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A tank holds 3,000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes.

A tank holds 3000 gallons of water which drains from the bottom of the tank in half an hour The values in the table show the volume V of water remaining in the class=

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a) The slope of the secant lines are: q = 5, f(q) = 2,076: m = - 131.1, q = 10, f(q) = 1,299: m = - 106.8, q = 20, f(q) = 357: m = - 81.6, q = 25, f(q) = 84: m = - 68.1, q = 30, f(q) = 0: m = - 51.

b) The estimated slope of the line tangent to the curve at the point (x, y) = (15, 765) is - 94.2 gallons per minute.

How to estimate the slope of a tangent line by averaging two adjacent secant lines

In this problem we must determine the slope of several lines based on the information given by the table and using the secant line formula:

m = [f(q) - f(p)] / (q - p)     (1)

If we know that a = 15 and p = 765, then the slope of the secant lines are:

q = 5, f(q) = 2,076

m = [2,076 - 765] / (5 - 15)

m = - 131.1

q = 10, f(q) = 1,299

m = [1,299 - 765] / (10 - 15)

m = - 106.8

q = 20, f(q) = 357

m = [357 - 765] / (20 - 15)

m = - 81.6

q = 25, f(q) = 84

m = [84 - 765] / (25 - 15)

m = - 68.1

q = 30, f(q) = 0

m = [0 - 765] / (30 - 15)

m = - 51

The slope of the line tangent to the curve at the point (x, y) = (15, 765) can be estimated by averaging the slopes of the closest secant lines:

m' = [(- 106.8) + (- 81.6)] / 2

m' = - 94.2

The estimated slope of the line tangent to the curve at the point (x, y) = (15, 765) is - 94.2 gallons per minute.

To learn more on tangent lines: https://brainly.com/question/23265136

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