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
__
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.
You might be interested in
I’m not sure if your answer is right but here’s how to solve it
This question is incomplete because it lacks the appropriate attachment containing the argument of Paul and Manuel
Kindly find attached the appropriate attachment necessary to solve this question.
The attachment was sourced online
Answer:
a) Paul's argument is correct, Manuel used the wrong formula to find the volume of square pyramid X
Step-by-step explanation:
From the above question, we are given two shapes: Cone W and Square Pyramid
Cone W: Radius = 8cm, Height = 5cm
Volume of a Cone = 1/3 πr²h
Where π = 3.14
Volume of Cone X = 1/3 × 3.14 × 8² × 5
= 334.93cm³
For Square pyramid X
Volume = 1/3 × Base Area × Height
Base Area of Cone W = Base Area of square pyramid X = πr²
Where π = 3.14
Base area = 3.14 × 8² = 200.96cm²
Height of Square pyramid X = Height of Cone W = 5cm
Volume of Square pyramid = 1/3 ×200.96cm² × 5cm
= 334.93cm³
Therefore, from my above calculation and compared with the arguments of Paul and Manuel, Option a) "Paul's argument is correct, Manuel used the wrong formula to find the volume of square pyramid X" is the correct option.
The reason why is because the correct formula for the volume of a square pyramid = 1/3 × Base area × Height
This was the formula Paul used.
Manuel on the other hand used the formula: Base Area × Height to find the volume of cone W. This formula is wrong.
Option a, is the correct answer.
