \(65-68\), evaluate. $$ \sqrt{16} $$
View solution