Mathematics

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).

2 Answer

  • 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.

  • 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:

    [tex]P = (15,1275)[/tex]

    (a) The slopes of the secant lines PQ

    The points are given as:

    [tex]Q = \{(5,3410),(10, 2210),(20, 550) ,(25, 145),(30, 0) \}[/tex]

    The slope (m) is calculated using:

    [tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

    So, we have:

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

    [tex]m_1 = \frac{1275 - 3410}{15 - 5}[/tex]

    [tex]m_1 = \frac{-2135}{10}[/tex]

    [tex]m_1 = -213.5[/tex]

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

    [tex]m_2 = \frac{1275 - 2210}{15 - 10}[/tex]

    [tex]m_2 = \frac{-935}{5}[/tex]

    [tex]m_2 = -187[/tex]

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

    [tex]m_3 = \frac{1275 - 550}{15 - 20}[/tex]

    [tex]m_3 = \frac{725}{-5}[/tex]

    [tex]m_3 = -145[/tex]

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

    [tex]m_4 = \frac{1275 - 140}{15 - 25}[/tex]

    [tex]m_4 = \frac{1135}{-10}[/tex]

    [tex]m_4 = -113.5[/tex]

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

    [tex]m_5 = \frac{1275 - 0}{15 - 30}[/tex]

    [tex]m_5 = \frac{1275}{-15}[/tex]

    [tex]m_5 = -85[/tex]

    (b) The slope of the tangent by average

    The closest secant lines to tangent P are

    [tex]Q = \{(10, 2210),(20, 550)\}[/tex]

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

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

    [tex]m_2 = -187[/tex]

    [tex]m_3 = -145[/tex]

    The average slope (m) is:

    [tex]m = \frac{m_2 + m_3}{2}[/tex]

    [tex]m = \frac{-187 - 145}{2}[/tex]

    [tex]m = \frac{-332}{2}[/tex]

    [tex]m = -166[/tex]

    Hence, the average slope is -166

    Read more about slopes of secant and tangent lines at:

    https://brainly.com/question/20356370

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