nitial value = 51, decreasing at a rate of 0.41% A) f(t) = 0.41 0.49t C) f(t) = 51 1.0041t per week B) f(t)=51 0.9959t D) f(t)=51 1.41t Initial population = 30,599, increasing at a rate of 2.8% per year A) P(t) = 30,599-0.028t C) P(t) = 2.8 30,599t B) P(t) = 30,599 2.8t D) P(t) = 30,599 1.028t Initial mass = 14 g, decreasing at a rate of 3.7% per day A) m(t) = 3.7 0.86t C) m(t) = 14 1.37t B) m(t) = 14 0.963t D) m(t) = 14 1.037t

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter3: The Derivative
Section3.3: Rates Of Change
Problem 30E: If the instantaneous rate of change of f(x) with respect to x is positive when x=1, is f increasing...
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Initial value = 51, decreasing at a rate of 0.41% per week
A) f(t)= 0.41 0.49t
C) f(t) = 51 1.0041t
B) f(t) = 51 0.9959t
D) f(t) = 51 1.41t
Initial population = 30,599, increasing at a rate of 2.8% per year
A) P(t) = 30,599-0.028t
C) P(t) = 2.8 30,599t
Initial mass = 14 g, decreasing at a rate of 3.7% per day
A) m(t) = 3.7 0.86t
C) m(t) = 14 1.37t
B) P(t) = 30,599-2.8t
D) P(t) = 30,599 1.028t
B) m(t) = 14 0.963t
D) m(t) = 14 1.037t
oro a
Transcribed Image Text:Initial value = 51, decreasing at a rate of 0.41% per week A) f(t)= 0.41 0.49t C) f(t) = 51 1.0041t B) f(t) = 51 0.9959t D) f(t) = 51 1.41t Initial population = 30,599, increasing at a rate of 2.8% per year A) P(t) = 30,599-0.028t C) P(t) = 2.8 30,599t Initial mass = 14 g, decreasing at a rate of 3.7% per day A) m(t) = 3.7 0.86t C) m(t) = 14 1.37t B) P(t) = 30,599-2.8t D) P(t) = 30,599 1.028t B) m(t) = 14 0.963t D) m(t) = 14 1.037t oro a
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