assignment applications 2 6a

View Answer Keys

View the correct answers for activities in the learning path.

This procedure is for activities that are not provided by an app in the toolbar.

Some MindTap courses contain only activities provided by apps.

• Click an activity in the learning path.

• Science & Math
• Mathematics

Buy new: .savingPriceOverride { color:#CC0C39!important; font-weight: 300!important; } .reinventMobileHeaderPrice { font-weight: 400; } #apex_offerDisplay_mobile_feature_div .reinventPriceSavingsPercentageMargin, #apex_offerDisplay_mobile_feature_div .reinventPricePriceToPayMargin { margin-right: 4px; } -76% $32.37 $ 32 . 37 FREE delivery May 22 - 23 Ships from: Taha Shop Sold by: Taha Shop

Save with used - good .savingpriceoverride { color:#cc0c39important; font-weight: 300important; } .reinventmobileheaderprice { font-weight: 400; } #apex_offerdisplay_mobile_feature_div .reinventpricesavingspercentagemargin, #apex_offerdisplay_mobile_feature_div .reinventpricepricetopaymargin { margin-right: 4px; } $13.05 $ 13 . 05 free delivery thursday, may 23 on orders shipped by amazon over $35 ships from: amazon sold by: canyonlands books & media, return this item for free.

Free returns are available for the shipping address you chose. You can return the item for any reason in new and unused condition: no shipping charges

• Go to your orders and start the return
• Select the return method

Download the free Kindle app and start reading Kindle books instantly on your smartphone, tablet, or computer - no Kindle device required .

Read instantly on your browser with Kindle for Web.

Using your mobile phone camera - scan the code below and download the Kindle app.

Follow the author

Image Unavailable

• To view this video download Flash Player

Financial Algebra: Advanced Algebra with Financial Applications 2nd Edition

Purchase options and add-ons.

• ISBN-10 1337271799
• ISBN-13 978-1337271790
• Edition 2nd
• Publisher Cengage Learning
• Publication date February 7, 2017
• Language English
• Dimensions 9.5 x 1.25 x 12 inches
• Print length 752 pages
• See all details

Frequently bought together

Customers who bought this item also bought

From the Publisher

This product is included in a cengage unlimited subscription.

What’s included in Cengage Unlimited?

• Get all your Cengage access codes for platforms like MindTap, WebAssign, CengageNowv2, SAM, OWLv2 and OpenNow
• Access to the online version of your textbook + our full library
• A lower cost hardcopy textbook rental with each access code, available within the 50 states
• New study tools including online homework, flashcards, test prep and study guides
• A career center where you can boost your job skills, explore career options and build your resume

Editorial Reviews

About the author, product details.

• Publisher ‏ : ‎ Cengage Learning; 2nd edition (February 7, 2017)
• Language ‏ : ‎ English
• Hardcover ‏ : ‎ 752 pages
• ISBN-10 ‏ : ‎ 1337271799
• ISBN-13 ‏ : ‎ 978-1337271790
• Reading age ‏ : ‎ 1 year and up
• Item Weight ‏ : ‎ 3.9 pounds
• Dimensions ‏ : ‎ 9.5 x 1.25 x 12 inches
• #108 in Business Finance
• #115 in Algebra & Trigonometry
• #122 in Business Investments

Videos for this product

Click to play video

Save with Cengage Unlimited

Merchant Video

About the author

Richard j. sgroi.

Discover more of the author’s books, see similar authors, read author blogs and more

Customer reviews

Customer Reviews, including Product Star Ratings help customers to learn more about the product and decide whether it is the right product for them.

To calculate the overall star rating and percentage breakdown by star, we don’t use a simple average. Instead, our system considers things like how recent a review is and if the reviewer bought the item on Amazon. It also analyzed reviews to verify trustworthiness.

• Sort reviews by Top reviews Most recent Top reviews

Top reviews from the United States

There was a problem filtering reviews right now. please try again later..

• Amazon Newsletter
• About Amazon
• Accessibility
• Sustainability
• Press Center
• Investor Relations
• Amazon Devices
• Amazon Science
• Sell on Amazon
• Sell apps on Amazon
• Supply to Amazon
• Protect & Build Your Brand
• Become an Affiliate
• Become a Delivery Driver
• Start a Package Delivery Business
• Advertise Your Products
• Self-Publish with Us
• Become an Amazon Hub Partner
• › See More Ways to Make Money
• Amazon Visa
• Amazon Store Card
• Amazon Secured Card
• Amazon Business Card
• Shop with Points
• Credit Card Marketplace
• Reload Your Balance
• Amazon Currency Converter
• Your Account
• Your Orders
• Shipping Rates & Policies
• Amazon Prime
• Returns & Replacements
• Manage Your Content and Devices
• Recalls and Product Safety Alerts
• Conditions of Use
• Privacy Notice
• Consumer Health Data Privacy Disclosure
• Your Ads Privacy Choices

6.1 Exponential Functions

g ( x ) = 0.875 x g ( x ) = 0.875 x and j ( x ) = 1095.6 − 2 x j ( x ) = 1095.6 − 2 x represent exponential functions.

5.5556 5.5556

About 1.548 1.548 billion people; by the year 2031, India’s population will exceed China’s by about 0.001 billion, or 1 million people.

( 0 , 129 ) ( 0 , 129 ) and ( 2 , 236 ) ; N ( t ) = 129 ( 1 .3526 ) t ( 2 , 236 ) ; N ( t ) = 129 ( 1 .3526 ) t

f ( x ) = 2 ( 1.5 ) x f ( x ) = 2 ( 1.5 ) x

f ( x ) = 2 ( 2 ) x . f ( x ) = 2 ( 2 ) x . Answers may vary due to round-off error. The answer should be very close to 1.4142 ( 1.4142 ) x . 1.4142 ( 1.4142 ) x .

y ≈ 12 ⋅ 1.85 x y ≈ 12 ⋅ 1.85 x

about $3,644,675.88

e − 0.5 ≈ 0.60653 e − 0.5 ≈ 0.60653

$3,659,823.44

3.77E-26 (This is calculator notation for the number written as 3.77 × 10 − 26 3.77 × 10 − 26 in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)

6.2 Graphs of Exponential Functions

The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 0 , ∞ ) ; ( 0 , ∞ ) ; the horizontal asymptote is y = 0. y = 0.

The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 3 , ∞ ) ; ( 3 , ∞ ) ; the horizontal asymptote is y = 3. y = 3.

x ≈ − 1.608 x ≈ − 1.608

f ( x ) = − 1 3 e x − 2 ; f ( x ) = − 1 3 e x − 2 ; the domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( − ∞ , −2 ) ; ( − ∞ , −2 ) ; the horizontal asymptote is y = −2. y = −2.

6.3 Logarithmic Functions

• ⓐ log 10 ( 1 , 000 , 000 ) = 6 log 10 ( 1 , 000 , 000 ) = 6 is equivalent to 10 6 = 1 , 000 , 000 10 6 = 1 , 000 , 000
• ⓑ log 5 ( 25 ) = 2 log 5 ( 25 ) = 2 is equivalent to 5 2 = 25 5 2 = 25
• ⓐ 3 2 = 9 3 2 = 9 is equivalent to log 3 ( 9 ) = 2 log 3 ( 9 ) = 2
• ⓑ 5 3 = 125 5 3 = 125 is equivalent to log 5 ( 125 ) = 3 log 5 ( 125 ) = 3
• ⓒ 2 − 1 = 1 2 2 − 1 = 1 2 is equivalent to log 2 ( 1 2 ) = − 1 log 2 ( 1 2 ) = − 1

log 121 ( 11 ) = 1 2 log 121 ( 11 ) = 1 2 (recalling that 121 = ( 121 ) 1 2 = 11 121 = ( 121 ) 1 2 = 11 )

log 2 ( 1 32 ) = − 5 log 2 ( 1 32 ) = − 5

log ( 1 , 000 , 000 ) = 6 log ( 1 , 000 , 000 ) = 6

log ( 123 ) ≈ 2.0899 log ( 123 ) ≈ 2.0899

The difference in magnitudes was about 3.929. 3.929.

It is not possible to take the logarithm of a negative number in the set of real numbers.

6.4 Graphs of Logarithmic Functions

( 2 , ∞ ) ( 2 , ∞ )

( 5 , ∞ ) ( 5 , ∞ )

The domain is ( 0 , ∞ ) , ( 0 , ∞ ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 0. x = 0.

The domain is ( − 4 , ∞ ) , ( − 4 , ∞ ) , the range ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the asymptote x = – 4. x = – 4.

The domain is ( 2 , ∞ ) , ( 2 , ∞ ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 2. x = 2.

The domain is ( − ∞ , 0 ) , ( − ∞ , 0 ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 0. x = 0.

x ≈ 3.049 x ≈ 3.049

x = 1 x = 1

f ( x ) = 2 ln ( x + 3 ) − 1 f ( x ) = 2 ln ( x + 3 ) − 1

6.5 Logarithmic Properties

log b 2 + log b 2 + log b 2 + log b k = 3 log b 2 + log b k log b 2 + log b 2 + log b 2 + log b k = 3 log b 2 + log b k

log 3 ( x + 3 ) − log 3 ( x − 1 ) − log 3 ( x − 2 ) log 3 ( x + 3 ) − log 3 ( x − 1 ) − log 3 ( x − 2 )

2 ln x 2 ln x

− 2 ln ( x ) − 2 ln ( x )

log 3 16 log 3 16

2 log x + 3 log y − 4 log z 2 log x + 3 log y − 4 log z

2 3 ln x 2 3 ln x

1 2 ln ( x − 1 ) + ln ( 2 x + 1 ) − ln ( x + 3 ) − ln ( x − 3 ) 1 2 ln ( x − 1 ) + ln ( 2 x + 1 ) − ln ( x + 3 ) − ln ( x − 3 )

log ( 3 ⋅ 5 4 ⋅ 6 ) ; log ( 3 ⋅ 5 4 ⋅ 6 ) ; can also be written log ( 5 8 ) log ( 5 8 ) by reducing the fraction to lowest terms.

log ( 5 ( x − 1 ) 3 x ( 7 x − 1 ) ) log ( 5 ( x − 1 ) 3 x ( 7 x − 1 ) )

log x 12 ( x + 5 ) 4 ( 2 x + 3 ) 4 ; log x 12 ( x + 5 ) 4 ( 2 x + 3 ) 4 ; this answer could also be written log ( x 3 ( x + 5 ) ( 2 x + 3 ) ) 4 . log ( x 3 ( x + 5 ) ( 2 x + 3 ) ) 4 .

The pH increases by about 0.301.

ln 8 ln 0.5 ln 8 ln 0.5

ln 100 ln 5 ≈ 4.6051 1.6094 = 2.861 ln 100 ln 5 ≈ 4.6051 1.6094 = 2.861

6.6 Exponential and Logarithmic Equations

x = − 2 x = − 2

x = − 1 x = − 1

x = 1 2 x = 1 2

The equation has no solution.

x = ln 3 ln ( 2 3 ) x = ln 3 ln ( 2 3 )

t = 2 ln ( 11 3 ) t = 2 ln ( 11 3 ) or ln ( 11 3 ) 2 ln ( 11 3 ) 2

t = ln ( 1 2 ) = − 1 2 ln ( 2 ) t = ln ( 1 2 ) = − 1 2 ln ( 2 )

x = ln 2 x = ln 2

x = e 4 x = e 4

x = e 5 − 1 x = e 5 − 1

x ≈ 9.97 x ≈ 9.97

x = 1 x = 1 or x = − 1 x = − 1

t = 703 , 800 , 000 × ln ( 0.8 ) ln ( 0.5 ) years  ≈ 226 , 572 , 993 years . t = 703 , 800 , 000 × ln ( 0.8 ) ln ( 0.5 ) years  ≈ 226 , 572 , 993 years .

6.7 Exponential and Logarithmic Models

f ( t ) = A 0 e − 0.0000000087 t f ( t ) = A 0 e − 0.0000000087 t

less than 230 years, 229.3157 to be exact

f ( t ) = A 0 e ln 2 3 t f ( t ) = A 0 e ln 2 3 t

6.026 hours

895 cases on day 15

Exponential. y = 2 e 0.5 x . y = 2 e 0.5 x .

y = 3 e ( ln 0.5 ) x y = 3 e ( ln 0.5 ) x

6.8 Fitting Exponential Models to Data

• ⓐ The exponential regression model that fits these data is y = 522.88585984 ( 1.19645256 ) x . y = 522.88585984 ( 1.19645256 ) x .
• ⓑ If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.
• ⓐ The logarithmic regression model that fits these data is y = 141.91242949 + 10.45366573 ln ( x ) y = 141.91242949 + 10.45366573 ln ( x )
• ⓑ If sales continue at this rate, about 171,000 games will be sold in the year 2015.
• ⓐ The logistic regression model that fits these data is y = 25.65665979 1 + 6.113686306 e − 0.3852149008 x . y = 25.65665979 1 + 6.113686306 e − 0.3852149008 x .
• ⓑ If the population continues to grow at this rate, there will be about 25,634   25,634   seals in 2020.
• ⓒ To the nearest whole number, the carrying capacity is 25,657.

6.1 Section Exercises

Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.

When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal .

exponential; the population decreases by a proportional rate. .

not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .

The forest represented by the function B ( t ) = 82 ( 1.029 ) t . B ( t ) = 82 ( 1.029 ) t .

After t = 20 t = 20 years, forest A will have 43 43 more trees than forest B.

Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.

exponential growth; The growth factor, 1.06 , 1.06 , is greater than 1. 1.

exponential decay; The decay factor, 0.97 , 0.97 , is between 0 0 and 1. 1.

f ( x ) = 2000 ( 0.1 ) x f ( x ) = 2000 ( 0.1 ) x

f ( x ) = ( 1 6 ) − 3 5 ( 1 6 ) x 5 ≈ 2.93 ( 0.699 ) x f ( x ) = ( 1 6 ) − 3 5 ( 1 6 ) x 5 ≈ 2.93 ( 0.699 ) x

$ 10 , 250 $ 10 , 250

$ 13 , 268.58 $ 13 , 268.58

P = A ( t ) ⋅ ( 1 + r n ) − n t P = A ( t ) ⋅ ( 1 + r n ) − n t

$ 4,572.56 $ 4,572.56

continuous growth; the growth rate is greater than 0. 0.

continuous decay; the growth rate is less than 0. 0.

$ 669.42 $ 669.42

f ( − 1 ) = − 4 f ( − 1 ) = − 4

f ( − 1 ) ≈ − 0.2707 f ( − 1 ) ≈ − 0.2707

f ( 3 ) ≈ 483.8146 f ( 3 ) ≈ 483.8146

y = 3 ⋅ 5 x y = 3 ⋅ 5 x

y ≈ 18 ⋅ 1.025 x y ≈ 18 ⋅ 1.025 x

y ≈ 0.2 ⋅ 1.95 x y ≈ 0.2 ⋅ 1.95 x

APY = A ( t ) − a a = a ( 1 + r 365 ) 365 ( 1 ) − a a = a [ ( 1 + r 365 ) 365 − 1 ] a = ( 1 + r 365 ) 365 − 1 ; APY = A ( t ) − a a = a ( 1 + r 365 ) 365 ( 1 ) − a a = a [ ( 1 + r 365 ) 365 − 1 ] a = ( 1 + r 365 ) 365 − 1 ; I ( n ) = ( 1 + r n ) n − 1 I ( n ) = ( 1 + r n ) n − 1

Let f f be the exponential decay function f ( x ) = a ⋅ ( 1 b ) x f ( x ) = a ⋅ ( 1 b ) x such that b > 1. b > 1. Then for some number n > 0 , n > 0 , f ( x ) = a ⋅ ( 1 b ) x = a ( b − 1 ) x = a ( ( e n ) − 1 ) x = a ( e − n ) x = a ( e ) − n x . f ( x ) = a ⋅ ( 1 b ) x = a ( b − 1 ) x = a ( ( e n ) − 1 ) x = a ( e − n ) x = a ( e ) − n x .

47 , 622 47 , 622 fox

1.39 % ; 1.39 % ; $ 155 , 368.09 $ 155 , 368.09

$ 35 , 838.76 $ 35 , 838.76

$ 82 , 247.78 ; $ 82 , 247.78 ; $ 449.75 $ 449.75

6.2 Section Exercises

An asymptote is a line that the graph of a function approaches, as x x either increases or decreases without bound. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small.

g ( x ) = 4 ( 3 ) − x ; g ( x ) = 4 ( 3 ) − x ; y -intercept: ( 0 , 4 ) ; ( 0 , 4 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

g ( x ) = − 10 x + 7 ; g ( x ) = − 10 x + 7 ; y -intercept: ( 0 , 6 ) ; ( 0 , 6 ) ; Domain: all real numbers; Range: all real numbers less than 7. 7.

g ( x ) = 2 ( 1 4 ) x ; g ( x ) = 2 ( 1 4 ) x ; y -intercept: ( 0 , 2 ) ; ( 0 , 2 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

y -intercept: ( 0 , − 2 ) ( 0 , − 2 )

Horizontal asymptote: h ( x ) = 3 ; h ( x ) = 3 ; Domain: all real numbers; Range: all real numbers strictly greater than 3. 3.

As x → ∞ x → ∞ , f ( x ) → − ∞ f ( x ) → − ∞ ; As x → − ∞ x → − ∞ , f ( x ) → − 1 f ( x ) → − 1

As x → ∞ x → ∞ , f ( x ) → 2 f ( x ) → 2 ; As x → − ∞ x → − ∞ , f ( x ) → ∞ f ( x ) → ∞

f ( x ) = 4 x − 3 f ( x ) = 4 x − 3

f ( x ) = 4 x − 5 f ( x ) = 4 x − 5

f ( x ) = 4 − x f ( x ) = 4 − x

y = − 2 x + 3 y = − 2 x + 3

y = − 2 ( 3 ) x + 7 y = − 2 ( 3 ) x + 7

g ( 6 ) = 800 + 1 3 ≈ 800.3333 g ( 6 ) = 800 + 1 3 ≈ 800.3333

h ( − 7 ) = − 58 h ( − 7 ) = − 58

x ≈ − 2.953 x ≈ − 2.953

x ≈ − 0.222 x ≈ − 0.222

The graph of G ( x ) = ( 1 b ) x G ( x ) = ( 1 b ) x is the refelction about the y -axis of the graph of F ( x ) = b x ; F ( x ) = b x ; For any real number b > 0 b > 0 and function f ( x ) = b x , f ( x ) = b x , the graph of ( 1 b ) x ( 1 b ) x is the the reflection about the y -axis, F ( − x ) . F ( − x ) .

The graphs of g ( x ) g ( x ) and h ( x ) h ( x ) are the same and are a horizontal shift to the right of the graph of f ( x ) ; f ( x ) ; For any real number n , real number b > 0 , b > 0 , and function f ( x ) = b x , f ( x ) = b x , the graph of ( 1 b n ) b x ( 1 b n ) b x is the horizontal shift f ( x − n ) . f ( x − n ) .

6.3 Section Exercises

A logarithm is an exponent. Specifically, it is the exponent to which a base b b is raised to produce a given value. In the expressions given, the base b b has the same value. The exponent, y , y , in the expression b y b y can also be written as the logarithm, log b x , log b x , and the value of x x is the result of raising b b to the power of y . y .

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation b y = x , b y = x , and then properties of exponents can be applied to solve for x . x .

The natural logarithm is a special case of the logarithm with base b b in that the natural log always has base e . e . Rather than notating the natural logarithm as log e ( x ) , log e ( x ) , the notation used is ln ( x ) . ln ( x ) .

a c = b a c = b

x y = 64 x y = 64

15 b = a 15 b = a

13 a = 142 13 a = 142

e n = w e n = w

log c ( k ) = d log c ( k ) = d

log 19 y = x log 19 y = x

log n ( 103 ) = 4 log n ( 103 ) = 4

log y ( 39 100 ) = x log y ( 39 100 ) = x

ln ( h ) = k ln ( h ) = k

x = 2 − 3 = 1 8 x = 2 − 3 = 1 8

x = 3 3 = 27 x = 3 3 = 27

x = 9 1 2 = 3 x = 9 1 2 = 3

x = 6 − 3 = 1 216 x = 6 − 3 = 1 216

x = e 2 x = e 2

14.125 14.125

2 . 7 0 8 2 . 7 0 8

0.151 0.151

No, the function has no defined value for x = 0. x = 0. To verify, suppose x = 0 x = 0 is in the domain of the function f ( x ) = log ( x ) . f ( x ) = log ( x ) . Then there is some number n n such that n = log ( 0 ) . n = log ( 0 ) . Rewriting as an exponential equation gives: 10 n = 0 , 10 n = 0 , which is impossible since no such real number n n exists. Therefore, x = 0 x = 0 is not the domain of the function f ( x ) = log ( x ) . f ( x ) = log ( x ) .

Yes. Suppose there exists a real number x x such that ln x = 2. ln x = 2. Rewriting as an exponential equation gives x = e 2 , x = e 2 , which is a real number. To verify, let x = e 2 . x = e 2 . Then, by definition, ln ( x ) = ln ( e 2 ) = 2. ln ( x ) = ln ( e 2 ) = 2.

No; ln ( 1 ) = 0 , ln ( 1 ) = 0 , so ln ( e 1.725 ) ln ( 1 ) ln ( e 1.725 ) ln ( 1 ) is undefined.

6.4 Section Exercises

Since the functions are inverses, their graphs are mirror images about the line y = x . y = x . So for every point ( a , b ) ( a , b ) on the graph of a logarithmic function, there is a corresponding point ( b , a ) ( b , a ) on the graph of its inverse exponential function.

Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

Domain: ( − ∞ , 1 2 ) ; ( − ∞ , 1 2 ) ; Range: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Domain: ( − 17 4 , ∞ ) ; ( − 17 4 , ∞ ) ; Range: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Domain: ( 5 , ∞ ) ; ( 5 , ∞ ) ; Vertical asymptote: x = 5 x = 5

Domain: ( − 1 3 , ∞ ) ; ( − 1 3 , ∞ ) ; Vertical asymptote: x = − 1 3 x = − 1 3

Domain: ( − 3 , ∞ ) ; ( − 3 , ∞ ) ; Vertical asymptote: x = − 3 x = − 3

Domain: ( 3 7 , ∞ ) ( 3 7 , ∞ ) ; Vertical asymptote: x = 3 7 x = 3 7 ; End behavior: as x → ( 3 7 ) + , f ( x ) → − ∞ x → ( 3 7 ) + , f ( x ) → − ∞ and as x → ∞ , f ( x ) → ∞ x → ∞ , f ( x ) → ∞

Domain: ( − 3 , ∞ ) ( − 3 , ∞ ) ; Vertical asymptote: x = − 3 x = − 3 ; End behavior: as x → − 3 + x → − 3 + , f ( x ) → − ∞ f ( x ) → − ∞ and as x → ∞ x → ∞ , f ( x ) → ∞ f ( x ) → ∞

Domain: ( 1 , ∞ ) ; ( 1 , ∞ ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 1 ; x = 1 ; x -intercept: ( 5 4 , 0 ) ; ( 5 4 , 0 ) ; y -intercept: DNE

Domain: ( − ∞ , 0 ) ; ( − ∞ , 0 ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 0 ; x = 0 ; x -intercept: ( − e 2 , 0 ) ; ( − e 2 , 0 ) ; y -intercept: DNE

Domain: ( 0 , ∞ ) ; ( 0 , ∞ ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 0 ; x = 0 ; x -intercept: ( e 3 , 0 ) ; ( e 3 , 0 ) ; y -intercept: DNE

f ( x ) = log 2 ( − ( x − 1 ) ) f ( x ) = log 2 ( − ( x − 1 ) )

f ( x ) = 3 log 4 ( x + 2 ) f ( x ) = 3 log 4 ( x + 2 )

x = 2 x = 2

x ≈ 2 .303 x ≈ 2 .303

x ≈ − 0.472 x ≈ − 0.472

The graphs of f ( x ) = log 1 2 ( x ) f ( x ) = log 1 2 ( x ) and g ( x ) = − log 2 ( x ) g ( x ) = − log 2 ( x ) appear to be the same; Conjecture: for any positive base b ≠ 1 , b ≠ 1 , log b ( x ) = − log 1 b ( x ) . log b ( x ) = − log 1 b ( x ) .

Recall that the argument of a logarithmic function must be positive, so we determine where x + 2 x − 4 > 0 x + 2 x − 4 > 0 . From the graph of the function f ( x ) = x + 2 x − 4 , f ( x ) = x + 2 x − 4 , note that the graph lies above the x -axis on the interval ( − ∞ , − 2 ) ( − ∞ , − 2 ) and again to the right of the vertical asymptote, that is ( 4 , ∞ ) . ( 4 , ∞ ) . Therefore, the domain is ( − ∞ , − 2 ) ∪ ( 4 , ∞ ) . ( − ∞ , − 2 ) ∪ ( 4 , ∞ ) .

6.5 Section Exercises

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, log b ( x 1 n ) = 1 n log b ( x ) . log b ( x 1 n ) = 1 n log b ( x ) .

log b ( 2 ) + log b ( 7 ) + log b ( x ) + log b ( y ) log b ( 2 ) + log b ( 7 ) + log b ( x ) + log b ( y )

log b ( 13 ) − log b ( 17 ) log b ( 13 ) − log b ( 17 )

− k ln ( 4 ) − k ln ( 4 )

ln ( 7 x y ) ln ( 7 x y )

log b ( 4 ) log b ( 4 )

log b ( 7 ) log b ( 7 )

15 log ( x ) + 13 log ( y ) − 19 log ( z ) 15 log ( x ) + 13 log ( y ) − 19 log ( z )

3 2 log ( x ) − 2 log ( y ) 3 2 log ( x ) − 2 log ( y )

8 3 log ( x ) + 14 3 log ( y ) 8 3 log ( x ) + 14 3 log ( y )

ln ( 2 x 7 ) ln ( 2 x 7 )

log ( x z 3 y ) log ( x z 3 y )

log 7 ( 15 ) = ln ( 15 ) ln ( 7 ) log 7 ( 15 ) = ln ( 15 ) ln ( 7 )

log 11 ( 5 ) = log 5 ( 5 ) log 5 ( 11 ) = 1 b log 11 ( 5 ) = log 5 ( 5 ) log 5 ( 11 ) = 1 b

log 11 ( 6 11 ) = log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 ) − log 5 ( 11 ) log 5 ( 11 ) = a − b b = a b − 1 log 11 ( 6 11 ) = log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 ) − log 5 ( 11 ) log 5 ( 11 ) = a − b b = a b − 1

2.81359 2.81359

0.93913 0.93913

− 2.23266 − 2.23266

x = 4 ; x = 4 ; By the quotient rule: log 6 ( x + 2 ) − log 6 ( x − 3 ) = log 6 ( x + 2 x − 3 ) = 1. log 6 ( x + 2 ) − log 6 ( x − 3 ) = log 6 ( x + 2 x − 3 ) = 1.

Rewriting as an exponential equation and solving for x : x :

6 1 = x + 2 x − 3 0 = x + 2 x − 3 − 6 0 = x + 2 x − 3 − 6 ( x − 3 ) ( x − 3 ) 0 = x + 2 − 6 x + 18 x − 3 0 = x − 4 x − 3 ​ x = 4 6 1 = x + 2 x − 3 0 = x + 2 x − 3 − 6 0 = x + 2 x − 3 − 6 ( x − 3 ) ( x − 3 ) 0 = x + 2 − 6 x + 18 x − 3 0 = x − 4 x − 3 ​ x = 4

Checking, we find that log 6 ( 4 + 2 ) − log 6 ( 4 − 3 ) = log 6 ( 6 ) − log 6 ( 1 ) log 6 ( 4 + 2 ) − log 6 ( 4 − 3 ) = log 6 ( 6 ) − log 6 ( 1 ) is defined, so x = 4. x = 4.

Let b b and n n be positive integers greater than 1. 1. Then, by the change-of-base formula, log b ( n ) = log n ( n ) log n ( b ) = 1 log n ( b ) . log b ( n ) = log n ( n ) log n ( b ) = 1 log n ( b ) .

6.6 Section Exercises

Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

x = − 1 3 x = − 1 3

n = − 1 n = − 1

b = 6 5 b = 6 5

x = 10 x = 10

No solution

p = log ( 17 8 ) − 7 p = log ( 17 8 ) − 7

k = − ln ( 38 ) 3 k = − ln ( 38 ) 3

x = ln ( 38 3 ) − 8 9 x = ln ( 38 3 ) − 8 9

x = ln 12 x = ln 12

x = ln ( 3 5 ) − 3 8 x = ln ( 3 5 ) − 3 8

no solution

x = ln ( 3 ) x = ln ( 3 )

10 − 2 = 1 100 10 − 2 = 1 100

n = 49 n = 49

k = 1 36 k = 1 36

x = 9 − e 8 x = 9 − e 8

n = 1 n = 1

x = ± 10 3 x = ± 10 3

x = 0 x = 0

x = 3 4 x = 3 4

x = 9 x = 9

x = e 2 3 ≈ 2.5 x = e 2 3 ≈ 2.5

x = − 5 x = − 5

x = e + 10 4 ≈ 3.2 x = e + 10 4 ≈ 3.2

x = 11 5 ≈ 2.2 x = 11 5 ≈ 2.2

x = 101 11 ≈ 9.2 x = 101 11 ≈ 9.2

about $ 27 , 710.24 $ 27 , 710.24

about 5 years

ln ( 17 ) 5 ≈ 0.567 ln ( 17 ) 5 ≈ 0.567

x = log ( 38 ) + 5 log ( 3 )    4 log ( 3 ) ≈ 2.078 x = log ( 38 ) + 5 log ( 3 )    4 log ( 3 ) ≈ 2.078

x ≈ 2.2401 x ≈ 2.2401

x ≈ − 44655 . 7143 x ≈ − 44655 . 7143

about 5.83 5.83

t = ln ( ( y A ) 1 k ) t = ln ( ( y A ) 1 k )

t = ln ( ( T − T s T 0 − T s ) − 1 k ) t = ln ( ( T − T s T 0 − T s ) − 1 k )

6.7 Section Exercises

Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.

Doubling time is a measure of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity to double in size.

An order of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about 10 2 10 2 times, or 2 orders of magnitude greater, than the mass of Earth.

f ( 0 ) ≈ 16.7 ; f ( 0 ) ≈ 16.7 ; The amount initially present is about 16.7 units.

exponential; f ( x ) = 1.2 x f ( x ) = 1.2 x

logarithmic

about 1.4 1.4 years

about 7.3 7.3 years

4 4 half-lives; 8.18 8.18 minutes

M = 2 3 log ( S S 0 ) log ( S S 0 ) = 3 2 M S S 0 = 10 3 M 2 S = S 0 10 3 M 2 M = 2 3 log ( S S 0 ) log ( S S 0 ) = 3 2 M S S 0 = 10 3 M 2 S = S 0 10 3 M 2

Let y = b x y = b x for some non-negative real number b b such that b ≠ 1. b ≠ 1. Then,

ln ( y ) = ln ( b x ) ln ( y ) = x ln ( b ) e ln ( y ) = e x ln ( b )             y = e x ln ( b ) ln ( y ) = ln ( b x ) ln ( y ) = x ln ( b ) e ln ( y ) = e x ln ( b )             y = e x ln ( b )

A = 125 e ( − 0.3567 t ) ; A ≈ 43 A = 125 e ( − 0.3567 t ) ; A ≈ 43 mg

about 60 60 days

A ( t ) = 250 e ( − 0.00822 t ) ; A ( t ) = 250 e ( − 0.00822 t ) ; half-life: about 84 84 minutes

r ≈ − 0.0667 , r ≈ − 0.0667 , So the hourly decay rate is about 6.67 % 6.67 %

f ( t ) = 1350 e ( 0.03466 t ) ; f ( t ) = 1350 e ( 0.03466 t ) ; after 3 hours: P ( 180 ) ≈ 691 , 200 P ( 180 ) ≈ 691 , 200

f ( t ) = 256 e ( 0.068110 t ) ; f ( t ) = 256 e ( 0.068110 t ) ; doubling time: about 10 10 minutes

about 88 88 minutes

T ( t ) = 90 e ( − 0.008377 t ) + 75 , T ( t ) = 90 e ( − 0.008377 t ) + 75 , where t t is in minutes.

about 113 113 minutes

log ( x ) = 1.5 ; x ≈ 31.623 log ( x ) = 1.5 ; x ≈ 31.623

MMS magnitude: 5.82 5.82

N ( 3 ) ≈ 71 N ( 3 ) ≈ 71

6.8 Section Exercises

Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic model best describes populations.

Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu.

The y -intercept on the graph of a logistic equation corresponds to the initial population for the population model.

P ( 0 ) = 22 P ( 0 ) = 22 ; 175

p ≈ 2.67 p ≈ 2.67

y -intercept: ( 0 , 15 ) ( 0 , 15 )

about 6.8 6.8 months.

About 38 wolves

About 8.7 years

f ( x ) = 776.682 ( 1.426 ) x f ( x ) = 776.682 ( 1.426 ) x

f ( x ) = 731.92 e -0.3038 x f ( x ) = 731.92 e -0.3038 x

When f ( x ) = 250 , x ≈ 3.6 f ( x ) = 250 , x ≈ 3.6

y = 5.063 + 1.934 log ( x ) y = 5.063 + 1.934 log ( x )

When f ( 10 ) ≈ 2.3 f ( 10 ) ≈ 2.3

When f ( x ) = 8 , x ≈ 0.82 f ( x ) = 8 , x ≈ 0.82

f ( x ) = 25.081 1 + 3.182 e − 0.545 x f ( x ) = 25.081 1 + 3.182 e − 0.545 x

When f ( x ) = 68 , x ≈ 4.9 f ( x ) = 68 , x ≈ 4.9

f ( x ) = 1.034341 ( 1.281204 ) x f ( x ) = 1.034341 ( 1.281204 ) x ; g ( x ) = 4.035510 g ( x ) = 4.035510 ; the regression curves are symmetrical about y = x y = x , so it appears that they are inverse functions.

f − 1 ( x ) = ln ( a ) - ln ( c x - 1 ) b f − 1 ( x ) = ln ( a ) - ln ( c x - 1 ) b

Review Exercises

exponential decay; The growth factor, 0.825 , 0.825 , is between 0 0 and 1. 1.

y = 0.25 ( 3 ) x y = 0.25 ( 3 ) x

$ 42 , 888.18 $ 42 , 888.18

continuous decay; the growth rate is negative.

domain: all real numbers; range: all real numbers strictly greater than zero; y -intercept: (0, 3.5);

g ( x ) = 7 ( 6.5 ) − x ; g ( x ) = 7 ( 6.5 ) − x ; y -intercept: ( 0 , 7 ) ; ( 0 , 7 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

17 x = 4913 17 x = 4913

log a b = − 2 5 log a b = − 2 5

x = 64 1 3 = 4 x = 64 1 3 = 4

log ( 0 .000001 ) = − 6 log ( 0 .000001 ) = − 6

ln ( e − 0.8648 ) = − 0.8648 ln ( e − 0.8648 ) = − 0.8648

Domain: x > − 5 ; x > − 5 ; Vertical asymptote: x = − 5 ; x = − 5 ; End behavior: as x → − 5 + , f ( x ) → − ∞ x → − 5 + , f ( x ) → − ∞ and as x → ∞ , f ( x ) → ∞ . x → ∞ , f ( x ) → ∞ .

log 8 ( 65 x y ) log 8 ( 65 x y )

ln ( z x y ) ln ( z x y )

log y ( 12 ) log y ( 12 )

ln ( 2 ) + ln ( b ) + ln ( b + 1 ) − ln ( b − 1 ) 2 ln ( 2 ) + ln ( b ) + ln ( b + 1 ) − ln ( b − 1 ) 2

log 7 ( v 3 w 6 u 3 ) log 7 ( v 3 w 6 u 3 )

x = log ( 125 ) log ( 5 ) + 17 12 = 5 3 x = log ( 125 ) log ( 5 ) + 17 12 = 5 3

x = − 3 x = − 3

x = ln ( 11 ) x = ln ( 11 )

a = e 4 − 3 a = e 4 − 3

x = ± 9 5 x = ± 9 5

about 5.45 5.45 years

f − 1 ( x ) = 2 4 x − 1 3 f − 1 ( x ) = 2 4 x − 1 3

f ( t ) = 300 ( 0.83 ) t ; f ( t ) = 300 ( 0.83 ) t ; f ( 24 ) ≈ 3.43     g f ( 24 ) ≈ 3.43     g

about 45 45 minutes

about 8.5 8.5 days

exponential

y = 4 ( 0.2 ) x ; y = 4 ( 0.2 ) x ; y = 4 e -1.609438 x y = 4 e -1.609438 x

about 7.2 7.2 days

logarithmic; y = 16.68718 − 9.71860 ln ( x ) y = 16.68718 − 9.71860 ln ( x )

Practice Test

About 13 dolphins.

$ 1,947 $ 1,947

y -intercept: ( 0 , 5 ) ( 0 , 5 )

8.5 a = 614.125 8.5 a = 614.125

x = ( 1 7 ) 2 = 1 49 x = ( 1 7 ) 2 = 1 49

ln ( 0.716 ) ≈ − 0.334 ln ( 0.716 ) ≈ − 0.334

Domain: x < 3 ; x < 3 ; Vertical asymptote: x = 3 ; x = 3 ; End behavior: x → 3 − , f ( x ) → − ∞ x → 3 − , f ( x ) → − ∞ and x → − ∞ , f ( x ) → ∞ x → − ∞ , f ( x ) → ∞

log t ( 12 ) log t ( 12 )

3 ln ( y ) + 2 ln ( z ) + ln ( x − 4 ) 3 3 ln ( y ) + 2 ln ( z ) + ln ( x − 4 ) 3

x = ln ( 1000 ) ln ( 16 ) + 5 3 ≈ 2.497 x = ln ( 1000 ) ln ( 16 ) + 5 3 ≈ 2.497

a = ln ( 4 ) + 8 10 a = ln ( 4 ) + 8 10

x = ln ( 9 ) x = ln ( 9 )

x = ± 3 3 2 x = ± 3 3 2

f ( t ) = 112 e − .019792 t ; f ( t ) = 112 e − .019792 t ; half-life: about 35 35 days

T ( t ) = 36 e − 0.025131 t + 35 ; T ( 60 ) ≈ 43 o F T ( t ) = 36 e − 0.025131 t + 35 ; T ( 60 ) ≈ 43 o F

exponential; y = 15.10062 ( 1.24621 ) x y = 15.10062 ( 1.24621 ) x

logistic; y = 18.41659 1 + 7.54644 e − 0.68375 x y = 18.41659 1 + 7.54644 e − 0.68375 x

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra/pages/chapter-6

© Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

Chapter 6 Review Exercises

• Last updated
• Save as PDF
• Page ID 10839

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

True or False? Justify your answer with a proof or a counterexample.

1) The amount of work to pump the water out of a half-full cylinder is half the amount of work to pump the water out of the full cylinder.

2) If the force is constant, the amount of work to move an object from \(x=a\) to \(x=b\) is \(F(b−a)\).

3) The disk method can be used in any situation in which the washer method is successful at finding the volume of a solid of revolution.

4) If the half-life of \(seaborgium-266\) is \(360\) ms, then \(k=\dfrac{\ln 2}{360}.\)

For exercises 5 - 8, use the requested method to determine the volume of the solid.

5) The volume that has a base of the ellipse \(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1\) and cross-sections of an equilateral triangle perpendicular to the \(y\)-axis. Use the method of slicing.

6) \(y=x^2−x\), from \(x=1\) to \(x=4\), rotated around the \(y\)-axis using the washer method

7) \(x=y^2\) and \(x=3y\) rotated around the \(y\)-axis using the washer method

8) \(x=2y^2−y^3,\; x=0\),and \(y=0\) rotated around the \(x\)-axis using cylindrical shells

For exercises 9 - 14, find

a. the area of the region,

b.the volume of the solid when rotated around the \(x\) -axis, and

c. the volume of the solid when rotated around the \(y\) -axis. Use whichever method seems most appropriate to you.

9) \(y=x^3,x=0,y=0\), and \(x=2\)

10) \(y=x^2−x\) and \(x=0\)

11) [T] \(y=\ln(x)+2\) and \(y=x\)

12) \(y=x^2\) and \(y=\sqrt{x}\)

13) \(y=5+x, y=x^2, x=0\), and \(x=1\)

14) Below \(x^2+y^2=1\) and above \(y=1−x\)

15) Find the mass of \(ρ=e^{−x}\) on a disk centered at the origin with radius \(4\).

16) Find the center of mass for \(ρ=\tan^2x\) on \(x\in (−\frac{π}{4},\frac{π}{4})\).

17) Find the mass and the center of mass of \(ρ=1\) on the region bounded by \(y=x^5\) and \(y=\sqrt{x}\).

For exercises 18 - 19, find the requested arc lengths.

18) The length of \(x\) for \(y=\cosh(x)\) from \(x=0\) to \(x=2\).

19) The length of \(y\) for \(x=3−\sqrt{y}\) from \(y=0\) to \(y=4\)

For exercises 20 - 21, find the surface area and volume when the given curves are revolved around the specified axis.

20) The shape created by revolving the region between \(y=4+x, \;y=3−x, \;x=0,\) and \(x=2\) rotated around the \(y\)-axis.

21) The loudspeaker created by revolving \(y=\dfrac{1}{x}\) from \(x=1\) to \(x=4\) around the \(x\)-axis.

For exercise 22, consider the Karun-3 dam in Iran. Its shape can be approximated as an isosceles triangle with height 205 m and width 388 m. Assume the current depth of the water is 180 m. The density of water is 1000 kg/m 3 .

22) Find the total force on the wall of the dam.

23) You are a crime scene investigator attempting to determine the time of death of a victim. It is noon and \(45\) °F outside and the temperature of the body is \(78\) °F. You know the cooling constant is \(k=0.00824\) °F/min. When did the victim die, assuming that a human’s temperature is \(98\) °F?

For the following exercise, consider the stock market crash in 1929 in the United States. The table lists the Dow Jones industrial average per year leading up to the crash.

Source : http:/stockcharts.com/freecharts/hi...a19201940.html

24) [T] The best-fit exponential curve to these data is given by \(y=40.71+1.224^x\). Why do you think the gains of the market were unsustainable? Use first and second derivatives to help justify your answer. What would this model predict the Dow Jones industrial average to be in 2014 ?

For exercises 25 - 26, consider the catenoid, the only solid of revolution that has a minimal surface, or zero mean curvature. A catenoid in nature can be found when stretching soap between two rings.

25) Find the volume of the catenoid \(y=\cosh(x)\) from \(x=−1\) to \(x=1\) that is created by rotating this curve around the \(x\) -axis, as shown here.

26) Find surface area of the catenoid \(y=\cosh(x)\) from \(x=−1\) to \(x=1\) that is created by rotating this curve around the \(x\) -axis.

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .

ENTC 2170—Assignment 2-6: Problem Solving with CADD

Factoring Calculator

Enter the expression you want to factor in the editor.

Difference of Squares : a 2   –   b 2   =   ( a   +   b ) ( a   –   b )

Please ensure that your password is at least 8 characters and contains each of the following:

• a special character: @$#!%*?&

English Language Arts and Reading.2.6.A

establish purpose for reading assigned and self-selected texts ;

A student expectation is directly related to the knowledge and skills statement, is more specific about how students demonstrate their learning, and always begins with a verb. Student expectations are further broken down into their component parts, often referred to as “breakouts.”

Knowledge and Skills Statement

A knowledge and skills statement is a broad statement of what students must know and be able to do. It generally begins with a learning strand and ends with the phrase “The student is expected to:” Knowledge and skills statements always include related student expectations.

Use an observational checklist. A teacher can prompt students to identify whether they can establish a purpose for reading both assigned and self-selected texts.

• What is the purpose for reading this story?
• Why did you choose to read this story?
• How will this story help you?

An observation checklist could be supported with a scale such as this:

- The student is unable to identify the purpose for reading a text.

* The student is able to identify the purpose for reading a self-selected text but unable to identify the purpose for reading an assigned text (beyond my teacher told me to ).

+ The student is able to identify the purpose for reading both self-selected texts and assigned texts (e.g., “I am reading this book so I can learn more about animal homes ” ).

This type of observation should occur over time with multiple books. 

Glossary Support for ELA.2.6.A

What Works Clearinghouse. (2010). Improving reading comprehension in kindergarten through 3rd grade: practice guide summary . Washington, DC: Institute of Education Science. Retrieved from https://ies.ed.gov/ncee/wwc/PracticeGuide/14#tab-summary

Summary:  The goal of this practice guide is to offer educators specific evidence-based recommendations that address the challenge of teaching reading comprehension to students in kindergarten through 3rd grade. The guide provides practical, clear information on critical topics related to teaching reading comprehension and is based on the best available evidence as judged by the authors.