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 v
olume 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).
1 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
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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.
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Answer:
Option A
Step-by-step explanation:
According to the distance formula , for any 2 points P & Q whose coordinates are (x¹ , y¹) and (x² , y²) respectively. So ,
NOTE = HERE x² , y² DOESN'T MEAN THE SQUARES OF x & y . THEY ARE JUST COORDINATES.
According to the question ,
Coordinate of C = (2 , -1)
Coordinate of D = (5 , 3)
Using distance formula ,
Answer:
Step-by-step explanation:
You know that the triangle ABC and the triangle DEF are similar.
Therefore, if the lenght of AC is 12 centimeters and the lenght of DF is 10 centimeters, then you can find the ratio as following:
Then, the calculte the length of EF, you must multiply the lenght BC of the triangle ABC by the ratio obtained above.
Therefore, the lenght EF is the following: