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Multiple Choice Question An engineer is designing a cylinder-shaped structure using the formula shown below. As shown, an appro...Asked by Butterball
Multiple Choice Question
An engineer is designing a cylinder-shaped structure using the formula shown below. As shown, an approximation of 22/7 for PIE is used in the formula.
The engineer is considering using different values for , the radius in meters, and , the height in meters. Which pair of values makes , the volume in cubic meters, an irrational number?
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GPT 3.5
To determine when the volume is an irrational number, we need to find a pair of values for the radius and height that result in an irrational volume.
The formula for the volume of a cylinder is given by V = πr^2h, where V represents the volume, π is a constant (approximated as 22/7), r is the radius, and h is the height.
For the volume to be irrational, both πr^2 and h must be irrational numbers.
Let's examine the given options:
A) r = √2 meters, h = √5 meters:
πr^2 = π(√2)^2 = π(2) = 2π, which is irrational (π is irrational).
h = √5 meters, which is irrational.
Therefore, this option could result in an irrational volume.
B) r = 3 meters, h = √7 meters:
πr^2 = π(3)^2 = 9π, which is irrational (π is irrational).
h = √7 meters, which is irrational.
Therefore, this option could result in an irrational volume.
C) r = 5 meters, h = 2 meters:
πr^2 = π(5)^2 = 25π, which is rational (π is irrational, but 25 is rational).
h = 2 meters, which is rational.
Therefore, this option will result in a rational volume.
D) r = 1 meter, h = π meters:
πr^2 = π(1)^2 = π, which is rational.
h = π meters, which is rational.
Therefore, this option will result in a rational volume.
From the given options, only options A) and B) have the potential to result in an irrational volume.
The formula for the volume of a cylinder is given by V = πr^2h, where V represents the volume, π is a constant (approximated as 22/7), r is the radius, and h is the height.
For the volume to be irrational, both πr^2 and h must be irrational numbers.
Let's examine the given options:
A) r = √2 meters, h = √5 meters:
πr^2 = π(√2)^2 = π(2) = 2π, which is irrational (π is irrational).
h = √5 meters, which is irrational.
Therefore, this option could result in an irrational volume.
B) r = 3 meters, h = √7 meters:
πr^2 = π(3)^2 = 9π, which is irrational (π is irrational).
h = √7 meters, which is irrational.
Therefore, this option could result in an irrational volume.
C) r = 5 meters, h = 2 meters:
πr^2 = π(5)^2 = 25π, which is rational (π is irrational, but 25 is rational).
h = 2 meters, which is rational.
Therefore, this option will result in a rational volume.
D) r = 1 meter, h = π meters:
πr^2 = π(1)^2 = π, which is rational.
h = π meters, which is rational.
Therefore, this option will result in a rational volume.
From the given options, only options A) and B) have the potential to result in an irrational volume.
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