https://file.aiquickdraw.com/mathbot/file/1727230483373vq8m7z84.jpg

We are given the function \( w = 150 - 12s \), where \( w \) is a function of \( s \). We need to determine how \( w \) changes as \( s \) changes between various values, specifically from 4 to 10, from 4 to 4.5, and from 4 to 4.1.

Let’s solve for the changes in \( w \) using the formula \( w = 150 - 12s \).

### 1. Change in \( w \) from \( s = 4 \) to \( s = 10 \)
For \( s = 4 \):
\[
w(4) = 150 - 12(4) = 150 - 48 = 102
\]
For \( s = 10 \):
\[
w(10) = 150 - 12(10) = 150 - 120 = 30
\]
So the change in \( w \), \( \Delta w \), from \( s = 4 \) to \( s = 10 \) is:
\[
\Delta w = w(10) - w(4) = 30 - 102 = -72
\]
Thus, \( \Delta w = -72 \) when \( s \) changes from 4 to 10.

### 2. Change in \( w \) from \( s = 4 \) to \( s = 4.5 \)
For \( s = 4.5 \):
\[
w(4.5) = 150 - 12(4.5) = 150 - 54 = 96
\]
So the change in \( w \), \( \Delta w \), from \( s = 4 \) to \( s = 4.5 \) is:
\[
\Delta w = w(4.5) - w(4) = 96 - 102 = -6
\]
Thus, \( \Delta w = -6 \) when \( s \) changes from 4 to 4.5.

### 3. Change in \( w \) from \( s = 4 \) to \( s = 4.1 \)
For \( s = 4.1 \):
\[
w(4.1) = 150 - 12(4.1) = 150 - 49.2 = 100.8
\]
So the change in \( w \), \( \Delta w \), from \( s = 4 \) to \( s = 4.1 \) is:
\[
\Delta w = w(4.1) - w(4) = 100.8 - 102 = -1.2
\]
Thus, \( \Delta w = -1.2 \) when \( s \) changes from 4 to 4.1.

### 4. Linear Relationship
The function \( w = 150 - 12s \) is linear, which means the change in \( w \) is directly proportional to the change in \( s \). In fact, the slope of this linear function is the coefficient of \( s \), which is \( -12 \).

Therefore, for any change \( \Delta s \) in \( s \), the corresponding change in \( w \), \( \Delta w \), is given by:
\[
\Delta w = m \Delta s
\]
where \( m = -12 \). This confirms that \( m \) is a constant.

### Conclusion
- The change in \( w \) from \( s = 4 \) to \( s = 10 \) is \( \Delta w = -72 \).
- The change in \( w \) from \( s = 4 \) to \( s = 4.5 \) is \( \Delta w = -6 \).
- The change in \( w \) from \( s = 4 \) to \( s = 4.1 \) is \( \Delta w = -1.2 \).
- The constant \( m = -12 \), showing a linear relationship between \( \Delta w \) and \( \Delta s \).

Would you like more details on any part? 

### Follow-up questions:
1. How do we derive the slope of a linear function from an equation?
2. How can we interpret negative slopes in linear equations?
3. How would the change in \( w \) differ if the coefficient of \( s \) were positive?
4. What are the geometric interpretations of \( w = 150 - 12s \)?
5. How do small changes in \( s \) affect linear functions compared to large changes?

**Tip:** In any linear equation, the slope tells you how much the dependent variable changes for each unit change in the independent variable.

Suppose that w = 150 - 12s. How much does w change when s changes from 4 to 10? From 4 to 4.5? From 4 to 4.1? Let Δs denote a change in s, and let Δw denote the corresponding change in w. Is Δw = mΔs for some constant m? If so, what is m?

Explore how the linear equation w = 150 - 12s changes when s changes from 4 to 10, 4 to 4.5, and 4 to 4.1. Calculate the corresponding changes in w and determine if the relationship between Δw and Δs is linear. Learn about slope and rates of change in algebra. Suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation