Algebra 1
Algebra 1
1st Edition
ISBN: 9780078884801
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
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Question
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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=y2y1x2x1

Calculation:

So the rate of change is:

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

Rate of change: m=y2y1x2x1

  m=y2y1x2x1Rateofchangeformulam=102001200021(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.

Chapter 3 Solutions

Algebra 1

Ch. 3.1 - Prob. 5ACYPCh. 3.1 - Prob. 5BCYPCh. 3.1 - Prob. 5CCYPCh. 3.1 - Prob. 1CYUCh. 3.1 - Prob. 2CYUCh. 3.1 - Prob. 3CYUCh. 3.1 - Prob. 4CYUCh. 3.1 - Prob. 5CYUCh. 3.1 - Prob. 6CYUCh. 3.1 - Prob. 7CYUCh. 3.1 - Prob. 8CYUCh. 3.1 - Prob. 9CYUCh. 3.1 - Prob. 10CYUCh. 3.1 - Prob. 11CYUCh. 3.1 - Prob. 12CYUCh. 3.1 - Prob. 13PPSCh. 3.1 - Prob. 14PPSCh. 3.1 - Prob. 15PPSCh. 3.1 - Prob. 16PPSCh. 3.1 - Prob. 17PPSCh. 3.1 - Prob. 18PPSCh. 3.1 - Prob. 19PPSCh. 3.1 - Prob. 20PPSCh. 3.1 - Prob. 21PPSCh. 3.1 - Prob. 22PPSCh. 3.1 - Prob. 23PPSCh. 3.1 - Prob. 24PPSCh. 3.1 - Prob. 25PPSCh. 3.1 - Prob. 26PPSCh. 3.1 - Prob. 27PPSCh. 3.1 - Prob. 28PPSCh. 3.1 - Prob. 29PPSCh. 3.1 - Prob. 30PPSCh. 3.1 - Prob. 31PPSCh. 3.1 - Prob. 32PPSCh. 3.1 - Prob. 33PPSCh. 3.1 - Prob. 34PPSCh. 3.1 - Prob. 36PPSCh. 3.1 - Prob. 37PPSCh. 3.1 - Prob. 38PPSCh. 3.1 - Prob. 39PPSCh. 3.1 - Prob. 40PPSCh. 3.1 - Prob. 41PPSCh. 3.1 - Prob. 42PPSCh. 3.1 - Prob. 43PPSCh. 3.1 - Prob. 44PPSCh. 3.1 - Prob. 45PPSCh. 3.1 - Prob. 46PPSCh. 3.1 - 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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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