Two water tanks are being drained at constant rates.
1 answer:
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This question is incomplete because it lacks the appropriate attachment containing the argument of Paul and Manuel
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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.
Remember
the deritivive of f(x)/g(x)=(f'(x)g(x)-g'(x)f(x))/((g(x))^2)
deritivive of lnx is 1/x
derivitive of t^2=2t
so
=
=
the deritivive of f(x)/g(x)=(f'(x)g(x)-g'(x)f(x))/((g(x))^2)
deritivive of lnx is 1/x
derivitive of t^2=2t
so