Math Problem Statement
Is the relationship linear, exponential, or neither?
Solution
The question asks whether the relationship between and is linear, exponential, or neither based on the data points:
x & y \\ \hline 9 & 13 \\ 14 & 16 \\ 19 & 19 \\ 24 & 22 \\ \end{array}$$ ### Step 1: Check for a Linear Relationship For a relationship to be linear, the rate of change (or slope) between consecutive points should be constant. We can calculate the differences in $$y$$-values divided by the differences in $$x$$-values (i.e., $$\Delta y / \Delta x$$) for each consecutive pair of points: 1. Between $$(9, 13)$$ and $$(14, 16)$$: $$\frac{16 - 13}{14 - 9} = \frac{3}{5} = 0.6$$ 2. Between $$(14, 16)$$ and $$(19, 19)$$: $$\frac{19 - 16}{19 - 14} = \frac{3}{5} = 0.6$$ 3. Between $$(19, 19)$$ and $$(24, 22)$$: $$\frac{22 - 19}{24 - 19} = \frac{3}{5} = 0.6$$ Since the rate of change is consistent (0.6) across each interval, this relationship is **linear**. ### Answer The correct answer is: - **A) Linear** Would you like further details on how to identify linear, exponential, or other relationships? --- ### Related Questions 1. How do you calculate the rate of change between two points? 2. What are the key differences between linear and exponential relationships? 3. Can a function be both linear and exponential? Why or why not? 4. How can you determine if a relationship is "neither" linear nor exponential? 5. How do real-world examples illustrate linear relationships? **Tip:** For linear relationships, always check if the differences in $$y$$-values divided by $$x$$-values remain constant!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Linear Relationships
Exponential Relationships
Formulas
Rate of change formula: (y2 - y1) / (x2 - x1)
Theorems
Slope formula for linear relationships
Suitable Grade Level
Grades 8-10
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