Problem 42
Question
Write an algebraic expression that represents the relationship in each table. $$\begin{array}{|c|c|} \hline \text { Wruming } & \text { Total coss } \\ \hline \text { Ofitems } & \\ \hline 5 & \$ 25 \\ \hline 6 & \$ 30 \\ \hline 8 & \$ 40 \\ \hline 10 & \$ 50 \\ \hline n & ? \\ \hline \end{array}$$
Step-by-Step Solution
Verified Answer
The algebraic expression for the relationship is \(5n\).
1Step 1: Observe the given data
Look at the table to understand the relationship between the number of items (Wruming Ofitems) and the total cost (Total coss).
2Step 2: Identify pattern of change
Observe how the Total cost changes as the number of items increases. As the number of items increases by 1, the Total cost increases by $5. This suggests a linear relationship.
3Step 3: Determine the formula
The total cost seems to be the product of the number of items and a fixed price per item. From 5 to 25, 6 to 30, 8 to 40, 10 to 50, each increment corresponds to multiplying the number of items by 5.
4Step 4: Write the algebraic expression
Let the number of items be represented by the variable \(n\). Each item costs $5, so the expression for Total cost is \(5n\).
5Step 5: Verify with given values
Substitute \(n = 5, 6, 8, 10\) into \(5n\) to make sure the total cost calculated matches the values in the table. For \(n = 5\), Total cost = 25; for \(n = 6\), Total cost = 30; similarly verify for other values.
Key Concepts
Linear RelationshipsVariables in MathematicsPatterns in DataCost Analysis
Linear Relationships
In mathematics, a linear relationship is one where variables change at a constant rate relative to each other. This means that if you plot the values of two related variables, they create a straight line on a graph. This exercise explored a linear relationship between the number of items and their total cost.
In this scenario, for every single item added, the cost increased by a constant amount of $5. This change in cost with respect to each item being added is indicative of a linear relationship. Let's break down this relationship a bit more:
In this scenario, for every single item added, the cost increased by a constant amount of $5. This change in cost with respect to each item being added is indicative of a linear relationship. Let's break down this relationship a bit more:
- Consistent Change: In the table, you can see that as the number of items increases, the cost increases by the same fixed amount every time — this confirms a linear pattern.
- Graphical Representation: If we plotted these points on a graph, the points would align perfectly along a line, underscoring a linear relationship.
- Rate of Change: The rate of growth here is $5 per item, which is the slope of the line in graphical terms.
Variables in Mathematics
Variables are symbols or letters used to represent numbers in equations and expressions. They hold an important place because they allow us to generalize problems and solve for unknowns.
In the given problem, the table mentioned 'Wruming Ofitems', which we can simplify and replace with a variable, say, 'n'. Here, the variable 'n' represents the number of items. By replacing values with variables, like 'n' in this case, we derive an algebraic expression that represents the relationship universally for any number of items.
Let's understand this further through some bullet points:
In the given problem, the table mentioned 'Wruming Ofitems', which we can simplify and replace with a variable, say, 'n'. Here, the variable 'n' represents the number of items. By replacing values with variables, like 'n' in this case, we derive an algebraic expression that represents the relationship universally for any number of items.
Let's understand this further through some bullet points:
- Types of Variables: Variables can be dependent or independent. In this expression, 'n' is the independent variable which influences the total cost, a dependent variable.
- Power of Variables: They allow for flexible manipulation, so instead of writing multiple equations for different values, we have one standardized form: in this case, 5n.
- Generalization: Use of variables like 'n' prepares expressions like 5n to apply universally, helping us calculate costs for any given number of items.
Patterns in Data
Identifying patterns in data is crucial in mathematics and its applications. By recognizing patterns, we can predict future trends, assess changes, and simplify complex problems. In this problem, the data pattern reveals the consistent increase in costs as the number of items purchased increases.
Here’s how patterns come into play:
Here’s how patterns come into play:
- Identifying Patterns: In the table, every time another item is added, there is a noticeable and consistent increase of $5 in cost, revealing a simplistic pattern.
- Predicting Outcomes: Once a pattern is established, you can predict the result of an unknown or untabulated outcome. For 'n' items, we predict the cost using 5n.
- Validating Predictions: By testing whether predicted outcomes match real data, as shown in step 5 of the solution, we confirm the pattern’s accuracy.
Cost Analysis
Cost analysis is a crucial financial assessment tool that helps in determining the cost per item when purchasing multiple units. It's used to manage expenses and plan budgets effectively. In the exercise, the cost analysis reveals the algebraic relationship between the number of items and the total expense incurred.
Here is a digest on how cost analysis works within this context:
Here is a digest on how cost analysis works within this context:
- Per Item Pricing: In this instance, each item costs a set price of $5, so per item pricing simplifies determining total expenses for any quantity.
- Fixed and Variable Costs: While prices here are consistent (fixed at $5 per item), in many scenarios, identifying variable and fixed costs are part of a broader cost analysis.
- Calculating Total Costs: Using the expression 5n, we determine the total cost for 'n' items, allowing for easy cost ascertainment without tedious calculations.
Other exercises in this chapter
Problem 42
Graph each ordered pair on a coordinate system. $$X(1,1.3)$$
View solution Problem 42
Simplify each expression. $$6 \cdot(y \cdot 2)$$
View solution Problem 43
Evaluate each expression. $$2+3 \cdot 5$$
View solution Problem 43
Copy each sentence. Then insert parentheses to make each sentence true. $$12 \times 3 \div 1+2=12$$
View solution