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
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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.
