Question

a tank holds 5000 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 gallon) after t minutes.a) If P is the point (15,1275) on the graph of V, find the slopes of hte secant lines PQ when Q is the point on the graph with the following values.(5, 3410) -427/2(10, 2210) -187(20, 550) -145(25, 145) -113(30, 0) -85B) estimate the slope of hte tangent line at P by averaging the slopes of two adjacent secant lines. (Round your answer to one decimal places).

Answer 1

Answer:

• secant slopes: -213.5, -187, -145, -113, -85
• tangent slope: -166

Step-by-step explanation:

A) the slope values you have put in your problem statement are correct. As you know, they are computed from ...

  (change in gallons)/(change in time)

where the reference point for changes is P. Using the first listed point Q as an example, the secant slope is ...

  (3410 -1275)/(5 -15) = 2135/-10 = -213.5 . . . . gallons per minute

__

B) The average of the secant slopes for points Q adjacent to P is ...

  (-187 +(-145))/2 = -332/2 = -166 . . . . gallons per minute

The tangent slope at point P is estimated at -166 gpm.

Answer 2

The secant line joins two points on the curve of a graph.

• The slopes of secant lines PQ are: -213.5, -187, -145, -113.5, -85
• The average slope of the tangent line is -166

Point P is given as:

$P = (15,1275)$

(a) The slopes of the secant lines PQ

The points are given as:

$Q = \{(5,3410),(10, 2210),(20, 550) ,(25, 145),(30, 0) \}$

The slope (m) is calculated using:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

So, we have:

For Q = (5,3410), the slope of the secant line is:

$m_1 = \frac{1275 - 3410}{15 - 5}$

$m_1 = \frac{-2135}{10}$

$m_1 = -213.5$

For Q = (10, 2210), the slope of the secant line is:

$m_2 = \frac{1275 - 2210}{15 - 10}$

$m_2 = \frac{-935}{5}$

$m_2 = -187$

For Q = (20, 550), the slope of the secant line is:

$m_3 = \frac{1275 - 550}{15 - 20}$

$m_3 = \frac{725}{-5}$

$m_3 = -145$

For Q = (25, 145), the slope of the secant line is:

$m_4 = \frac{1275 - 140}{15 - 25}$

$m_4 = \frac{1135}{-10}$

$m_4 = -113.5$

For Q = (30, 0), the slope of the secant line is:

$m_5 = \frac{1275 - 0}{15 - 30}$

$m_5 = \frac{1275}{-15}$

$m_5 = -85$

(b) The slope of the tangent by average

The closest secant lines to tangent P are

$Q = \{(10, 2210),(20, 550)\}$

This is so, because point P (15, 1275) is between the above points.

The slopes of secant lines at $Q = \{(10, 2210),(20, 550)\}$ are:

$m_2 = -187$

$m_3 = -145$

The average slope (m) is:

$m = \frac{m_2 + m_3}{2}$

$m = \frac{-187 - 145}{2}$

$m = \frac{-332}{2}$

$m = -166$

Hence, the average slope is -166

Read more about slopes of secant and tangent lines at:

https://brainly.com/question/20356370