Solve the problem. 1) A deep sea diving bell is being lowered at a constant rate. After 10 minutes, the bell is at a depth of 300 ft. After 30 minutes the bell is at a depth of 1900 ft. Use the ordered pairs (10, 300) and (30, 1900) to determine the average rate to lower the bell per minute. Q Search hp meh
Solve the problem. 1) A deep sea diving bell is being lowered at a constant rate. After 10 minutes, the bell is at a depth of 300 ft. After 30 minutes the bell is at a depth of 1900 ft. Use the ordered pairs (10, 300) and (30, 1900) to determine the average rate to lower the bell per minute. Q Search hp meh
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Solve the problem.**
1) A deep sea diving bell is being lowered at a constant rate. After 10 minutes, the bell is at a depth of 300 ft. After 30 minutes, the bell is at a depth of 1900 ft. Use the ordered pairs (10, 300) and (30, 1900) to determine the average rate to lower the bell per minute.
**Explanation:**
The problem involves calculating the average rate of descent for a deep sea diving bell using the given time and depth data points. The data points are represented as ordered pairs:
- (10, 300) represents the time in minutes and depth in feet after 10 minutes.
- (30, 1900) represents the time in minutes and depth in feet after 30 minutes.
The average rate can be calculated using the formula for the slope of a line:
\[
\text{Slope} = \frac{y_2 - y_1}{x_2 - x_1}
\]
Where:
- \(y_2\) and \(y_1\) are the depths at times \(x_2\) and \(x_1\), respectively.
- Here, \(x_1 = 10\), \(y_1 = 300\), \(x_2 = 30\), and \(y_2 = 1900\).
Substituting the given values:
\[
\text{Slope} = \frac{1900 - 300}{30 - 10} = \frac{1600}{20} = 80
\]
Thus, the average rate to lower the bell is 80 feet per minute.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f680d4f-94aa-4d7a-935b-bbf3923d5869%2F4c177b2b-bcef-4399-b68d-be94c59a09f6%2Fxsfgqmp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Solve the problem.**
1) A deep sea diving bell is being lowered at a constant rate. After 10 minutes, the bell is at a depth of 300 ft. After 30 minutes, the bell is at a depth of 1900 ft. Use the ordered pairs (10, 300) and (30, 1900) to determine the average rate to lower the bell per minute.
**Explanation:**
The problem involves calculating the average rate of descent for a deep sea diving bell using the given time and depth data points. The data points are represented as ordered pairs:
- (10, 300) represents the time in minutes and depth in feet after 10 minutes.
- (30, 1900) represents the time in minutes and depth in feet after 30 minutes.
The average rate can be calculated using the formula for the slope of a line:
\[
\text{Slope} = \frac{y_2 - y_1}{x_2 - x_1}
\]
Where:
- \(y_2\) and \(y_1\) are the depths at times \(x_2\) and \(x_1\), respectively.
- Here, \(x_1 = 10\), \(y_1 = 300\), \(x_2 = 30\), and \(y_2 = 1900\).
Substituting the given values:
\[
\text{Slope} = \frac{1900 - 300}{30 - 10} = \frac{1600}{20} = 80
\]
Thus, the average rate to lower the bell is 80 feet per minute.
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