Atank 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).

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

Point P is given as:

(a) The slopes of the secant lines PQ

The points are given as:

The slope (m) is calculated using:

So, we have:

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

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

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

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

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

(b) The slope of the tangent by average

The closest secant lines to tangent P are

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

The slopes of secant lines at are:

The average slope (m) is:

Hence, the average slope is -166

Read more about slopes of secant and tangent lines at:

link

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

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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.

ok then what happened.