PRECALCULUS:...COMMON CORE ED.-W/ACCESS
PRECALCULUS:...COMMON CORE ED.-W/ACCESS
10th Edition
ISBN: 9780134781839
Author: Demana
Publisher: PEARSON
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Chapter P.4, Problem 45E
Solution

a.

The value of m and b .

  m=3187.5, b=42000

Given:

The person purchased a house 8 years ago for $42000 and this year it was appraised at $67500 .

Concept Used:

  • The slope-intercept form of a line is y=mx+b, where m is the slope of the line and b is the y -intercept of the line.
  • Slope of the line passing through the points x1,y1 and x2,y2 is given by m=y2y1x2x1
  • The slope of a line is defined as rise over run

Calculation:

Let initially at t=0 , the value of the house was V=$42000 , so the corresponding point is t,V=0,42000 .

Since now after 8 years the value of house is $67500 , the corresponding ordered pair is t,V=8,67500 .

In order to find the slope-intercept form the of line passing through the points 0,42000 and 8,67500 , first find the slope of the line using the slope formula as shown below:

  m=y2y1x2x1=675004200080=255008=3187.5

Now substitute m=3187.5 and x,y=0,42000 in the general equation of line and solve for b as shown below:

  y=mx+b42000=3187.50+b42000=0+bb=42000

Thus, in slope of the given linear equation is m=3187.5 and the y -intercept is b=42000 .

b.

The years after purchase this house will be worth $72500 .

  After 9.57 years

Given:

The person purchased a house 8 years ago for $42000 and this year it was appraised at $67500 .

Concept Used:

  • The slope-intercept form of a line is y=mx+b, where m is the slope of the line and b is the y -intercept of the line.
  • Slope of the line passing through the points x1,y1 and x2,y2 is given by m=y2y1x2x1
  • The slope of a line is defined as rise over run

Calculation:

In order to plot the graph of the function, plot the points 0,42000 and 8,67500 , and connect them with a straight line.

Then to trace the years after purchase when the house will be worth $72500 , draw a horizontal line at y=72500 as shown in below graph,

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter P.4, Problem 45E

Thus, the house will be worth $72500 about 9.57 years after purchase.

c.

The number of years after purchase this house will be worth $74000

  3187.5t+42000=74000t=10.04

Given:

The person purchased a house 8 years ago for $42000 and this year it was appraised at $67500 .

Concept Used:

  • The slope-intercept form of a line is y=mx+b, where m is the slope of the line and b is the y -intercept of the line.
  • Slope of the line passing through the points x1,y1 and x2,y2 is given by m=y2y1x2x1
  • The slope of a line is defined as rise over run

Calculation:

Since the slope of the given linear equation is m=3187.5 and the y -intercept is b=42000 , the linear equation is

  V=3187.5t+42000

Now in order to find the year when the house will be worth $74000 , substitute V=74000 into above equation and simplify further as shown below

  V=3187.5t+4200074000=3187.5t+420003187.5t=74000420003187.5t=32000t=320003187.5t10.04

Thus, the house will be worth $74000 about 10.04 years after purchase.

d.

The number of years after purchase this house will be worth $80250

  12 years

Given:

The person purchased a house 8 years ago for $42000 and this year it was appraised at $67500 .

Concept Used:

  • The slope-intercept form of a line is y=mx+b, where m is the slope of the line and b is the y -intercept of the line.
  • Slope of the line passing through the points x1,y1 and x2,y2 is given by m=y2y1x2x1
  • The slope of a line is defined as rise over run

Calculation:

Since the slope of the given linear equation is m=3187.5 and the y -intercept is b=42000 , the linear equation is

  V=3187.5t+42000

Now in order to find the year when the house will be worth $80250 , substitute V=80250 into above equation and simplify further as shown below

  V=3187.5t+4200080250=3187.5t+420003187.5t=80250420003187.5t=38250t=382503187.5t=12

Thus, the house will be worth $80250 about 12 years after purchase.

Chapter P Solutions

PRECALCULUS:...COMMON CORE ED.-W/ACCESS

Ch. P.1 - Prob. 1ECh. P.1 - Prob. 2ECh. P.1 - Prob. 3ECh. P.1 - Prob. 4ECh. P.1 - Prob. 5ECh. P.1 - Prob. 6ECh. P.1 - Prob. 7ECh. P.1 - Prob. 8ECh. P.1 - Prob. 9ECh. P.1 - Prob. 10ECh. P.1 - Prob. 11ECh. P.1 - Prob. 12ECh. P.1 - Prob. 13ECh. P.1 - Prob. 14ECh. P.1 - Prob. 15ECh. P.1 - Prob. 16ECh. P.1 - Prob. 17ECh. P.1 - Prob. 18ECh. P.1 - Prob. 19ECh. P.1 - Prob. 20ECh. P.1 - Prob. 21ECh. P.1 - Prob. 22ECh. P.1 - Prob. 23ECh. P.1 - Prob. 24ECh. P.1 - Prob. 25ECh. P.1 - Prob. 26ECh. P.1 - Prob. 27ECh. P.1 - Prob. 28ECh. P.1 - Prob. 29ECh. P.1 - Prob. 30ECh. P.1 - Prob. 31ECh. P.1 - Prob. 32ECh. P.1 - Prob. 33ECh. P.1 - Prob. 34ECh. P.1 - Prob. 35ECh. P.1 - Prob. 36ECh. P.1 - Prob. 37ECh. P.1 - Prob. 38ECh. P.1 - Prob. 39ECh. P.1 - Prob. 40ECh. P.1 - Prob. 41ECh. P.1 - Prob. 42ECh. P.1 - Prob. 43ECh. P.1 - Prob. 44ECh. P.1 - Prob. 45ECh. P.1 - Prob. 46ECh. P.1 - Prob. 47ECh. P.1 - In Exercises 4752, simplify the expression. Assume...Ch. P.1 - In Exercises 4752, simplify the expression. Assume...Ch. P.1 - Prob. 50ECh. P.1 - Prob. 51ECh. P.1 - Prob. 52ECh. P.1 - Prob. 53ECh. P.1 - Prob. 54ECh. P.1 - Prob. 55ECh. P.1 - Prob. 56ECh. P.1 - Prob. 57ECh. P.1 - Prob. 58ECh. P.1 - Prob. 59ECh. P.1 - Prob. 60ECh. P.1 - Prob. 61ECh. P.1 - Prob. 62ECh. P.1 - Prob. 63ECh. P.1 - In Exercises 63 and 64, use scientific notation to...Ch. P.1 - Prob. 65ECh. P.1 - Prob. 66ECh. P.1 - Prob. 67ECh. P.1 - Prob. 68ECh. P.1 - Prob. 69ECh. P.1 - Prob. 70ECh. P.1 - Prob. 71ECh. P.1 - Prob. 72ECh. P.1 - Prob. 73ECh. P.1 - Prob. 74ECh. P.1 - Prob. 75ECh. P.1 - Prob. 76ECh. P.2 - Prob. 1QRCh. P.2 - Prob. 2QRCh. P.2 - Prob. 3QRCh. P.2 - Prob. 4QRCh. P.2 - Prob. 5QRCh. P.2 - Prob. 6QRCh. P.2 - Prob. 7QRCh. P.2 - Prob. 8QRCh. P.2 - Prob. 9QRCh. P.2 - Prob. 10QRCh. P.2 - Prob. 1ECh. P.2 - Prob. 2ECh. P.2 - Prob. 3ECh. P.2 - Prob. 4ECh. P.2 - Prob. 5ECh. P.2 - Prob. 6ECh. P.2 - Prob. 7ECh. P.2 - Prob. 8ECh. P.2 - Prob. 9ECh. P.2 - Prob. 10ECh. 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