a. Answer Linear model is V = − 7527.5 t + 41820. Explanation Given information: A new 2018 luxury sedan that sold for $41,820 has a book value V . of $26,765 after 2 years . Calculation: Find the slope through the points (0, 41820) and (2, 26765). m = 41820 − 26765 0 − 2 = − 7527.5. The V −intercept is (0, 41820). So the linear model is V = − 7527.5 t + 41820. b. To determine To find: an exponential model for the value V of the sedan and round the numbers in the model to four decimal places. Answer An exponential model is V = 41820 e − 0.2231 t . Explanation Given information: A new 2018 luxury sedan that sold for $41,820 has a book value V . of $26,765 after 2 years . Calculation: Use the point (0, 41820) to find a . V = a e k t 41820 = a e k ( 0 ) 41820 = a S o , V = 41820 e k t Use the point (2 , 26765) to find k V = 41820 e k t 26765 = 41820 e 2 k e 2 k = 26765 41820 ln ( e 2 k ) = ln ( 26765 41820 ) k = 1 2 ln ( 26765 41820 ) k ≈ − 0.2231 An exponential model is V = 41820 e − 0.2231 t . c. To determine To graph: the two models in the same viewing window. Answer Explanation Given information: Given models are V = − 7527.5 t + 41820 , V = 41820 e − 0.2231 t . Calculation: The graph of the two models in the same window is shown below. d. To determine To find: which model represents a greater depreciation rate in the first year. Answer The exponential model has a greater depreciation rate in the first year. Explanation Given information: Calculation: The exponential model has a greater depreciation rate in the first year. e. To determine To find: for what years is the value of the sedan greater using the linear model, the exponential model. Answer Using the graph, the linear model predicts a greater value for almost two full years (0 < t < 2). The exponential model predicts a greater value after about two years ( t ≥ 2). Explanation Given information: Calculation: Using the graph, the linear model predicts a greater value for almost two full years (0 < t < 2). The exponential model predicts a greater value after about two years ( t ≥ 2).

8th Edition

ISBN: 9781337904285

Author: Larson

Publisher: CENGAGE L

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Question

Chapter 3.5, Problem 41E

To determine

To find: a linear model for the value V of the sedan.

Expert Solution

Answer to Problem 41E

Linear model is V=−7527.5t+41820.

Explanation of Solution

Given information: A new 2018 luxury sedan that sold for $41,820 has a book value V .

of $26,765 after 2 years .

Calculation:

Find the slope through the points (0, 41820) and (2, 26765).

m=41820−267650−2=−7527.5.

The V −intercept is (0, 41820). So the linear model is

V=−7527.5t+41820.

To determine

To find: an exponential model for the value V of the sedan and round the numbers in the model to four decimal places.

Expert Solution

Answer to Problem 41E

An exponential model is V=41820e−0.2231t.

Explanation of Solution

Given information: A new 2018 luxury sedan that sold for $41,820 has a book value V .

of $26,765 after 2 years.

Calculation:

Use the point (0, 41820) to find a .

V=aekt41820=aek(0)41820=aSo, V=41820ektUse the point (2 , 26765) to find kV=41820ekt26765=41820e2ke2k=2676541820ln(e2k)=ln(2676541820)k=12ln(2676541820)k≈−0.2231An exponential model is V=41820e−0.2231t.

To determine

To graph: the two models in the same viewing window.

Expert Solution

Answer to Problem 41E

Explanation of Solution

Given information: Given models are V=−7527.5t+41820 , V=41820e−0.2231t.

Calculation:

The graph of the two models in the same window is shown below.

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.5, Problem 41E

To determine

To find: which model represents a greater depreciation rate in the first year.

Expert Solution

Answer to Problem 41E

The exponential model has a greater depreciation rate in the first year.

Explanation of Solution

Given information:

Calculation:

The exponential model has a greater depreciation rate in the first year.

To determine

To find: for what years is the value of the sedan greater using the linear model, the exponential model.

Expert Solution

Answer to Problem 41E

Using the graph, the linear model predicts a greater value for almost two full years

(0 < t < 2).

The exponential model predicts a greater value after about two years ( t ≥ 2).

Explanation of Solution

Given information:

Calculation:

Using the graph, the linear model predicts a greater value for almost two full years

(0 < t < 2).

The exponential model predicts a greater value after about two years ( t ≥ 2).

Chapter 3.5, Problem 40E

Chapter 3.5, Problem 42E

Chapter 3 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

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Prob. 78E Ch. 3.6 - Prob. 1E Ch. 3.6 - Prob. 2E Ch. 3.6 - Prob. 3E Ch. 3.6 - Prob. 4E Ch. 3.6 - Prob. 5E Ch. 3.6 - Prob. 6E Ch. 3.6 - Prob. 7E Ch. 3.6 - Prob. 8E Ch. 3.6 - Prob. 9E Ch. 3.6 - Prob. 10E Ch. 3.6 - Prob. 11E Ch. 3.6 - Prob. 12E Ch. 3.6 - Prob. 13E Ch. 3.6 - Prob. 14E Ch. 3.6 - Prob. 15E Ch. 3.6 - Prob. 16E Ch. 3.6 - Prob. 17E Ch. 3.6 - Prob. 18E Ch. 3.6 - Prob. 19E Ch. 3.6 - Prob. 20E Ch. 3.6 - Prob. 21E Ch. 3.6 - Prob. 22E Ch. 3.6 - Prob. 23E Ch. 3.6 - Prob. 24E Ch. 3.6 - Prob. 25E Ch. 3.6 - Prob. 26E Ch. 3.6 - Prob. 27E Ch. 3.6 - Prob. 28E Ch. 3.6 - Prob. 29E Ch. 3.6 - Prob. 30E Ch. 3.6 - Prob. 31E Ch. 3.6 - Prob. 32E Ch. 3.6 - Prob. 33E Ch. 3.6 - Prob. 34E Ch. 3.6 - Prob. 35E Ch. 3.6 - Prob. 36E Ch. 3.6 - Prob. 37E Ch. 3.6 - Prob. 38E Ch. 3.6 - Prob. 39E Ch. 3.6 - Prob. 40E Ch. 3.6 - Prob. 41E Ch. 3.6 - Prob. 42E Ch. 3.6 - Prob. 43E Ch. 3.6 - Prob. 44E Ch. 3.6 - Prob. 45E Ch. 3.6 - Prob. 46E Ch. 3 - Prob. 1CR Ch. 3 - Prob. 2CR Ch. 3 - Prob. 3CR Ch. 3 - Prob. 4CR Ch. 3 - Prob. 5CR Ch. 3 - Prob. 6CR Ch. 3 - Prob. 7CR Ch. 3 - Prob. 8CR Ch. 3 - Prob. 9CR Ch. 3 - Prob. 10CR Ch. 3 - Prob. 11CR Ch. 3 - Prob. 12CR Ch. 3 - Prob. 13CR Ch. 3 - Prob. 14CR Ch. 3 - Prob. 15CR Ch. 3 - Prob. 16CR Ch. 3 - Prob. 17CR Ch. 3 - Prob. 18CR Ch. 3 - Prob. 19CR Ch. 3 - Prob. 20CR Ch. 3 - Prob. 21CR Ch. 3 - Prob. 22CR Ch. 3 - Prob. 23CR Ch. 3 - Prob. 24CR Ch. 3 - Prob. 25CR Ch. 3 - Prob. 26CR Ch. 3 - Prob. 27CR Ch. 3 - Prob. 28CR Ch. 3 - Prob. 29CR Ch. 3 - Prob. 30CR Ch. 3 - Prob. 31CR Ch. 3 - Prob. 32CR Ch. 3 - Prob. 33CR Ch. 3 - Prob. 34CR Ch. 3 - Prob. 35CR Ch. 3 - Prob. 36CR Ch. 3 - Prob. 37CR Ch. 3 - Prob. 38CR Ch. 3 - Prob. 39CR Ch. 3 - Prob. 40CR Ch. 3 - Prob. 41CR Ch. 3 - Prob. 42CR Ch. 3 - Prob. 43CR Ch. 3 - Prob. 44CR Ch. 3 - Prob. 45CR Ch. 3 - Prob. 46CR Ch. 3 - Prob. 47CR Ch. 3 - Prob. 48CR Ch. 3 - Prob. 49CR Ch. 3 - Prob. 50CR Ch. 3 - Prob. 51CR Ch. 3 - Prob. 52CR Ch. 3 - Prob. 53CR Ch. 3 - Prob. 54CR Ch. 3 - Prob. 55CR Ch. 3 - Prob. 56CR Ch. 3 - Prob. 57CR Ch. 3 - Prob. 58CR Ch. 3 - Prob. 59CR Ch. 3 - Prob. 60CR Ch. 3 - Prob. 61CR Ch. 3 - Prob. 62CR Ch. 3 - Prob. 63CR Ch. 3 - Prob. 64CR Ch. 3 - Prob. 65CR Ch. 3 - Prob. 66CR Ch. 3 - Prob. 67CR Ch. 3 - Prob. 68CR Ch. 3 - Prob. 69CR Ch. 3 - Prob. 70CR Ch. 3 - Prob. 71CR Ch. 3 - Prob. 72CR Ch. 3 - Prob. 73CR Ch. 3 - Prob. 74CR Ch. 3 - Prob. 75CR Ch. 3 - Prob. 76CR Ch. 3 - Prob. 77CR Ch. 3 - Prob. 78CR Ch. 3 - Prob. 79CR Ch. 3 - Prob. 80CR Ch. 3 - Prob. 81CR Ch. 3 - Prob. 82CR Ch. 3 - Prob. 83CR Ch. 3 - Prob. 84CR Ch. 3 - Prob. 85CR Ch. 3 - Prob. 86CR Ch. 3 - Prob. 87CR Ch. 3 - Prob. 88CR Ch. 3 - Prob. 89CR Ch. 3 - Prob. 90CR Ch. 3 - Prob. 91CR Ch. 3 - Prob. 92CR Ch. 3 - Prob. 93CR Ch. 3 - Prob. 94CR Ch. 3 - Prob. 95CR Ch. 3 - Prob. 96CR Ch. 3 - Prob. 97CR Ch. 3 - Prob. 98CR Ch. 3 - Prob. 99CR Ch. 3 - Prob. 100CR Ch. 3 - Prob. 101CR Ch. 3 - Prob. 102CR Ch. 3 - Prob. 103CR Ch. 3 - Prob. 104CR Ch. 3 - Prob. 105CR Ch. 3 - Prob. 106CR Ch. 3 - Prob. 107CR Ch. 3 - Prob. 108CR Ch. 3 - Prob. 109CR Ch. 3 - Prob. 110CR Ch. 3 - Prob. 111CR Ch. 3 - Prob. 112CR Ch. 3 - Prob. 113CR Ch. 3 - Prob. 114CR Ch. 3 - Prob. 115CR Ch. 3 - Prob. 116CR Ch. 3 - Prob. 117CR Ch. 3 - Prob. 118CR Ch. 3 - Prob. 119CR Ch. 3 - Prob. 120CR Ch. 3 - Prob. 121CR Ch. 3 - Prob. 122CR Ch. 3 - Prob. 123CR Ch. 3 - Prob. 124CR Ch. 3 - Prob. 125CR Ch. 3 - Prob. 126CR Ch. 3 - Prob. 1CT Ch. 3 - Prob. 2CT Ch. 3 - Prob. 3CT Ch. 3 - Prob. 4CT Ch. 3 - Prob. 5CT Ch. 3 - Prob. 6CT Ch. 3 - Prob. 7CT Ch. 3 - Prob. 8CT Ch. 3 - Prob. 9CT Ch. 3 - Prob. 10CT Ch. 3 - Prob. 11CT Ch. 3 - Prob. 12CT Ch. 3 - Prob. 13CT Ch. 3 - Prob. 14CT Ch. 3 - Prob. 15CT Ch. 3 - Prob. 16CT Ch. 3 - Prob. 17CT Ch. 3 - Prob. 18CT Ch. 3 - Prob. 19CT Ch. 3 - Prob. 20CT Ch. 3 - Prob. 21CT Ch. 3 - Prob. 22CT Ch. 3 - Prob. 23CT Ch. 3 - Prob. 24CT Ch. 3 - Prob. 25CT Ch. 3 - Prob. 26CT Ch. 3 - Prob. 27CT Ch. 3 - Prob. 28CT Ch. 3 - Prob. 29CT Ch. 3 - Prob. 30CT Ch. 3 - Prob. 1STP Ch. 3 - Prob. 2STP Ch. 3 - Prob. 3STP Ch. 3 - Prob. 4STP Ch. 3 - Prob. 5STP Ch. 3 - Prob. 6STP Ch. 3 - Prob. 7STP Ch. 3 - Prob. 8STP Ch. 3 - Prob. 9STP Ch. 3 - Prob. 10STP Ch. 3 - Prob. 11STP Ch. 3 - Prob. 12STP Ch. 3 - Prob. 13STP Ch. 3 - Prob. 14STP Ch. 3 - Prob. 15STP Ch. 3 - Prob. 16STP Ch. 3 - Prob. 17STP Ch. 3 - Prob. 18STP Ch. 3 - Prob. 19STP Ch. 3 - Prob. 20STP Ch. 3 - Prob. 21STP Ch. 3 - Prob. 22STP Ch. 3 - Prob. 23STP Ch. 3 - Prob. 24STP Ch. 3 - Prob. 25STP Ch. 3 - Prob. 26STP

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