Algebra 1
Algebra 1
1st Edition
ISBN: 9780078884801
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 3.4, Problem 29PPS

(a)

To determine

Write an equation that relates the age x of the car to the value y that is lost after each year.

(a)

Expert Solution
Check Mark

Answer to Problem 29PPS

The equation: y=−1800x+13800

Explanation of Solution

Given:

  Algebra 1, Chapter 3.4, Problem 29PPS , additional homework tip 1

The ordered pairs are: (1,12000),   (2,  10200),   (3,  8400),   (4,6600),   (5,  4800)

Concept Used:

The age x of the car to the value y that is lost after each year

So the rate of change is:

Rate of change: m=  y2−y1x2−x1

Calculation:

So the rate of change is:

The points are: (  x1,  y1 )  ⇒  (1,12000);    (  x2,  y2 )  ⇒  (2,10200)

Rate of change: m=  y2−y1x2−x1

  m=  y2−y1x2−x1          Rate  of  change  formulam=   10200−120002−1            (  x1,  y1 )  ⇒  (1,12000)    and    (  x2,  y2 )  ⇒  (2,10200)m  =   −1800

Rate of change is negative means the cost if depreciates $1800 per year.

The difference in y − values is 1800 times the difference of x − values.

This suggest the equation y=1800x

The cost of the cars depreciates $1800 per year..

Thus, the relationship is y=1800x

(b)

To determine

Find the age of the car if the value is $300

(b)

Expert Solution
Check Mark

Answer to Problem 29PPS

Age of the car will be 7.5 years or 7 years 6 months.

Explanation of Solution

Given:

  Algebra 1, Chapter 3.4, Problem 29PPS , additional homework tip 2

The ordered pairs are: (1,12000),   (2,  10200),   (3,  8400),   (4,6600),   (5,  4800)

Concept Used:

The equation: y=1800x

First find the values of the car when it was purchased. If after 1 year the car worth was $12,000 and it had decreased in values by $1800, its original worth was $13,800.

Next subtract its current its current value to find how much its value has decreased by

Calculation:

First find the values of the car when it was purchased. If after 1 year the car worth was $12,000 and it had decreased in values by $1800, its original worth was $13,800.

Next subtract its current its current value to find how much its value has decreased by

  $13,800−$300=$13500.

Substitute this value into the equation for y=1800x

  y=1800x13500=1800x135001800  =  1800x1800x=7.5

So, this means that the car will be worth $300 in 7.5 years or 7 years and 6 months.

Thus,age of the car will be 7.5 years or 7 years 6 months.

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Chapter 3.4, Problem 28PPS
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Chapter 3.4, Problem 30PPS

Chapter 3 Solutions

Algebra 1

Ch. 3.1 - Prob. 35PPSCh. 3.1 - Prob. 58PPSCh. 3.1 - Prob. 60HPCh. 3.1 - Prob. 68STPCh. 3.1 - Prob. 1ACYPCh. 3.1 - Prob. 1BCYPCh. 3.1 - Prob. 2CYPCh. 3.1 - Prob. 3CYPCh. 3.1 - Prob. 4ACYPCh. 3.1 - Prob. 4BCYP
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Prob. 15HPCh. 3.6 - Prob. 16HPCh. 3.6 - Prob. 17HPCh. 3.6 - Prob. 18HPCh. 3.6 - Prob. 1ACYPCh. 3.6 - Prob. 1BCYPCh. 3.6 - Prob. 1CCYPCh. 3.6 - Prob. 2ACYPCh. 3.6 - Prob. 2BCYPCh. 3.6 - Prob. 19STPCh. 3.6 - Prob. 20STPCh. 3.6 - Prob. 21STPCh. 3.6 - Prob. 22STPCh. 3.6 - Prob. 23SRCh. 3.6 - Prob. 24SRCh. 3.6 - Prob. 25SRCh. 3.6 - Prob. 26SRCh. 3.6 - Prob. 27SRCh. 3.6 - Prob. 28SRCh. 3.6 - Prob. 29SRCh. 3.6 - Prob. 30SCh. 3.6 - Prob. 31SCh. 3.6 - Prob. 32SCh. 3.6 - Prob. 33SCh. 3.6 - Prob. 34SCh. 3.6 - Prob. 35SCh. 3 - Prob. 1QCCh. 3 - Prob. 2QCCh. 3 - Prob. 3QCCh. 3 - Prob. 4QCCh. 3 - Prob. 5QCCh. 3 - Prob. 6QCCh. 3 - Prob. 13QCCh. 3 - Prob. 14QCCh. 3 - Prob. 15QCCh. 3 - Prob. 16QCCh. 3 - Prob. 17QCCh. 3 - Prob. 18QCCh. 3 - Prob. 19QCCh. 3 - Prob. 20QCCh. 3 - Prob. 21QCCh. 3 - Prob. 22QCCh. 3 - Prob. 23QCCh. 3 - Prob. 1MCQCh. 3 - Prob. 2MCQCh. 3 - Prob. 3MCQCh. 3 - Prob. 4MCQCh. 3 - Prob. 5MCQCh. 3 - Prob. 6MCQCh. 3 - Prob. 9MCQCh. 3 - Prob. 10MCQCh. 3 - Prob. 11MCQCh. 3 - Prob. 12MCQCh. 3 - 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