Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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### Understanding Linear Equations: Practical Application

**Question 4:**

A drill is used to dig down into the earth. The depth to which the drill has reached is given at two different time intervals. 

- After 2 hours, the drill is 12 meters below the surface.
- After 5 hours, the drill is 30 meters below the surface.

**Problem Statement:**
Which equation represents the depth of the drill in meters, \( y \), after \( x \) hours of drilling?

**Options:**

a. \( y = 12x \)

b. \( y = -12x \)

c. \( y = 6x \)

d. \( y = -6x \)

**Solution Explanation:**

To find the correct equation, consider the following:

1. **Determine the Depth Change:** 
   Compare the depth at different times; from 2 hours to 5 hours, the depth changes from 12 meters to 30 meters.
   - Depth change = 30 meters - 12 meters = 18 meters.
   
2. **Determine the Time Change:**
   - Time change = 5 hours - 2 hours = 3 hours.
   
3. **Calculate the Rate of Change (Slope):**
   - Slope = Depth change / Time change = 18 meters / 3 hours = 6 meters per hour.
   
   However, since the drill is moving downwards, this should be considered a negative slope, representing descent.
   - Thus, slope = -6 meters/hour.
   
4. **Verify the Equation:**
   - The equation with this slope consistent with the given changes in depth aligns with option d:
   \[
   y = -6x
   \]

Therefore, the solution to the problem is **d. \( y = -6x \)**. This equation accurately represents the depth of the drill after \( x \) hours of drilling.
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Transcribed Image Text:### Understanding Linear Equations: Practical Application **Question 4:** A drill is used to dig down into the earth. The depth to which the drill has reached is given at two different time intervals. - After 2 hours, the drill is 12 meters below the surface. - After 5 hours, the drill is 30 meters below the surface. **Problem Statement:** Which equation represents the depth of the drill in meters, \( y \), after \( x \) hours of drilling? **Options:** a. \( y = 12x \) b. \( y = -12x \) c. \( y = 6x \) d. \( y = -6x \) **Solution Explanation:** To find the correct equation, consider the following: 1. **Determine the Depth Change:** Compare the depth at different times; from 2 hours to 5 hours, the depth changes from 12 meters to 30 meters. - Depth change = 30 meters - 12 meters = 18 meters. 2. **Determine the Time Change:** - Time change = 5 hours - 2 hours = 3 hours. 3. **Calculate the Rate of Change (Slope):** - Slope = Depth change / Time change = 18 meters / 3 hours = 6 meters per hour. However, since the drill is moving downwards, this should be considered a negative slope, representing descent. - Thus, slope = -6 meters/hour. 4. **Verify the Equation:** - The equation with this slope consistent with the given changes in depth aligns with option d: \[ y = -6x \] Therefore, the solution to the problem is **d. \( y = -6x \)**. This equation accurately represents the depth of the drill after \( x \) hours of drilling.
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