Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership
Page 219 Problem 1 Answer
Given that sales tax rate in Mary Ann′s state is 4%.
Also given the price of the car is x dollars.
We have to find the total cost of the car with sales tax algebraically.
As we know that the sales tax is the product of the price of the item with the sales tax rate,
So we have it as Sales tax= Price of item × Sales tax rate
=x×4/100
Sales tax=0.04x dollars
Hence we have the Sales tax algebraically as 0.04x dollars
Cengage Financial Algebra Chapter 5.1 Automobile Ownership Guide
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 219 Problem 2 Answer
Given that Ramon plans to sell his car and places an ad with x lines.
The news paper charges y dollars for the first g lines and p dollars per extra line to run the ad for a week.
Also given that x>g
We have that The news paper charges y dollars for the first g lines and p dollars per extra line to run the ad for a week.
As we have given that x>g
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Now we have the extra numbers that are extra chargable is x−g
So that we have the cost of running the ad for a week as gy+(x−g)p
cost of running the ad for a week=gy+xp−gp dollars.
Hence we have the cost of running the ad for a week as gy+xp−gp dollars.
Page 220 Problem 3 Answer
Given that Smithtown News charges $38 for a classified ed ad that is 4 or fewer lines long.
Each line above four lines costs an additional $6.25.
We can express the piece wise function
Now let us assume that c(x) represent the cost.
Now let us consider the number of lines of the ad as x
Now we have the piece wise function as
c(x)={38 when x≤4
{ 38+6.25(x−4) when x>4
Hence we have the piecewise function as
c(x)={38 when x≤4
{ 38+6.25(x−4) when x>4
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 221 Problem 4 Answer
We are given : The piece-wise function :
p(w)={60 when w≤5
{ 60+8(w−5) when w>5
We have to: Translate the function into words.
Answer: Translating we get: The classified ad charges $60 for the first 5−lines and $60+8(w−5) for each additional line w
Translating we get: The classified ad charges $60 for the first 5−lines and $60+8(w−5) for each additional line w
Page 221 Problem 5 Answer
We have to : Find the cusp of the graph of the piece-wise function :
c(x)={42.50 when x≤5
{ 42.50+7(x−5) when x>5
The cost function changes from c(x)=38 to c(x)=38+6.25(x-4) when x becomes laeger than 4
x-coordinates cusp=4
y-coordinates cusp=c(4)=38
The coordinates of the cusp are:(4,38)
Solutions For Exercise 5.1 Financial Algebra 1st Edition
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 222 Problem 6 Answer
We have to find :
How much does a six-line ad cost.
Cost 4 additional lines = Number of additional lines × Cost per additional line
= 4 × 7
= $28
Total cost = Cost first 2 lines + Cost 4 additional line
= $19.50 + $28
= $47.50
Total cost = $47.50
Page 222 Problem 7 Answer
We have to find :
How much would a four-line ad with a photo cost.
Total cost = Cost first 3 lines + Cost 1 additional line + Cost Photo
= $45 + $8.50 + $40
= $93.50
The total cost of a four-line ad with a photo cost : $93.50
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 222 Problem 8 Answer
We are given that : Cost first 4 lines : g dollars
Cost per additional line : d dollars
We have to : Write an expression for the cost of a7 line advertisement.
We have find the cost of the 7−line ad, thus we are placing 7−4=3 additional lines beside the basic 4−line ad
Cost 3 additional lines = Number of additional lines × Cost per additional line
= 3 × d
= 3d
Total cost = Cost first 4 lines + Cost 3 additional line
= g + 3d
The cost of a 7 line advertisement : g + 3d
Page 222 Problem 9 Answer
We have to :
Express the cost of an x−line ad algebraically.
Cost x − m additional lines = Number of additional lines × Cost per additional line
= (x − m) × d
= d(x − m) dollars
Total cost = Cost first m lines + Cost x − m additional line
= g + d(x − m) dollars
Total cost = g + d(x − m) dollars.
Chapter 5 Exercise 5.1 Automobile Ownership Walkthrough Cengage
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 222 Problem 10 Answer
We have to find : How much does Samantha need to pay for these extra charges, not including the price of the car.
State tax = State tax rate × Purchase price
= 4% × $4, 200
= 0.04 × $4, 200
= $168
Total extra charges = State tax + Charge license plates + Charge state safety and emissions inseption
= $168 + $47 + $35
= $250
Total extra charges = $250
Page 222 Problem 11 Answer
We have to find : What would the new price be if Ralph reduced it according to the suggestion.
Discount = Discount rate × Selling price
= 5% × $18, 500
= 0.05 × $18, 500
= $925
New selling price = Selling price − Discount
= $18, 500 − $925
= $17, 575
New selling price : $17,575
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 222 Problem 12 Answer
We are given,
Cost first 200 characters: $46
Cost per additional character: $0.15
We have to,express the cost c(x) of an ad as a piece wise function.
Let us assume : x = number of characters in ad
c(x) = cost of ad
Cost x − 200 additional characters = Number of additional characters × Cost per additional character
= (x − 200) × 0.15
= 0.15(x − 200)
c(x) = Total cost = Cost first 200 characters + Cost x − 200 additional characters
= 46 + 0.15(x − 200) if x > 200
c(x) = {46 if x ≤ 200
{46 + 0.15(x − 200) if x > 200
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 222 Problem 13 Answer
We have to, Graph the function :
c(x)={46 if x≤200
46+0.15(x−200) if x>200
Sketching the graph we get :
The required graph :
Page 222 Problem 14 Answer
We have to: Find the coordinates of the cusp in the graph.
We can see that : the cost function changes from c(x) = 46 to c(x) = 46 + 0.15(x − 200) when x becomes larger than 200
So we have : The x−coordinate of the cusp is at x = 200
y−coordinates cusp = c(x) = 46
The coordinates of the cusp are : (46, 200)
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 222 Problem 15 Answer
We have to find : How much would the magazine charge to print a2x1/2−inch ad
The amount to be paid for the 2×1/2 inch ad :
21/2 × 67 = $167.50
The amount to be paid for the 2×1/2 inch ad :
$167.50
Page 223 Problem 16 Answer
We have to : Find if the newspaper charges$48 for the first three lines and $5
for each extra line, how much will this ad cost.
Cost 2 additional lines = Number of additional lines × Cost per additional line
= 2 × 5
= $10
Total cost = Cost first 3 lines + Cost 2 additional line
= $48 + $10
= $58
Total cost = $58
Cengage Financial Algebra Automobile Ownership Exercise 5.1 Solutions
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 223 Problem 17 Answer
We have to find : If Ruth buys the car for 8% less than the advertised price, how much does she pay
We are given that :
Selling price = $52, 900
Discount rate = 8% = 0.08
Discount = Discount rate × Selling price
= 8% × $52, 900
= 0.08 × $52, 900
= $4, 232
New selling price= Selling price − Discount
= $52, 900 − $4, 232
= $48, 668
New selling price = $48, 668
Page 223 Problem 18 Answer
We have to : How much sales tax must be paid.
Discount = Discount rate × Selling price
= 8% × $52, 900
= 0.08 × $52, 900
= $4, 232
New selling price = Selling price − Discount
= $52, 900 − $4, 232
= $48, 668
Sales tax = Sales tax rate × New selling price
= 6% × $48, 668
= 0.06 × $48, 668
= $2, 920.08
Sales tax = $2, 920.08
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 223 Problem 19 Answer
We are given that : Advertised price=$12,000
Selling price=$11,200
Commission rate=4%=0.04
We have to find : How much must be paid for the ad.
Commission = Commission rate × Advertised price
= 4% × $12, 000
= 0.04 × $12, 000
= $480
Commission = $480
Page 223 Exercise 1 Answer
We are given the Piecewise function :
c(x)={38 when x≤4
{ 38+6.25(x−4) when x>4
We have to find the cost of a three-line ad.
Answer: We know that, cost is 38 if no. of lines is less than or equal to 4
Therefore,cost of a 3−line ad =38 cost of a 3−line ad =38
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 223 Exercise 2 Answer
We are given the Piece wise function :
c(x)={38 when x≤4
{ 38+6.25(x−4) when x>4
We have to find the difference in cost between a one-line ad and a four-line ad.
We know that, the cost is 38 is the number of lines is less than or equal to 4.
Therefore, there is no difference in cost between a one-line ad and a four-line ad.
There is no difference in cost between a one-line ad and a four-line ad.
Page 223 Exercise 3 Answer
We are given the Piece wise function :
c(x)={38 when x≤4
{ 38+6.25(x−4) when x>4
We have to: Find the cost of a seven-line ad.
Cost of a 7−line ad :
c(7) = 38 + 6.25(7 − 4)
= 38 + 6.25(3)
= 56.75
Cost of a 7−line ad = 56.75
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 223 Exercise 4 Answer
We are given the Piecewise function :
c(x) = { 38 when x ≤ 4
{ 38 + 6.25(x − 4) when x > 4
We have to: Graph this function on your graphing calculator
Sketching the graph we get :
Sketching the graph we get
Page 223 Exercise 5 Answer
We are given the Piecewise function :
c(x) = { 38 when x ≤ 4
{ 38 + 6.25(x − 4) when x > 4
We have to: Find the coordinates of the cusp from the graph.
The cost function changes from c(x) = 38 to c(x) = 38 + 6.25(x − 4) when x become larger than 4
x−coordinates cusp = 4
y−coordinates cusp = c(4) = 38
The coordinates of the cusp are : (4, 38)
How To Solve Cengage Financial Algebra Chapter 5.1 Automobile Ownership
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 223 Exercise 6 Answer
We are given that: The piecewise function describes a newspaper’s classified ad rates :
y={21.50 when x≤3
{ 21.50+5(x−3) when x>3
We have to: Translate the function into words.
Answer : Translating we get :
A newspaper charges $21.50 for the first 3−lines and $5 for each additional line
Translating we get :
A newspaper charges $21.50 for the first 3−lines and $5 for each additional line
Page 223 Exercise 7 Answer
Given
y={ 21.50 when x≤3
{ 21.50+5(x−3) when x>3
If the function is graphed,
we need to find the coordinates of the cusp.
Given:
c(x) = { 21.50 if x ≤ 3
{ 21.50 + 5(x − 3) if x > 3
The cusp is the point in the graph where the two straight lines meet,where each straight line will represent one ”piece” of the piecewise function.
The x−coordinate of the cusp is at x = 3,because the cost function changes from c(x) = 21.50 to c(x) = 21.50 + 5(x − 3) when x becomes larger than 3 x−coordinates cusp = 3
The y−coordinates of the cusp is then the cost function evaluated at x34 : y−coordinates cusp = c(3) = 21.50
Thus the coordinates of the cusp are then (3, 21.50)
Thus the coordinates of the cusp are then (3, 21.50)
Step-By-Step Solutions For Chapter 5.1 Automobile Ownership Exercise
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership Page 223 Exercise 8 Answer
In the above question they have given
Charge: $15
Cost first 5 lines: $2.5 per line
Cost per additional line: $8
We need find If x is the number of lines in the ad, write a piecewise function for the cost of the ad, c(x)
Let x represent the number of lines in the ad and let c(x) represent the cost of the ad.
When x ≤ 5, then the cost is $2.5 per line for each of the first 5 lines (thus you need to pay for 5 lines even if you use less than 5 lines and $8 per additional line
The cost of the lines is then the product of the price per line and the number of lines x :
Cost x lines = Number of lines × Cost per line ad additional line beside the basic5 ad line
= 5 × 2.5
= 12.5 if x ≤ 5
The total cost is then the sum of the charge and the cost of the x lines:
c(x) = Total cost x lines
= Charge + Cost per line
= 15 + 12.5
= 27.5 if x ≤
When x > 5, then the costs is $2.5 per line for the first 5 lines and $8 per additional line.
The cost of the first 5 lines is 1
Cost 5 lines = Number of lines × Cost per line 2.5 perline.
= 5 × 2.5
We are interested in placing a
= 12.5
x−line ad, thus we are placing x−5
Cost x − 5 additional tines = Number of additional lines × Cost per ad additionall ines beside the basic5 − line ad.
= (x − 5) × 8
= 8(x − 5)
x−line ad then includes the charge, the cost of the first 3 lines and the cost of the additional x−3
c(x) = Total cost lines.
= Charge + Cost first 3 tines + Cost x − 3 additional line
= 15 + 12.5 + 8(x − 5) Combining this in for
= 27.5 + 8(x − 5) if x > 3
c(x) = {27.5 if x ≤ 5
{27.5 + 8(x − 5) if x > 5
piecewise function for the cost of the ad, c(x) is
{27.5 if x ≤ 5
{27.5 + 8(x − 5) if x > 5