50 How should this asymptote be interpreted? A. As time goes by the gerbil population will level off at around 205,000 B. As the number of gerbils increases, the maximum population will be attained in 205,000 years C. As the number of gerbils increases, the maximum population will be attained in 205 years D. As time goes by the gerbil population will level off at around 205

50 How should this asymptote be interpreted? A. As time goes by the gerbil population will level off at around 205,000 B. As the number of gerbils increases, the maximum population will be attained in 205,000 years C. As the number of gerbils increases, the maximum population will be attained in 205 years D. As time goes by the gerbil population will level off at around 205

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

How should this asymptote be interpreted?

A. As time goes by the gerbil population will level off at around 205,000

B. As the number of gerbils increases, the maximum population will be attained in 205,000 years

C. As the number of gerbils increases, the maximum population will be attained in 205 years

D. As time goes by the gerbil population will level off at around 205

205
A gerbil population can be estimated with the function P(t)
where t = 0 represents the gerbil population, P, measured in thousands in the year 2020, and tis measured in years. The graph of this function is given below.
1+
P(t)
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90

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A graphing calculator is recommended. Straight-line depreciation is a method for estimating the value of an asset (such as a piece of machinery) as it loses value ("depreciates") through use. Given the original price of an asset, its useful lifetime, and its scrap value (its value at the end of its useful lifetime), the value of the asset after t years is given by the formula for 0 ≤ t ≤ (Useful lifetime). (a) A farmer buys a harvester for $40,000 and estimates its useful life to be 20 years, after which its scrap value will be $4000. Use the formula above to find a formula for the value V of the harvester after t years, for 0 ≤ t ≤ 20. (Do not use commas in your answer.) V = Value = Price - (b) Use your formula to find the value of the harvester after 5 years. V(5) = No Solution (c) Graph the function found in part (a) on a graphing calculator on the window [0, 20] by [0, 40,000]. (Hint: Use x instead of t.) Help 40000 30000 Price (Scrap value) (Useful Lifetime) 20000 10000 10 Fill…