Problem 59
Question
As the cost of a purchase that is less than \(\$ 5\) increases, the amount of change received from a five-dollar bill decreases. Is this inverse variation? Explain.
Step-by-Step Solution
Verified Answer
No, this is not inverse variation, as the relationship follows a quadratic equation, not a constant product.
1Step 1: Understanding Inverse Variation
Inverse variation describes a relationship where one variable increases while the other decreases, such that the product of the two variables remains constant. In mathematical terms, for two variables, say x and y, the relationship is defined as: \( xy = k \), where \( k \) is a constant.
2Step 2: Defining Variables
Let x represent the cost of the purchase and y represent the amount of change received from a five-dollar bill. The relationship between them can be expressed as \( y = 5 - x \), since the change is calculated by subtracting the cost from five dollars.
3Step 3: Analyzing the Relationship
Substituting \( y = 5 - x \) into the inverse variation equation \( xy = k \), we have \( x(5-x) = k \). Expanding this gives \( 5x - x^2 = k \). This equation is not a product of \( xy = k \) but rather a quadratic equation.
4Step 4: Concluding the Relationship Type
Since the equation \( 5x - x^2 = k \) is not a direct product of a constant times \( x \) and \( y \), the relationship between the cost of a purchase and change received is not inverse variation.
Key Concepts
Algebraic RelationshipsQuadratic EquationsMathematical Concepts
Algebraic Relationships
Algebraic relationships describe how two or more variables interact mathematically. In these relationships, a change in one variable causes a corresponding change in another. This can take various forms such as direct variation, inverse variation, and quadratic relationships.
One common form is inverse variation, characterized by a situation where as one variable increases, the other decreases, maintaining a constant product. Mathematically, if variable \( x \) is inversely proportional to \( y \), then this relationship is expressed as \( xy = k \), where \( k \) is a constant. This contrasts with direct variation, where both variables increase or decrease together, expressed as \( y = kx \).
In our example, we looked at the cost of a purchase and the amount of change received when paying with a five-dollar bill. Initially, one might think it’s an inverse relationship, but upon deeper understanding, the relationship was more complex and didn’t fit the inverse variation model.
One common form is inverse variation, characterized by a situation where as one variable increases, the other decreases, maintaining a constant product. Mathematically, if variable \( x \) is inversely proportional to \( y \), then this relationship is expressed as \( xy = k \), where \( k \) is a constant. This contrasts with direct variation, where both variables increase or decrease together, expressed as \( y = kx \).
In our example, we looked at the cost of a purchase and the amount of change received when paying with a five-dollar bill. Initially, one might think it’s an inverse relationship, but upon deeper understanding, the relationship was more complex and didn’t fit the inverse variation model.
Quadratic Equations
Quadratic equations are mathematical statements where the highest degree of any variable involved is squared. These equations take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable.
In the problem of determining the relationship between the cost of a purchase and the change received, we derived the equation \( 5x - x^2 = k \). This is a quadratic equation, as it involves \( x^2 \). Quadratic equations can often be solved by factoring, completing the square, or using the quadratic formula.
In the problem of determining the relationship between the cost of a purchase and the change received, we derived the equation \( 5x - x^2 = k \). This is a quadratic equation, as it involves \( x^2 \). Quadratic equations can often be solved by factoring, completing the square, or using the quadratic formula.
- Factoring involves rewriting the equation as a product of its roots.
- Completing the square reconfigures the equation into a perfect square trinomial.
- The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides solutions for the values of \( x \).
Mathematical Concepts
Understanding mathematical concepts is crucial for deciphering relationships between variables in algebra and other areas of mathematics. In particular, identifying the nature of relationships—whether linear, quadratic, or inverse—is essential.
Mathematical concepts such as equations, functions, and variations enable us to represent real-world situations abstractly and predict outcomes under various conditions. These representations form the basis for much of algebraic analysis and problem-solving.
Mathematical concepts such as equations, functions, and variations enable us to represent real-world situations abstractly and predict outcomes under various conditions. These representations form the basis for much of algebraic analysis and problem-solving.
- Equations like \( y = 5 - x \) model real-life scenarios by describing the relationship between two quantities.
- Functions extend this, providing a systematic way to assess how changes in one variable affect another.
- Variations such as direct, inverse, and more complex forms like quadratic show the diverse ways quantities can relate.
Other exercises in this chapter
Problem 58
Solve each inequality. Graph the solution set and write it using interval notation. \(|x-8| \leq 12\)
View solution Problem 59
Find \(h(5)\) and \(h(-2) .\) See Example 4. $$ h(x)=|x|+2 $$
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Solve for the specified variable or expression. $$ b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2} \text { for } a^{2} $$
View solution Problem 59
Factor difference of cubes. \(x^{3}-216 y^{6}\)
View solution