MTH 09 Week 3
Due for this week Homework 3 (on MyMathLab via the Materials Link) The fifth night after class at 11:59pm. Read Chapter 6.6, 8.4 and 11.1-11.5 Do the MyMathLab Self-Check for week 3. Learning team hardest problem assignment. Complete the Week 3 study plan after submitting week 3 homework. Participate in the Chat Discussions in the OLS Copyright 009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide
Section 7.1 Introduction to Rational Epressions Copyright 013, 009, and 005 Pearson Education, Inc.
Objectives Basic Concepts Simplifying Rational Epressions Applications
Basic Concepts Rational epressions can be written as quotients (fractions) of two polynomials. Eamples include: 5,, 3 4 4 61 3 4 8
If possible, evaluate each epression for the given value of the variable. a. b. w 4 w c. w 1 ; 3 3 ; w 4 3w 4 Solution a. b. w c. 4 1 ; 3 3 1 1 3 3 6 ; w 4 3w 4 ( 4) 3( 4) 4 16 1 4 16 8 ; 5 w 4 w ; w 5 w 4 4 ( 5) ( 5) 4 9 9 1 Try Q 7,11,13,17 pg. 47
Try Q 5,7,31,33 pg. 48 Find all values of the variable for which each epression is undefined. 1 a. b. w 6 c. w 4 Solution a. 1 b. w c. w 4 w 4 6 w 4 Undefined when = 0 or when = 0. Undefined when w 4 = 0 or when w = 4. Undefined when w 4 = 0 or when w =.
Try Q 39,43 pg. 48 Simplify each fraction by applying the basic principle of fractions. 9 0 45 a. b. c. 15 8 Solution a. The GCF of 9 and 15 is 3. 135 9 33 3 15 3 5 5 b. The GCF of 0 and 8 is 4. c. The GCF of 45 and 135 is 45. 0 45 5 8 4 7 7 45 135 451 453 1 3
Simplify each epression. a. 16y b. 3 1 c. 4y 4 16 Solution 3 1 a. 16y b. c. 4y 4y 4 4y y 4 y 4 16 3( 4) 4( 4) 3 4 5 715 5 715 ( 5)( 5) ( 3)( 5) 5 3
Try Q 51,55,61,79 pg. 48 Simplify each epression. a. y 7 b. y 14 Solution a. y 7 1( y 7) b. y 14 ( y 7) 1 8 8 8 (8 ) 8 8 8 8 8 1 8
Try Q 105 pg. 49 Suppose that n balls, numbered 1 to n, are placed in a container and two balls have the winning number. a. What is the probability of drawing the winning ball at random? b. Calculate this probability for n = 100, 1000 and 10,000. c. What happens to the probability of drawing the winning ball as the number of balls increases? Solution a. There are chances of drawing the winning ball. n
(cont) Try Q 105 pg. 49 b. Calculate this probability for n = 100, 1000 and 10,000. 1 1 1 100 50 1000 500 10,000 5000 c. What happens to the probability of drawing the winning ball as the number of balls increases? The probability decreases.
Section 7. Multiplication and Division of Rational Epressions Copyright 013, 009, and 005 Pearson Education, Inc.
Objectives Review of Multiplication and Division of Fractions Multiplication of Rational Epressions Division of Rational Epressions
Try Q 5,7,9 pg. 435 Multiply and simplify your answers to lowest terms. 4 5 a. b. c. 9 4 5 15 7 5 7 8 Solution a. 4 5 0 9 7 63 b. c. 4 15 4 60 15 1 5 1 5 5 5 5 5 1 5 7 8 78 7 4 8
Try Q 13,15,17 pg. 435 Divide and simplify your answers to lowest terms. 1 3 6 a. b. 18 c. 6 5 7 4 11 5 15 Solution a. 1 3 1 5 5 b. 6 18 6 5 6 3 18 7 6 1 16 7 18 7 18 1 1 c. 4 11 5 15 4 15 5 11 60 55 15 115 1 11
Multiply and simplify to lowest terms. Leave your answers in factored form. 6 5 3 4 a. b. 10 1 Solution a. 6 5 6 5 b. 10 1 10 1 30 10 1 4 1 39 3 4 1 39 ( 3)( 4) ( 1) ( 3 9) ( 3)( 4) 3( 1)( 3) 4 3( 1)
Try Q 9,31,41,45 pg. 435 Multiply and simplify to lowest terms. Leave your answer in factored form. 16 3 Solution 16 3 9 4 9 4 ( 16) ( 3) ( 9) ( 4) ( 4)( 4)( 3) ( 3)( 3)( 4) ( 4)( 3)( 4) ( 3) ( 3 )( 4) ( 4) ( 3)
Try Q 49,57,65 pg. 435 Divide and simplify to lowest terms. a. 3 1 b. 6 Solution a. 3 1 3 6 b. 6 1 18 (1) 18 1 16 4 8 16 8 4 16 8 4 ( 4)( 4) ( )( 4) 4 ( 4)( 4)( ) ( )( 4)( 4) 1
Section 7.3 Addition and Subtraction with Like Denominators Copyright 013, 009, and 005 Pearson Education, Inc.
Objectives Review of Addition and Subtraction of Fractions Rational Epressions Having Like Denominators
Simplify each epression to lowest terms. a. 4 1 b. 1 5 7 7 Solution 9 9 a. 4 1 7 7 41 7 5 7 b. 1 5 9 9 1 5 9 6 9 3
Try Q 7,9,11,13 pg. 44 Simplify each epression to lowest terms. a. 13 7 b. 18 18 Solution 15 11 30 30 a. 13 7 18 18 13 7 18 6 18 1 3 b. 15 11 30 30 15 11 30 4 30 15
SUMS OF RATIONAL EXPRESSIONS To add two rational epressions having like denominators, add their numerators. Keep the same denominator. A B A B C C C C is nonzero
Try Q 19,5,33,35 pg. 44 Add and simplify to lowest terms. a. 41 b. Solution a. 4 b. 3 3 1 41 3 3 3 5 7 10 7 10 5 7 10 7 10 5 1 3 5 7 10 5 5 1
Try Q 51,53,55 pg. 443 Add and simplify to lowest terms. 7 4 a. b. w ab ab Solution a. 7 4 ab ab 7 4 ab 11 ab y w y w y b. w y w y w y w y w y w y ( w y)( w y) 1 w y
DIFFERENCES OF RATIONAL EXPRESSIONS To subtract two rational epressions having like denominators, subtract their numerators. Keep the same denominator. A B A B C C C C is nonzero
Try Q 1,7,67 pg. 44-3 Subtract and simplify to lowest terms. 6 6 3 4 a. b. Solution a. 6 6 6 6 1 1 6 6 1 b. 3 4 1 1 3 4 1 1 1 1 1 1 1 1
Try Q 31,59 pg. 44-3 Subtract and simplify to lowest terms. Solution 7a a a a 7a a a a 7 a( a) a 6a a
Section 7.4 Addition and Subtraction with Unlike Denominators Copyright 013, 009, and 005 Pearson Education, Inc.
Objectives Finding Least Common Multiples Review of Fractions Having Unlike Denominators Rational Epressions Having Unlike Denominators
FINDING THE LEAST COMMON MULTIPLE The least common multiple (LCM) of two or more polynomials can be found as follows. Step 1: Factor each polynomial completely. Step : List each factor the greatest number of times that it occurs in either factorization. Step 3: Find the product of this list of factors. The result is the LCM.
Find the least common multiple of each pair of epressions. a. 6, 9 4 b. + 7 + 1, + 8 + 16 Solution Step 1: Factor each polynomial completely. 6 = 3 9 4 = 3 3 Step : List each factor the greatest number of times. 3 3 Step 3: The LCM is 18 4.
(cont) b. + 7 + 1, + 8 + 16 Step 1: Factor each polynomial completely. + 7 + 1 = ( + 3)( + 4) + 8 + 16 = (+ 4)( + 4) Step : List each factor the greatest number of times. ( + 3), ( + 4), and ( + 4) Step 3: The LCM is ( + 3)( + 4). Try Q 15,19,7,9 pg. 451
Try Q 45,47 pg. 45 Simplify each epression. a. 4 1 b. 7 6 5 11 1 30 Solution a. The LCD is the LCM, 4. 4 1 7 6 4 6 1 7 7 6 6 7 4 7 4 4 31 4 b. The LCD is 60. 5 11 1 30 5 5 11 1 5 30 5 60 60 3 60 1 0
Try Q 53,65,71 pg. 45 Find each sum and leave your answer in factored form. 5 4 3 a. b. Solution a. The LCD is. 5 1 1 5 5 5 b. 4 3 1 1 4 3 1 1 1 1 4 3 1 1 1 1
Simply the epression. Write your answer in lowest terms and leave it in factored form. 3 5 Solution 7 The LCD is ( + 7). 3 5 3 7 5 7 7 7 37 5 7 7 3 7 5 4 1 5 7 7 7 1
Try Q 63,77,79,81 pg. 45 Simplify the epression. Write your answer in lowest terms and leave it in factored form. 6 5 Solution 6 9 9 6 5 6 9 9 6 5 3 3 3 3 3 3 3 3 6 5 3 3 3 3 6 3 5 3 3 3 3 3 3 3 618 515 33 33 3 33 3
Try Q 101 pg. 453 Add 1 1, R S and then find the reciprocal of the result. Solution The LCD is RS. 1 1 1 S 1 R S R S S S R RS RS S R RS R R The reciprocal is RS. S R
Section 7.6 Rational Equations and Formulas Copyright 013, 009, and 005 Pearson Education, Inc.
Objectives Solving Rational Equations Rational Epressions and Equations Graphical and Numerical Solutions Solving a Formula for a Variable Applications
Rational Equations If an equation contains one or more rational epressions, it is called a rational equation.
Try Q 9,15,31,41 pg. 473 Solve each equation. a. 9 4 b. 7 Solution a. 9 4 b. 7 9 8 8 9 The solution is 8. 9 6 1 6 1 6 1 1 1 6(1) 6 6 0 3 0 3 0 3 The solutions are 3 and. 0
Determine whether you are given an epression or an equation. If it is an epression, simplify it and then evaluate it for = 4. If it is an equation, solve it. a. b. 3 10 4 3 3 Solution a. There is an equal sign, so it is an equation. 3 3 4 1 3 4 3 3 3 1 The answer checks. 4 The solution is 4.
(cont) 3 10 b. There is no equals sign, so it is an epression. The common denominator is, so we can add the numerators. 3 10 310 5 5 When = 4, the epression evaluates 4 + 5 = 9. Try Q 49,55 pg. 473
Solve 1 graphically and numerically. Solution Graph y 1 and y 1. 3 1 0 1 3 1 1 3 1 (, ) (1, 1) The solutions are and 1.
(cont) Try Q 75 pg. 474 Solve 1 graphically and numerically. Solution Numerical Solution y 1 y 3 1 0 1 3 1 1 1 3 1 0 1 3 3 1 The solutions are and 1.
Solve the equation for the specified variable. C r for r Solution C r for r C r C r
Solve the equation for the specified variable. A h for A B b Solution A h B b h( B b) A h( B b) A
Try Q 89,91,97 pg. 474 Solve the equation for the specified variable. S B sl for s Solution S B sl for S B B B sl S B sl S B sl l l S B s l s
Try Q 10 pg. 474 A pump can fill a swimming pool ¾ full in 6 hours, another can fill the pool ¾ full in 9 hours. How long would it take the pumps to fill the pool ¾ full, working together? t t 3 Solution 6 9 4 t t 3 36 36 6 9 4 6t4t 7 7 t 10 The two pumps can fill the pool ¾ full in hours. 7 10
Section 7.7 Proportions and Variation Copyright 013, 009, and 005 Pearson Education, Inc.
Objectives Proportions Direct Variation Inverse Variation Analyzing Data Joint Variation
Proportions A proportion is a statement (equation) that two ratios (fractions) are equal. The following property is a convenient way to solve proportions: a b c is equivalent to ad bc, d provided b 0 and d 0.
Try Q 65 pg. 488 On an elliptical machine, Francis can burn 370 calories in 5 minutes. If he increases his work time to 30 minutes, how many calories will he burn? Solution Let be the equivalent amount of calories. 5 30 370 5 11,100 Minutes = Calories Minutes Calories 444 Thus, in 30 minutes, Francis will burn 444 calories.
Try Q 56 pg. 488 A 6-foot tall person casts a shadow that is 8-foot long. If a nearby tree casts a 3-foot long shadow, estimate the height of the tree. Solution 6 ft h The triangles are similar because the measures of its corresponding 8 ft 3 ft angles are equal. Therefore corresponding sides are proportional. h 3 Height Shadow length = 6 8 Height Shadow length 8h 19 h 4 The tree is 4 feet tall.
Try Q 33 pg. 487 Let y be directly proportional to, or vary directly with. Suppose y = 9 when = 6. Find y when = 13. Solution y k Step 1 The general equation is y = k. 9 6k Step Substitute 9 for y and 6 for in 9 k y = k. Solve for k. 6 Step 3 Replace k with 9/6 in the equation y = 9/6. Step 4 To find y, let = 13. y 9 6 y y 9 (13) 6 19.5
Pay (dollars) Eample The table lists the amount of pay for various hours worked. Hours Pay 6 $138 11 $53 15 $345 3 $59 31 $713 6, 138 11, 53 15, 345 3, 59 a. Find the constant of proportionality. b. Predict the pay for 19 hours of work. 800 700 600 500 400 300 00 100 0 31, 713 0 10 0 30 40 Hours
(cont) Try Q 73 pg. 488 The slope of the line equals the proportionality, k. If we use the first and last data points (6, 138) and (31, 713), the slope is k 713 138 31 6 3 The amount of pay per hour is $3. The graph of the line y = 3, models the given graph. To find the pay for 19 hours, substitute 19 for. y = 3, y = 3(19) y = 437 19 hours of work would pay $437.00
Try Q 39 pg. 487 Let y be inversely proportional to, or vary inversely with. Suppose y = 6 when = 4. Find y when = 8. k Solution y Step 1 The general equation is y = k/. k Step Substitute 6 for y and 4 for in 6 4 Solve for k. 4 k Step 3 Replace k with 4 in the equation y = k/. k Step 4 To find y, let = 8. y y y 4 8 3
Try Q 51a,53a,55a pg. 487-488 Determine whether the data in each table represent direct variation, inverse variation, or neither. For direct and inverse variation, find the equation. a. b. c. 3 7 9 1 y 1 8 3 48 5 10 1 15 y 1 6 5 4 8 11 14 1 y 48 66 84 16 Neither the product y nor the ratio y/ are constant in the data in the table. Therefore there is neither direct variation nor indirect variation in this table. As increases, y decreases. Because y = 60 for each data point, the equation y = 60/ models the data. This represents an inverse variation. The equation y = 6 models the data. The data represents direct variation.
, 0 and 1, JOINT VARIATION f a a a Let, y, and z denote three quantities. Then z varies jointly with and y if there is a nonzero number k such that z ky.
Try Q 83 pg. 488 The strength S of a rectangular beam varies jointly as its width w and the square of its thickness t. If a beam 5 inches wide and inches thick supports 80 pounds, how much can a similar beam 4 inches wide and 3 inches thick support? Solution The strength of the beam is 80 k 5 modeled by S = kwt. 80 k 54 k 14
(cont) Try Q 83 pg. 488 Thus S = 14wt models the strength of this type of beam. When w = 4 and t = 3, the beam can support S = 14 4 3 = 504 pounds
Section 10.1 Radical Epressions and Functions Copyright 013, 009, and 005 Pearson Education, Inc.
Objectives Radical Notation The Square Root Function The Cube Root Function
Radical Notation Every positive number a has two square roots, one positive and one negative. Recall that the positive square root is called the principal square root. The symbol is called the radical sign. The epression under the radical sign is called the radicand, and an epression containing a radical sign is called a radical epression. Eamples of radical epressions: 7, 6, and 5 3 4
Try Q 15,17,19,1 pg. 641 Evaluate each square root. a. 36 6 b. 0.64 0.8 c. 16 5 4 5
Try Q 39 pg. 641 Approimate Solution 38 to the nearest thousandth. 6.164
Try Q 3,5,7,41 pg. 641 Evaluate the cube root. a. 3 64 4 b. 3 15 5 c. 3 1 8 1
Try Q 33,35,37 pg. 641 Find each root, if possible. a. 4 5 56 b. 43 c. Solution 4 196 a. 4 56 4 because 4444 56. b. 5 43 5 3 because ( 3) 43. c. 4 196 An even root of a negative number is not a real number.
Try Q 45,49,51 pg. 641 Write each epression in terms of an absolute value. ( 5) ( 3) a. b. c. w 6w9 Solution a. ( 5) 5 5 b. ( 3) 3 c. w 6w9 ( w 3) w3
Try Q 61,63 pg. 641 If possible, evaluate f(1) and f() for each f(). a. f ( ) 5 1 b. f ( ) 4 Solution a. f ( ) 5 1 b. f ( ) 4 f (1) 5(1) 1 6 f ( ) 5( ) 1 9 undefined f (1) 1 4 5 f ( ) ( ) 4 8
Try Q 75,89 pg. 64 Calculate the hang time for a ball that is kicked 75 feet into the air. Does the hang time double when a ball is kicked twice as high? Use the formula Solution The hang time is The hang time is 1 T( ) 1 T(75) 75 4.3 sec 1 T(150) 150 6.1 sec The hang times is less than double.
Try Q 75,89 pg. 641 Find the domain of each function. Write your answer in interval notation. a. f ( ) 3 4 b. f ( ) 4 Solution Solve 3 4 0. 3 4 0 4 3 3 4 The domain is 3,. 4 b. Regardless of the value of ; the epression is always positive. The function is defined for all real numbers, and it domain,. is
Section 10. Rational Eponents Copyright 013, 009, and 005 Pearson Education, Inc.
Objectives Basic Concepts Properties of Rational Eponents
Try Q 37,54,59,63 pg. 650 Write each epression in radical notation. Then evaluate the epression and round to the nearest hundredth when appropriate. 1/ 1/5 a. 49 b. c. 6 6 1/ Solution 1/ a. b. 49 49 7 1/5 6 c. 1/ (6 ) 6
Write each epression in radical notation. Evaluate the epression by hand when possible. a. /3 3/4 8 b. 10 Solution a. 8 /3 3 8 4 b. 3/4 10 3 4 10 4 1000
Try Q 47,51 pg. 650 Write each epression in radical notation. Evaluate the epression by hand when possible. 3/4 4/5 a. 81 b. 14 Solution 3/4 a. 81 4/5 b. 14 3/4 (81) Take the fourth root of 81 and then cube it. Take the fifth root of 14 and then fourth it. 3 3 3 4 81 7 4/5 14 4 5 14 Cannot be evaluated by hand.
Try Q 53,55 pg. 650 Write each epression in radical notation and then evaluate. 1/4 a. 81 b. 64 /3 Solution 1/4 1 a. b. 64 81 /3 1/4 81 4 1 3 1 81 1 64 /3 1 3 64 1 4 1 16
Try Q 53,55 pg. 650 Use rational eponents to write each radical epression. 3/7 a. 7 3 b. 1 b 3 3/ b c. 5 ( 1) ( 1) /5 d. 4 a b ( a b ) 1/4
Write each epression using rational eponents and simplify. Write the answer with a positive eponent. Assume that all variables are positive numbers. 4 a. 1/ 1/4 b. 4 56 3 3 1/4 (56 ) 1/ 1/4 3/4 1/4 3 1/4 56 ( ) 3/4 4
(cont) Try Q 77,83,91,97 pg. 650 Write each epression using rational eponents and simplify. Write the answer with a positive eponent. Assume that all variables are positive numbers. 5 3 c. 4 1/4 d. (3 ) 1/5 3 1/4 1/5 1/5 1/5 1/4 1/0 1/0 3 7 1/3 7 3 7 ( ) 3 1/3 1/3 3 1/3
Try Q 85,89,95 pg. 650 Write each epression with positive rational eponents and simplify, if 1/4 possible. y 4 a. b. 1/5 Solution a. 4 1/ ( ) 1/4 ( ) 1/8 b. y 1/4 1/5 y 1/5 1/4
Section 10.3 Simplifying Radical Epressions Copyright 013, 009, and 005 Pearson Education, Inc.
Objectives Product Rule for Radical Epressions Quotient Rule for Radical Epressions
Product Rule for Radicals Consider the following eample: 4 5 5 10 45 100 10 Note: the product rule only works when the radicals have the same inde.
Try Q 13,15,1 pg. 659 Multiply each radical epression. a. 36 4 364 144 1 b. 3 3 8 7 3 3 8 7 16 6 c. 1 1 1 4 16 4 4 4 4 1 1 1 1 1 4 16 4 56 4 4 4
Try Q 3,51,57,61 pg. 659-60 Multiply each radical epression. a. 4 4 6 3 b. 3 5a 10a 3 3 3 3 3 5a 10a 50a a 50 c. 4 3 7y 4 y 3 7y 1y y y 4 4 4 1
Try Q 73,75,77,79 pg. 660 Simplify each epression. a. 500 100 5 10 5 b. 3 40 3 3 3 8 5 5 c. 7 36 6
Try Q 45,85,89,91 pg. 660 Simplify each epression. Assume that all variables are positive. 5 a. 4 4 b. 75y 4 5y 3y 49 49 7 5y 4 3y c. 3 3 3a 9a w 3a 9a w 3 5y 3y 3 3 7a w 3 7a 3 3 w 3 3a w
Try Q 101,103,107 pg. 660 Simplify each epression. 3 a. 1/ 1/3 b. 7 7 7 7 1/ 1/3 7 5/6 7 a 3 5 a a 1/3 1/5 a 8/15 a a 1/3 1/5
Quotient Rule Consider the following eamples of dividing radical epressions: 4 9 3 3 3 4 4 9 9 3
Try Q 5,7,9 pg. 659 Simplify each radical epression. Assume that all variables are positive. 7 7 3 a. 3 7 b. 3 7 5 3 5 5 3 3 7 3 5
Try Q 33,39,41 pg. 659 Simplify each radical epression. Assume that all variables are positive. 90 a. b. 10 90 10 4 y y 4 y y 9 4 3
Try Q 95,97 pg. 660 Simplify the radical epression. Assume that all variables are positive. 5 3 5 y 4 3 5 4 5 y 5 5 3 5 y 5 4 5 5 4 y
Simplify the epression. 1 1 Solution 1 1 ( 1)( 1) 1
Try Q 63,67 pg. 660 Simplify the epression. Solution 3 3 56 3 3 56 ( 3)( ) 3 ( ) 3 3
End of week 3 You again have the answers to those problems not assigned Practice is SOOO important in this course. Work as much as you can with MyMathLab, the materials in the tet, and on my Webpage. Do everything you can scrape time up for, first the hardest topics then the easiest. You are building a skill like typing, skiing, playing a game, solving puzzles.