CHAPTER Equations, Inequalities, and Mathematical Modeling Section. Graphs of Equations.................... 9 Section. Linear Equations in One Variable............. Section. Modeling with Linear Equations.............. Section. Quadratic Equations and Applications........... 8 Section. Comple Numbers..................... 67 Section.6 Other Types of Equations................. 7 Section.7 Linear Inequalities in One Variable............ 8 Section.8 Other Types of Inequalities................ 87 Review Eercises............................. 97 Problem Solving............................. 0 Practice Test.............................. 07
CHAPTER Equations, Inequalities, and Mathematical Modeling Section. Graphs of Equations You should be able to use the point-plotting method of graphing. You should be able to find - and y-intercepts. (a) To find the -intercepts, let y 0 and solve for. (b) To find the y-intercepts, let 0 and solve for y. You should be able to test for symmetry. (a) To test for -ais symmetry, replace y with y. (b) To test for y-ais symmetry, replace with. (c) To test for origin symmetry, replace with and y with y. You should know the standard equation of a circle with center h, k and radius r: h y k r Vocabulary Check. solution or solution point. graph. intercepts. y-ais. circle; h, k; r 6. point-plotting. y. (a) 0, :? 0 Yes, the point is on the graph. (b), :? 9 Yes, the point is on the graph. y (a), :? (b) No, the point is not on the graph. 6, 0: 0? 6 0 Yes, the point is on the graph.. y y 0 y 7 0, y, 7 0,,,, 0 7 9
0 Chapter Equations, Inequalities, and Mathematical Modeling 7. y 0 y 0 0 y, y, 0, 0,,, 0 9. y 6. -intercepts: y-intercept: 0 6 6 ±, 0,, 0 y 6 0 6 0, 6 y -intercepts: 0 0 0 or 0, 0,, 0 y-intercept: y 0 0 y 0 0, 0. y-ais symmetry. Origin symmetry y y 7. y 0 9. y 0 y 0 y-ais symmetry y 0 y 0 No -ais symmetry y 0 y 0 No origin symmetry y y y No y-ais symmetry y y No -ais symmetry y y y Origin symmetry. y y y y y y y No y-ais symmetry No -ais symmetry y Origin symmetry. y 0 0 y 0 0 y 0 0 No y-ais symmetry y 0 0 y 0 0 -ais symmetry y 0 0 y 0 0 No origin symmetry
( Section. Graphs of Equations. y y 7., 0 -intercept: y-intercept: 0, No ais or origin symmetry ( (0, ), 0 ( y Intercepts: 0, 0,, 0 No ais or origin symmetry 0 y 0 0 y (0, 0) (, 0) 9. y Intercepts: 0,,, 0 No ais or origin symmetry 0 y (, 0 7 6 y (0, ). y Domain:, Intercept:, 0 No ais or origin symmetry 7 y 0 y (, 0) 6. y 6 Intercepts: 0, 6, 6, 0 No ais or origin symmetry 0 6 8 0 y 0 8 6 (0, 6) y 8 6 0 (6, 0) 6 8 0. y 7. y Intercepts: 0,, 0,,, 0 0 -ais symmetry y 0 0 0 y 0 ± ± (, 0) (0, ) (0, ) 0 Intercepts: 6, 0, 0, 9. y. 0 y 0. y 0 0 0 0 0 0 0 0 Intercepts:, 0,, 0, 0, 0 Intercept: 0, 0 0 Intercept: 0, 0
Chapter Equations, Inequalities, and Mathematical Modeling. y 6 7. 0 y 9. Center: 0, 0; radius: 0 Standard form: 0 0 0 0 0 y 0 y 6 0 Intercepts: 0, 0, 6, 0 0 Intercepts:, 0, 0,. Center:, ; radius: Standard form: y y 6. Center:, ; solution point: 0, 0 y r 0 0 r r Standard form: y. Endpoints of a diameter: Center: 0 6, 0 8, y r 0 0 r r Standard form: y 0, 0, 6, 8 7. y Center: 0, 0, radius: 6 6 y (0, 0) 6 9. y 9 6. Center:,, radius: y (, ) 6 7 y 9 6. Center: y,, (, ) radius: y,000 0,000t, 0 t 8 Depreciated value 0,000 00,000 0,000 00,000 0,000 y 6 7 8 Year t 6. (a) (c) y 8000 0 80 0 (b) y 00 y 00 y 0 A y 0 (d) When y 86 yards, the area is a maimum of 7 9 square yards. (e) A regulation NFL playing field is 0 yards long and yards wide. The actual area is 600 square yards.
Section. Linear Equations in One Variable 67. y 0.00t 0.7t., 0 t 00 (a) and (b) Life epectancy y 00 80 60 0 0 t 0 0 60 80 00 Year (0 90) (c) For the year 98, let t 8: y 66.0 years. (d) For the year 00, let t 0: y 77.0 years. For the year 00, let t 0: y 77. years. (e) No. The graph reaches a maimum of y 77. years when t.8, or during the year 0. After this time, the model has life epectancy decreasing, which is not realistic. 69. False. A graph is symmetric with respect to the -ais if, whenever, y is on the graph,, y is also on the graph. 7. The viewing window is incorrect. Change the viewing window. Eamples will vary. For eample, y 0 will not appear in the standard window setting. 7. 9 7 Terms: 9,, 7 7. 8 77. 6 t t 6 t t Section. Linear Equations in One Variable You should know how to solve linear equations. a b 0, a 0 An identity is an equation whose solution consists of every real number in its domain. To solve an equation you can: (a) Add or subtract the same quantity from both sides. (b) Multiply or divide both sides by the same nonzero quantity. (c) Remove all symbols of grouping and all fractions. (d) Combine like terms. (e) Interchange the two sides. Check the answer! A solution that does not satisfy the original equation is called an etraneous solution. Be able to find intercepts algebraically. Vocabulary Check. equation. solve. identities; conditional. a b 0. etraneous
Chapter Equations, Inequalities, and Mathematical Modeling. (a) 0? 0 (b)? 8 0 0 is not a solution. is not a solution. (c)? (d) 0? 0 7 7 7 is a solution. 0 is not a solution.. (b)? (a)?? 7 6? 8 0 0 6 6 is a solution. is a solution. (d)? (c)? 7 0? 0 8 8? 60 8 0 is not a solution. is not a solution.. (a)? 8? is a solution. (c) is undefined. 0 0 0 is not a solution. (b) (d)? 8? 8 is not a solution.? 0 6? 6 is not a solution. 7. (a)? (b)? (c) 9? (d) 6? 7 0 is not a solution. is not a solution. 9 is not a solution. 6is not a solution.
Section. Linear Equations in One Variable 9. 6 0 (a) 6? 0 (b) 6 7 7? 0 0 0? 0 0 0 7 9 9 9? 0 7 9 0 is a solution. 7 is not a solution. (c) 6 7 7? 0 7 77 70? 0 7 is a solution. 0 0 (d) 6? 0 0 0? 0 0 0 is not a solution.. is an identity by the Distributive. 6 0 is conditional. There are Property. It is true for all real values of. real values of for which the equation is not true.. 7. This is an identity by simplification. It is true for all real values of. 8 6 8 Thus, 8 is an identity by simplification. It is true for all real values of. 9. is conditional. There are real values of for which the equation is not true.. 8 Original equation 8 Subtract from both sides. Simplify. Divide both sides by. Simplify... 7 7. 8 0 7 7 7 8 0 8 0 8 9 0 9. 7. 0 7 6 6 6 6 6 9 8. 6 9 8 9 8 9 8 No solution 9 8 9
6 Chapter Equations, Inequalities, and Mathematical Modeling. z z 0 z z 0 z z 0 6z z 0 6z 0 z 0 z 6 7. 0. 0.70 0. 7. 0.7 0.0 7. 0.0. 9 z 6 9. or. or 7 7 7 The second way is easier since you are not working with fractions until the end of the solution. The first way is easier. The fraction is eliminated in the first step.. 8. 8 8 8 Contradiction; no solution 00 00 6 6 00 6 7 00 6 8 7 0 0 6 6 7. 9. 0 0 8 0 0 0 6 8. z z z z z 6 z z 0. 0 0 0 0 Contradiction; no solution
Section. Linear Equations in One Variable 7. Multiply both sides by. A check reveals that 8 0 is an etraneous solution it makes the denominator zero. There is no real solution. 7. 0 9 0 0 0 Multiply both sides by. 9. 9 A check reveals that Multiply both sides by. is an etraneous solution since it makes the denominator zero, so there is no solution. 6. 6. 6 9 9 6 9 0 0 The equation is an identity; every real number is a solution. 6. y 0 67. 6 8 0 0 6 6 The -intercept is at. The solution to 0 and the -intercept of y are the same. They are both. The -intercept is, 0. y 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The -intercept is at 0. The solution to 0 0 0 and the -intercept of y 0 0 are the same. They are both 0. The -intercept is 0, 0.
8 Chapter Equations, Inequalities, and Mathematical Modeling 69. y 8 9 0 8 9 7. y 0 0 8 -intercept: 0 6 0 0 7 7 y-intercept: The -intercept is y 0 y, 0 and the y-intercept is 0,. 7 7 The -intercept is at. The solution to 0 8 9 and the -intercept of y 8 9 are the same. They are both 7 The -intercept is 7., 0. 7. y 7. -intercept: 0 0 y-intercept: y 0 y The -intercept is, 0 and the y-intercept is 0,. y 0 -intercept: 0 0 0 y-intercept: 0 y 0 y 0 y 0 The -intercept is, 0 and the y-intercept is 0, 0. 77. 8 y 0 0 y 0 Multiply both sides by. -intercept: y-intercept: 0 0 0 0 0 0 0 0 y 0 0 y 0 y 0 8 The -intercept is 0, 0 and the y-intercept is 0, 8. 79. y 0.7. 0 -intercept: 0 0.7. 0 0.7. 0..6 0.7 y-intercept: y 0.70. 0 y. 0 y. 0. The -intercept is.6, 0 and the y-intercept is 0, 0.. 8. a 8. a a a a, a 6 a a a a, a 8. 9 a 9 87. 8 a 9 6 a 8 6 a 8 8 6 a, a 6 Multiply both sides by. a 6 a 6 8 a 0 8 a 0 7 a 0 7 7 a 0 7 0 a, a
Section. Linear Equations in One Variable 9 89. 0.7 0.700 00 0.7 6. 0.7 00 0. 6. 6. 0. 8.889.0 9. Multiply both sides by 7.98. 7.98.07.98 7.98.07.98 7.98.07.98.07.98 9.99 9. 7 h 7 0 0h 7 0 0h 7 0 h 0 h 0 feet 7 0(.) 0(.) 0 9. (a) Female: y 0. 0. (c) For y 6: 6 0. 0. 6. 0. 6. 0. 6. inches Height Female Femur Male Femur Length Length 60.8.79 70 9.80 9.8 80..77 90 8. 8.6 00.76.7 0 7.08 7. (b) Male: (d) y 0.9. For y 9: 9 0.9.. 0.9 69. Yes, it is likely that both bones came from the same person because the estimated height of a male with a 9-inch thigh bone is 69. inches. 0. 0. 0.9..7 0.07 00.9 inches It is unlikely that a female would be over 8 feet tall, so if a femur of this length was found, it most likely belonged to a very tall male. The lengths of the male and female femurs are approimately equal when the lengths are 00 inches. 97. y.6t 6.8, t (a) Consumption of gas (in billions of gallons) The 60 6 8 0 y 0 6 8 0 Year (0 990) y-intercept is 0, 6.8. t (b) Let t 0: y.60 6.8 6.8 (c) y-intercept: 0, 6.8 6.6t 6.8 8..6t t 8. 7..6 This corresponds with the year 07. Eplanations will vary.
0 Chapter Equations, Inequalities, and Mathematical Modeling 99. 0,000 0.m 00 0. False. 700 0.m 700 0. m 0 0 This is a quadratic equation. The equation cannot be written in the form a b 0. m,7. miles 0. Equivalent equations are derived from the substitution principle and simplification techniques. They have the same solution(s). 8 and are equivalent equations. 0. (a) 0..8 9.8.6 0.6.8 7 (b) Since the sign changes from negative at to positive at, the root is somewhere between and. < < (c)..6.7.8.9..8 0.68 0.6 0.0 0.8 0.6 (d) Since the sign changes from negative at.8 to positive at.9, the root is somewhere between.8 and.9..8 < <.9 To improve accuracy, evaluate the epression at subintervals within this interval and determine where the sign changes. 07. 6 9 7 9 9, 9 09. y. y y y 7 6 7 6 Intercepts: 0,,, 0 Intercepts: 0, 0,, 0
Section. Modeling with Linear Equations Section. Modeling with Linear Equations You should be able to set up mathematical models to solve problems. You should be able to translate key words and phrases. (a) Equality: Equals, equal to, is, are, was, will be, represents (c) Subtraction: Difference, minus, less than, decreased by, subtracted from, reduced by, the remainder (e) Division: Quotient, divided by, ratio, per You should know the following formulas: (b) Addition: Sum, plus, greater, increased by, more than, eceeds, total of (d) Multiplication: Product, multiplied by, twice, times, percent of (f) Consecutive: Net, subsequent (a) Perimeter: (b) Area:. Square: P s. Square: A s. Rectangle: P l w. Rectangle: A lw. Circle: C r. Circle: A r. Triangle: P a b c. Triangle: (c) Volume. Cube: V s. Rectangular solid: V lwh. Cylinder: V r h. Sphere: V r (d) Simple Interest: I Prt A bh (e) Compound Interest: A P r n nt (f) Distance: d rt (g) Temperature: F 9 C You should be able to solve word problems. Study the eamples in the tet carefully. Vocabulary Check. mathematical modeling. verbal model;. A r. P l w algebraic equation. V s 6. V r h 7. A P r 8. I Prt t.. The sum of a number and A number increased by u The ratio of a number and The quotient of a number and A number divided by. y The difference of a number and is divided by. A number decreased by is divided by.
Chapter Equations, Inequalities, and Mathematical Modeling 7. b 9. The product of and the sum of a number and. Negative is multiplied by a number increased by. The difference of a number and is multiplied by times the number. is multiplied by a number and that product is multiplied by the number decreased by.. Verbal Model: (Sum) (first number) (second number) Labels: Sum S, first number n, second number n Epression: S n n n. Verbal Model: Product (first odd integer)(second odd integer) Labels: Product P, first odd integer n, second odd integer n n Epression: P n n n. Verbal Model: (Distance) (rate) (time) Labels: Distance d, rate 0 mph, time t Epression: d 0t 7. Verbal Model: (Amount of acid) 0% (amount of solution) Labels: Amount of acid (in gallons) A, amount of solution (in gallons) Epression: A 0.0 9. Verbal Model: Perimeter width length Labels: Perimeter P, width, length (width) Epression: P 6. Verbal Model: (Total cost) (unit cost)(number of units) (fied cost) Labels: Total cost C, fied cost $00, unit cost $, number of units Epression: C 00. Verbal Model: Thirty percent of the list price L Epression: 0.0L. Verbal Model: percent of 00 that is represented by the number N Equation: N p00, p is in decimal form 7. 8 Area Area of top rectangle Area of bottom rectangle A 8
Section. Modeling with Linear Equations 9. Verbal Model: Sum (first number) (second number) Labels: Sum, first number n, second number n Equations: n n n n n 6 Answer: First number n 6, second number n 6. Verbal Model: Difference (one number) (another number) Labels: Difference 8, one number, another number Equation: 8 8 7 8 Answer: The two numbers are 7 and 8.. Verbal Model: Product (smaller number) (larger number) smaller number) Labels: Smaller number n, larger number n Equation: nn n Answer: Smaller number n n n n n, larger number n. percent number 7. 0% 0.0. percent 600 9. p600 600 p p 0.7 7% % number 0.00 0.00 00. Verbal Model: Loan payments 8.6% Annual income Labels: Loan payments,077.7 Annual income I Equation:,077.7 0.86I I,6.98 The family s annual income is $,6.98.
Chapter Equations, Inequalities, and Mathematical Modeling. Verbal Model: 00 price for a gallon of unleaded gasoline percentage increase990 price for a gallon of unleaded gasoline 990 price for a gallon of unleaded gasoline Labels: 00 price for a gallon of unleaded gasoline: $.6 990 price for a gallon of unleaded gasoline: $.6 Percentage increase: p Equation:.6.6p.6 0.0.6p 0.7 p Answer: Percentage increase 7.%. Verbal Model: 00 price for a pound of tomatoes percentage increase990 price for a pound of tomatoes 990 price for a pound of tomatoes Labels: 00 price for a pound of tomatoes: $.66 990 price for a pound of tomatoes: $0.86 Percentage increase: p Equation:.66 0.86p 0.86 0.80 0.86p 0.90 p Answer: Percentage increase 9% 7. Verbal Model: (Sale price) (list price) (discount) Labels: Sale price $0.7, list price L, discount 0.6L Equation: 0.7 L 0.6L 0.7 0.8L 0 L Answer: The list price of the pool is $0. 9. (a) (b) l.w (c) w w P l w w l.w w w Width: w meters Length: l.w 7. meters Dimensions: 7. meters meters. Verbal Model: Average Labels: Equation: Average 90 60 6 97 test # test # test # test # 90, test # 87, test # 9, test # 8, test # 87 9 8 Answer: You must score 97 (or better) on test # to earn an A for the course.
Section. Modeling with Linear Equations. Rate distance time Total time 0 kilometers hour 00 kilometershour total distance rate 00 kilometers hours 00 kilometershour. time on first part time on second part) Total time t t T d r d r T 7.7 8 Multiply both sides by 8. 87.78 8,86 8 7, 6 0 7 miles t 7 8 t hours 7 7.7 hours The salesman averaged 8 miles per hour for hours and miles per hour for hours and minutes. 7. (a) Time for the first family: Time for the other family: (b) t d r 00 00.08 hours 0 9 (c) d rt 60 t d r 60 t d r 60 0 60 0.6 miles.8 hours. hours 9. Verbal Model: time distance rate Labels: Let wind speed, then the rate to the city 600, the rate from the city 600, the distance to the city 00 kilometers, the distance traveled so far in the return trip 00 00 00 kilometers. Equation: 00 00 600 600 00600 00600 900,000 00 70,000 00 80,000 700 6. Verbal Model: Equation: time distance rate t.8 0 8 meters.0 0 8 meters per second t.8 seconds The radio wave travels from Mission Control to the moon in.8 seconds. 66 Wind speed: 66 kilometers per hour
6 Chapter Equations, Inequalities, and Mathematical Modeling 6. Verbal Model: height of building length of building's shadow height of stake length of stake's shadow Label: Let h height of the building in feet. Equation: h feet feet 87 feet foot h 87 h 8 h 0 feet The Chrysler building is 0 feet tall. 6. Verbal Model: height of silo length of silo s shadow height of person height of person s shadow Label: Let length of person s shadow. Equation: 0 6 0 6 0 9 6 9.6 feet 0 ft 6 ft ft 67. Verbal Model: Interest in % fund interest in % fund (total interest) Labels: Let amount in the % fund. Then,000 amount in the % fund. Equation: 000 0.0 0.0,000 000 0.0 0.0 0.0 $8. at %,000 $6,666.67 at % 69. Verbal Model: (profit on minivans) (profit on SUVs) (total profit) Labels: Let amount invested in minivans. Then, 600,000 amount invested in SUVs. Equation: 0. 0.8600,000 0.600,000 0. 68,000 0.8 0,000 0.0 8,000 0,000 $0,000 is invested in minivans and 600,000 $0,000 is invested in SUVs. 7. Verbal Model: Final concentrationamount Solution concentrationamount Solution concentrationamount Label: Let amount of 00% concentrate Equation: 0.7 0.0.00. 0.0.00. 0.60. gallons Approimately. gallons of the 00% concentrate will be needed.
Section. Modeling with Linear Equations 7 7. Verbal Model: (price per pound of peanuts)(number of pounds of peanuts) (price per pound of walnuts)(number of pounds of walnuts) (price per pound of nut miture)(number of pounds of nut miture) Labels: Let number of pounds of $.9 peanuts. Then 00 number of pounds of $.89 walnuts. Equation:.9.8900.900.9 89.89 9.0 70 70.0 00 0 pounds of $.89 walnuts Use 0 pounds of peanuts and 0 pounds of walnuts. 0 pounds of $.9 peanuts 7. Verbal Model: Total cost Fied cost Variable cost per unit Number of units Labels: Equation: Total cost $8,000, variable costs $9.0, fied costs $0,000 $8,000 $0,000 $9.0 7,000 9. 806. units At most the company can manufacture 806 units. 77. A bh 79. A bh A b h S C RC 8. S C R S R C V a b8. V a b V a b V a b h v 0 t at h v 0 t at h v 0 t t h v 0 t at a 8. C 87. L a n d 89. C C L a nd d C C C C C C C C CC C L a d nd L a d d n W W L 0 70 0 70 7 70 6 feet from 0-pound child. CC C C C
8 Chapter Equations, Inequalities, and Mathematical Modeling 9. V r.96 r 7.88 r 7.88 r r.7. inches 9. C 9F 9. F 9 C When F 6., When C 0, 9 96. 8C. 0 F. 97. False, it should be written as z 8 z 9. 99. a b 0 b a (a) If ab > 0, then a and b have the same sign and ba is negative. (b) If ab < 0, then a and b have opposite signs and ba is positive. 0., 0 0. 0. 0 0 0 7 7 07. 6 6 6 6 6 6 6 Section. Quadratic Equations and Applications You should be able to solve a quadratic equation by factoring, if possible. You should be able to solve a quadratic equation of the form u d by etracting square roots. You should be able to solve a quadratic equation by completing the square. You should know and be able to use the Quadratic Formula: For a b c 0, a 0, b ± b ac. a You should be able to determine the types of solutions of a quadratic equation by checking the discriminant b ac. (a) If b ac > 0, there are two distinct real solutions. The graph has two -intercepts. (b) If b ac 0, there is one repeated real solution. The graph has one -intercept. (c) If b ac < 0, there is no real solution. The graph has no -intercepts. You should be able to use your calculator to solve quadratic equations. You should be able to solve applications involving quadratic equations. Study the eamples in the tet carefully.
Section. Quadratic Equations and Applications 9 Vocabulary Check. quadratic equation. factoring, etracting square roots, completing the square, and the Quadratic Formula. discriminant. position equation; 6t v 0 t s 0 ; velocity of the object; initial height of the object. Pythagorean Theorem. 8. General form: 8 0. 6 9 General form: 6 6 0 0 8 0 90 General form: 90 0 0 7. 6 0 9. 8 0. 0 0 0 0 0 0 or 0 0 or 0 0 0 or or. 0. 7. 8 0 0 0 0 8 0 0 0 or 0 6 0 80 0 or 6 0 or 0 0 0 6 or 0 0 or 0 0 or 9. a a 0. 9. a 0 ±7 ± a 0 a. 8 7. 6 9. 7 ± ± ± ± ± 6 or 8. 8. 7 ±8 ± ± 7 ± 7 or 7 7 or The only solution to the equation is.
60 Chapter Equations, Inequalities, and Mathematical Modeling. 0 7. 6 ±6 ± 6 or 8 0 6 6 6 6 ± 6 ± 9. 9 8. 8 0. 8 0 ± ± ± 6 8 0 8 0 8 8 ± ± 8 89 6 ±89 ± 89 ± 89. 7. 7 9. 6 6 9 9 9. (a) y 0 6 (b) The -intercepts are, 0 and, 0. (c) 0 ± ± or (d) The -intercepts of the graph are solutions to the equation 0.
Section. Quadratic Equations and Applications 6. (a) y (c) 0 6 ± ± (b) The -intercepts are, 0 and, 0. or (d) The -intercepts of the graph are solutions to the equation 0.. (a) y (c) 0 7 (b) The -intercepts are and, 0. (d) The -intercepts of the graph are solutions to the equation 0., 0 ± ± or 7. (a) y (c) 0 0 8 0 or 0 or 7 (b) The -intercepts are, 0 and, 0. (d) The -intercepts of the graph are solutions to the equation 0. 9. 0 6. b ac < 0 No real solution 0 b ac 9 > 0 Two real solutions 6. 0 6. b ac < 0 0.. 8 0 b ac. 0.8 7.8 > 0 No real solution Two real solutions 67. 0 69. b ± b ac a ± ±, 6 8 0 7. 0 b ± b ac a 8 ± 8 6 6 8 ± 6, 0 b ± b ac a ± ± ±
6 Chapter Equations, Inequalities, and Mathematical Modeling 7. 0 7. 8 0 77. 9 b ± b ac a ± ) ± 7 ± b ± b ac a 8 ± 8 8 ± ± 9 0 b ± b ac a ± 9 9 ± 67 8 ± 7 79. 9 6 0 8. 7 8. 8 9 b ± b ac a ± 96 9 ± 0 8 7 0 b ± b ac a ± 7 ± 8 8 ± 9 8 0 b ± b ac a 8 ± 8 9 9 8 ± 0 98 7 8. t 8t 0 t b ± b ac a 8 ± 8 8 ± 6 8t t 87. ± 6 y y 0 y b ± b ac a ± y y ± 6 ± 89. 8 6 6 0 b ± b ac a ± 6 ± 6 8 8 ± 6 8 9...7. 0 9..7 ±.7... 0.976, 0.6 0.067 0.8.77 0 0.8 ± 0.8 0.067.77 0.067.07,. 9. 06 7 0 97..67. 8.09 0 06 ± 06 7.687, 0.88. ±..678.09.67.00, 0.90
Section. Quadratic Equations and Applications 6 99. 0 Complete the square. 0. ± ± 8 Etract square roots. ±9 9 or 9 6 or 0. 0 Complete the square. 0. Etract square roots. ± ± ± F or : 0 No solution F or : 07. 7 Quadratic Formula 0 ± ± 97 ± 97 09. (a) (b) w w + ww 6 (c) w w 6 0 w 8w 0 w 8 or w Since w must be greater than zero, we have w feet and the length is w 8 feet.. S h 8 0 8 8 0 6 or 6 Since must be positive, we have 6 inches. The dimensions of the bo are 6 inches 6 inches inches.. 00 00 0000 0,000 600 0,000 600 0,000 0 00 ft 00 ft 00 0 00 0 Thus, a, b 0, and c 00. 00 CONTINUED
6 Chapter Equations, Inequalities, and Mathematical Modeling. CONTINUED 0 ± 0 00 0.80 0.80 0.90 feet, 9.098 feet The person must go around the lot not possible since the lot is only 00 feet wide. 9.098 feet inches 0 ±.80 9.098 feet feet 9. times.. s 6t,000 7. (a) 6t,000 0 (b) Model: Labels: Rate 00 miles per hour Equation: t 000 t 000 0 seconds.7 seconds Rate time distance Time 0 600 Distance d 0.0 hour d 000.0 6. miles The bomb will travel approimately 6. miles horizontally. s 6t v 0 t s 0 (a) v 0 00 mph 0080 600 s 0 6 feet s 6t 6 t 6 (b) When t : s 0. feet When t : s 6.9 feet When t : s 9.8 feet 6 ftsec During the interval t, the baseball s speed decreased due to gravity. (c) The ball hits the ground when s 0. 6t 6 t 6 0 By the Quadratic Formula, t 0.0 or t 9.09. Assuming that the ball is not caught and drops to the ground, it will be in the air for approimately 9.09 seconds. 9. P 0.008t 0.7t.99, 7 t (a) t 7 8 9 0 P $. $.8 $.09 $. $.60 $.8 $6.0 (b) The average admission price reached or surpassed $.00 in 999. 0.008t 0.7t.99.00 0.008t 0.7t.0 0 By the Quadratic Formula, t 8.68 or.80. Since 7 t, we choose t 8.68 9 which corresponds to 999. (c) For 008, let t 8: P8 $6.87. Answers will vary.
Section. Quadratic Equations and Applications 6. Pythagorean Theorem. centimeters Each leg in the right triangle is approimately. centimeters.. d N hoursr 0 mph d E hoursr mph d N d E 0 9r 0 9r 0 8r 900r,9,00 0 r 900 ± 900 8,9,00 8 Using the positive value for r, we have one plane moving northbound at r 0 600 miles per hour and one plane moving eastbound at r 0 miles per hour. 900 ± 608,87 6 (r + 0) r 0 miles. 0 0.000 00,000 7. 0 0.000 0 00,000 0 00,000,00,000,000 0 0,000 0,000 units Using the positive value for, we have 0. 0 00,000 0. 0,00 0 0 ± 0 0.,00 0. 0 70 0. 8 units. 9. 800 0.0 0.00 680 0.00 0.0 880 0 0.0 ± 0.0 0.00880 0.00 Choosing the positive value for, we have 0.0 ± 7.06 0.00 0.0 7.06 0.00 6 units.
66 Chapter Equations, Inequalities, and Mathematical Modeling. M.8t.8t.0, t (a) t 6 7 8 9 0 M (in billions) $96.78 $0. $7.98 $79.08 $.86 $.0 $9. $60.0 $689.66 The total money in circulation reached or surpassed $600 billion in 00. (b).8t.8t.0 600.8t.8t 67.0 0 By the Quadratic Formula, t. or.. Since t, we choose t. which corresponds to 00. (c) For 008, let t 8: M8 $99.98 billion Answers will vary.. Distance from Oklahoma City to New Orleans Distance from Oklahoma City to Austin Distance from Distance from Oklahoma City to New Orleans 60 Distance from Oklahoma City to Austin Distance from New Orleans to Austin 8 60 788 New Orleans to Austin 60 (788,600 60,9 76 0 76 07, 76 ± 76 07,. or. The other two distances are. miles and. miles.. False b ac 0 < 0, so, the quadratic equation has no real solutions. 7. The student should have subtracted from both sides so that the equation is equal to zero. By factoring out an, there are two solutions, 0 and 6. 9. 0 (a) Let u u u 0 u u 0 u 0 u 0 or u 0 u or (b) 8 6 0 8 0 0 0 0 0 0 0 0 or 0 (c) The method of part (a) reduces the number of algebraic steps.. and 6. 8 and One possible equation is: One possible equation is: 6 0 6 0 8 0 Any non-zero multiple of this equation would also have these solutions. 8 0 0 Any non-zero multiple of this equation would also have these solutions.
Section. Comple Numbers 67. and 7. 0y 0y by the Associative Property of One possible equation is: Multiplication. 0 0 0 0 0 Any non-zero multiple of this equation would also have these solutions. 9. 7 7 0 by the Additive Inverse Property.. 6 6 8 8.. 8 8 7 7 9 7. 0 9. Answers will vary. Section. Comple Numbers Standard form: a bi. If b 0, then a bi is a real number. If a 0 and b 0, then a bi is a pure imaginary number. Equality of Comple Numbers: a bi c di if and only if a c and b d Operations on comple numbers (a) Addition: (b) Subtraction: (c) Multiplication: (d) Division: The comple conjugate of a bi is a bi: a bia bi a b The additive inverse of a bi is a bi. a a i for a > 0. a bi c di a c b di a bi c di a c b di a bic di ac bd ad bci a bi a bi c di c di c di c di ac bd bc ad c d c d i
68 Chapter Equations, Inequalities, and Mathematical Modeling Vocabulary Check. (a) iii (b) i (c) ii. ;. principal square. comple conjugates. a bi 0 6i. a b i 8i. 9 i a 0 a a 6 b 6 b 8 b 7. 7 7 i 9. 7 7 i i. 8 8 0i 8 i. 6i i 6i. 0.09 0.09 i 7. i 6 i i 6i 0.i 9. 8 i i 8 i i. 8 0 i i i. i 7i i 7i. 0i i i i i 9 6 6 i 0 6 6 i 6 7 6 i 7. i i i i i 9. i i 6i i 0i i 0i 0i. 0 i 0 i 0i. i 6 0i i 0 6 0i 9 0i. i i i 9i i 9i 7. The comple conjugate of 6 i is 6 i. i 9 i 9 6 i6 i 6 i 6 9 0 9. The comple conjugate of i is i.. The comple conjugate of 0 i is i. i i i ii 0i 0 6. The comple conjugate of 8 is 8.. 88 8 i i i i i i
Section. Comple Numbers 69 7. i i i i i 6 8 0i 8 0 i 9. i i i i i i 9 6i i 9 8 6i 0 i. 6 i i 6 i i i i 6i i 6i. i i i 6 0i i i 9 0i 9 0i 9 0i 7i 0i 8 600 0 7 68 68 i 0 7i 68. i i i i i i i i i i 7. i i i 8i i i i 8i i 8i i 8i 6i i 9 i 6i 6i i 9i 9 8i 6 9i 8i 8i 8i 00 7i i 6i 6 00 97i 6 99 6 97i 99 6 97 99 99 i 9. 6 6ii i 6. 0 0i 0i 0 6. 7 0 i7 0i 0i 7i 0i 0 7 0 i 7 0 i 6. 0; a, b, c 67. 6 7 0; a, b 6, c 7 ± ± ± i ± i 6 ± 6 7 6 ± 6 8 6 ± i 8 ± i
70 Chapter Equations, Inequalities, and Mathematical Modeling 69. 6 0; a, b 6, c 7. Multiply both sides by. 6 9 0 6 ± 6 6 ± 6 8 8 6 ± 8 or 0 8 8 0 ± 8 ± 7 6 ± 6i 6 ± i 7.. 0 0 Multiply both sides by. 7. 7 0 0 0 0 ± 0 70 7 0 ± 00 ± 7 0 ± 0 7 ± 7 6i i 6i i i 6i 6i 6i 77. i i i i 79. i i 7 i 8. i i 7 i i i i i i i i i i 8. (a) (b) z 9 6i, z 0 0i z z z 9 6i 0 0i z 0 0i 9 6i 9 6i 0 0i 9 6i 9 6i0 0i,0 60i 877,0 877 9 6i 0 0i 60 877 i 8. (a) (b) 6 6 87. False, if b 0 then a bi a bi a. That is, if the comple number is real, the number equals (c) i i 6 6 its conjugate. (d) i i 6 6 89. False i i 0 i 7 i 09 i 6 i i 7 i i 8 i i 7 i i i 7 8 7 i i i i
Section.6 Other Types of Equations 7 9. a b ia b i a a a b i a b i b b i a a b b a b a b i The comple conjugate of this product is a a b b a b a b i. The product of the comple conjugates is: a b ia b i a a a b i a b i b b i a a b b a b a b i Thus, the comple conjugate of the product of two comple numbers is the product of their comple conjugates. 9. 8 6 9. 97. 9 99. 6 6 0 0. V a b 0 8 0 7 0 7 V a b V a b 7 V b a a V Vb b b 0. Let # liters withdrawn and replaced. 0.0.00 0.60.0 0.0.00.00 0.0 0.0 liter Section.6 Other Types of Equations You should be able to solve certain types of nonlinear or nonquadratic equations by rewriting them in a form in which you can factor, etract square roots, complete the square, or use the Quadratic Formula. For equations involving radicals or rational eponents, raise both sides to the same power. For equations that are of the quadratic type, au bu c 0, a 0, use either factoring, the Quadratic Formula, or completing the square. For equations with fractions, multiply both sides by the least common denominator to clear the fractions. For equations involving absolute value, remember that the epression inside the absolute value can be positive or negative. Always check for etraneous solutions. Vocabulary Check. polynomial. etraneous. quadratic type
7 Chapter Equations, Inequalities, and Mathematical Modeling. 8 0. 9 0 0 0 9 0 ± 8 0 9 0 9 0 ±i 0 0. 6 0 6 0 6 6 6 0 6 0 6 6 6 0 ± i 7. 0 0 9. 6 9 0 0 0 0 0 0 0 0 0 0 0 0. 0. 0 0 0 0 0 0 0 0 0 0 0 0 ± i 0. 6 6 0 6 0 0 0 0 0 0
Section.6 Other Types of Equations 7 7. 6 7 8 0 8 0 0 0 0 ± i 0 0 ± i 9. 8 0. 8 0 0 9 9 0 9 0 0 0 0 is not a solution.. 0 0 0 0 8 0. y (a) 9 9 7 (b) -intercepts:, 0, 0, 0,, 0 (c) (d) The 0 0 0 0 0 -intercepts of the graph are the same as the solutions to the equation. 7. y 0 9 (a) 0 0 (b) -intercepts: ±, 0, ±, 0 (c) 0 0 9 0 9 0 0 0 0 0 (d) The -intercepts of the graph are the same as the solutions to the equation.
7 Chapter Equations, Inequalities, and Mathematical Modeling 9. 0 0. 0 0. 0 0 0 00 0 6 7 0 6 6. 6 7. 6 6 8 6 0 0 0 0 0 0 0 9.. 9 0 0 0 00 0 90 0 9 8 8 0 0 0.. 8 8 9 8 6 8 6 8 6 8 0 0 0, etraneous 0 6
Section.6 Other Types of Equations 7 7. 8 9. 8 ±8 ± ± 6 7 7 7 9 ±. 0 0 0 0 0 0, etraneous. y 0. y 7 6 6 (a) (a) 0. (b) -intercepts:, 0, 6, 0 (c) 6 0 0 6 0 0 0 0 0 0 6 0 6 (d) The -intercepts of the graph are the same as the solutions to the equation. (b) -intercepts: 0, 0,, 0 (c) 0. 7 6 6 7 6 6 7 6 6 7 6 6 6 7 6 0 6 6 6 8 6 8 6 6 6 6 6 6 0 6 0 0 0 7 6 6 0 0 (d) The -intercepts of the graph are the same as the solutions to the equation.
76 Chapter Equations, Inequalities, and Mathematical Modeling 7. 9. 6 6 0 0 0 a, b, c 0 0 ± ± 6 6. 0 0 6. 0 0 0 0 0 a, b, c 0 ± 0 ± 6 ± 6 0 0 ± 6. 67. First equation: 0 ± Second equation: 0 0 0 0 Only and are solutions to the original equation. and are etraneous.
Section.6 Other Types of Equations 77 69. First equation: 6 0 0 0 0 Second equation: 0 7 7 Only and are solutions to the original equation. and are etraneous. ± 7 7. y (a) 7. y (a) 8 0 8 (b) -intercept:, 0 (c) 0 0 0 0 0 0 0 (d) The -intercept of the graph is the same as the solution to the equation. (b) -intercepts:, 0,, 0 (c) 0 OR OR (d) The -intercepts of the graph are the same as the solutions to the equation. 7.... 0. ±.... Using the positive value for,. 9. we have ± ±.08. 6. 77..8 6.6 0 Given equation 79. and.8 6.6 0 One possible equation is: Use the Quadratic Formula with a.8, b 6, and c.6. Considering only the positive value for, we have:.09 6 ± 6.8.6.8 6.76. 6 ± 8.76.6 0 0 0 0 Any non-zero multiple of this equation would also have these solutions.
78 Chapter Equations, Inequalities, and Mathematical Modeling 8. 7 and 8.,, and One possible equation is: One possible equation is: 7 6 7 6 7 7 7 is a factor. 7 6 7 6 is a factor. 77 6 0 0 Any non-zero multiple of this equation would also have these solutions. 0 0 0 0 Any non-zero multiple of this equation would also have these solutions. 8.,, i, and i One possible equation is: i i 0 i i 0 0 0 Any non-zero multiple of this equation would also have these solutions. 700 87. Let the number of students in the original group. Then, the original cost per student. 700 When si more students join the group, the cost per student becomes 7.0. Model: Cost per student Number of students Total cost 700 7. 6 700 00 6 00 90 0,00 0 Multiply both sides by to clear fractions. 90 ± 90 0,00 Using the positive value for we conclude that the original number was students. 90 ± 0 0 89. Model: Time Distance Rate Labels: Let average speed of the plane. Then we have a travel time of t. If the average speed is increased by 0 mph, then t 60 0 t 0. Now, we equate these two equations and solve for. Equation: 0 0 0 7 9,000 7 0 0 0 9,000 Using the positive value for found by the Quadratic Formula, we have. mph and 0 9. mph. The airspeed required to obtain the decrease in travel time is 9. miles per hour.
Section.6 Other Types of Equations 79 9..0996 r 60.0996 60 r.0996 60 r A P r n nt 9. 0.9 00 r r 0.0 % M 6.97.6t, t (a) (b) 6.97.6t 86.6t.0 t.0.6 t 9.9 The number of medical doctors reached 86,000 late during the year 999. 6.97.6t 900.6t 6.0 t 6.0.6 t.7 The model predicts the number of medical doctors will reach 900,000 during the year 00. The actual number of medical doctors in 00 is about 800,000. 9. T 7.8.., 0 (a) Absolute Pressure, Temperature T (c) 7.8.. 0.. 6.8 6.6 By the Quadratic Formula, we have 0 9. 6.7798 8..68.8787.806 0 8.0 0.6 0 0.8 9.8 (d) Since is restricted to 0, let.806 pounds per square inch. 00 (.8069, ) 0 66.60 0 0 (b) T when pounds per square inch. 97. 7. 0 0.0 0.0. 0.0 6.00 0.0.00 0.0 00. Rounding to the nearest whole unit yields 00 units.
80 Chapter Equations, Inequalities, and Mathematical Modeling 99. Model: Labels: Equation: Distance from home to st distance from st to nd distance from Distance from home to st, distance from st to nd, distance from home to nd 7. 7. 6,6. 6,6. ±88. ±90 The distance between bases is approimately 90 feet. home to nd 0. S 86 h (a) 0 0 0 When S 0, h.. (b) h 8 9 0 S 8. 0.6.9.8 6. 8.6 S 0 when h is between and inches. (c) 0 86 h 0 86 h,00 6 6 h 6 h 9.9 h 9.9 h. (d) Solving graphically or numerically yields an approimate solution. An eact solution is obtained algebraically. 0. Model: Portion done by first person portion done by second person work done Labels: Work done, rate of first person time worked by first person, r, Equation: rate of second person r r, r rr r rr rr r r 6 r r r r time worked by second person 0 r r 6 r ± 6 It would take approimately hours and 6 hours individually. (Choose the positive value for r.) ± 8 0. v s R v gr s v gr s v s gr g 07. False See Eample 7 on page 7.
Section.6 Other Types of Equations 8 09. The distance between, and, 0 is.. The distance between 0, 0 and 8, y is 7. 0 8 0 y 0 7 8 y 7 69 6 y 89 0 y 6 0 y ± 0 ± 6 0 6 Both, 0 and 6,0 are a distance of Both 8, and 8, are a distance of 7 from 0, 0. from,.. 9 9 a b 9 a b 9 9 a b 9 OR 9 a b 9. 0 0 a b 0 a b 0 0 a b 0b 00 a b 8 9 a b 9 a b 0b 80 a 8 b a b a b 0b 80 a b Thus, a 8 b or a b. From the original equation we know that b 9. Some possibilities are: b 9, a 9 b 0, a 8 or a 0 b, a 7 or a b, a 6 or a b, a or a b, a or a This formula gives the relationship between a and b. From the original equation we know that a 0 and b 0. Choose a b value, where b 0 and then solve for a, keeping in mind that a 0. Some possibilities are: b 0, a 0 b, a 9 b, a 6 b, a b, a b, a 7. 8 6 6 9 6 6 9. z z z z z zz z zz z z 6z z zz z z zz. 0 0 0
8 Chapter Equations, Inequalities, and Mathematical Modeling Section.7 Linear Inequalities in One Variable You should know the properties of inequalities. (a) Transitive: a < b and b < c implies a < c. (b) Addition: a < b and c < d implies a c < b d. (c) Adding or Subtracting a Constant: a ± c < b ± c if a < b. (d) Multiplying or Dividing a Constant: For a < b,. If c < 0, a then ac > bc and c > b c. You should be able to solve absolute value inequalities. (a). If c > 0, a then ac < bc and c < b c. < a > a if and only if a < < a. (b) if and only if < a or > a. Vocabulary Check. solution set. graph. negative. equivalent. double 6. union. Interval:, (a) Inequality: (b) The interval is bounded.. Interval:, (a) Inequality: > (b) The interval is unbounded.. Interval:, (a) Inequality: < (b) The interval is unbounded. 7. < 9. Matches (b). < Matches (d).. < < < Matches (e).. > 0 (a) >? 0 (b) >? 0 (c) >? 0 (d) >? 0 > 0 7 > 0 > 0 9 > 0 Yes, is a solution. No, is not a solution. Yes, is a solution. No, is not a solution.. 0 < < (a) (b) 0 (c) 0 (d) 7 0 <? 0 < < <? Yes, is a solution.? 0 0 < <? 0 < < No, 0 is not a solution. 0 <? 0 0 < < <? No, 0 is not a solution.? 7 0 < <? 0 < 8 < Yes, 7 is a solution.
Section.7 Linear Inequalities in One Variable 8 7. 0 (a) 0? (b) 0? (c) 0? (d) 9 9 0? Yes, is a solution. Yes, is a solution. Yes, solution. is a No, 9 is not a solution. 9. <. >. 7 < < < < 0 0. 7 < 7. 9. < < 7 < 9 > 7 < 0 7 0 6 7. 6 7. 8..6..6. 6 0 6 7. < < 9 9. < < 6 < < < < < 9 < < <. > > > > 0 9 < < < < 9 6 0 6 8 0.. 0.... 0.. 0.. 0.. 0 < 6 7. 6 < < 6 6 0 6 > < or > < > 0
8 Chapter Equations, Inequalities, and Mathematical Modeling 9. <. No solution. The absolute value of a number cannot be less than a negative number. 0 6 6 0 6 6 6 0 0 0. 9. 9 or 0 9 6 8 or 0 0 0 8 7. 9 < 9. 9 < < 9 < 0 < < 8 > > < < 0 9 0 9 0 9 6 9 9 8 or 0 9 6 6. 6 > 6. 6 > 6. 8 8 0 6 0 0 0 0 0 0 0 0 0 0 67. 7 69. y (a) 7 7 7 or 7 y 8 0 (b) y 0 0 0
Section.7 Linear Inequalities in One Variable 8 7. y 6 7. (a) (b) 0 y 0 y 0 0 6 6 y (a) y (b) y or 0 or 7 8 7. 0 77., 0 79., 7 0 8. 7 7, 7 0 < 8 All real numbers within 8 units of 0. 8. The midpoint of the interval, is 0. The interval represents all real numbers no more than units from 0. 0 8. The graph shows all real numbers at least units from 7. 7 87. All real numbers within 0 units of < 0 89. All real numbers more than units from > > 9. Let the number of checks written in a month. Type A account charges: Type B account charges: 6.00 0. <.0 0.0.0 < 0. 6 < 6.00 0..0 0.0 If you write more than si checks a month, then the charges for the type A account are less than the charges for the type B account. 9. 000 r > 06.0 9..9R > C r >.06.9 > 9 70 r > 0.06 0.9 > 70 r > 0.0 0.9 >.799 r >.% 0.9 6 units
86 Chapter Equations, Inequalities, and Mathematical Modeling 97. Let daily sales level (in dozens) of doughnuts. Revenue: R.9 Cost: C 0. Profit: P R C.9 0..0 0 0 P 00 0.0 0 00 00.0 0 In whole dozens,. 99. (a) y 0.067.68 7 0 0 (b) From the graph we see that y when 9. Algebraically we have: 0.067.68 8.68 0.067 9 IQ scores are not a good predictor of GPAs. Other factors include study habits, class attendance, and attitude. 0. S.0t.0, 0 t 0. s 0. 6 (a).0t 6 s 0. 6 0. (b).0t 0.9 t 0.8 Rounding to the nearest year, t 0. The average salary was at least $,000 but not more than $,000 between 99 and 000..0t > 8.0t > 7 t > 6 According to the model, the average salary will eceed $8,000 in 006. 0 07. 0 0.9. gallons 0$.89 $0.9 You might have been undercharged or overcharged by $0.9. 0.06 s 0. 0.06 0.7 s 0.6 Since A s, we have 0.7 area 0.6 06.86 area 09.6. < t.6.9 < t.6.9 <.9 < t.6 <.9.7 < t < 7..7 7. 6 7 8 9 Two-thirds of the workers could perform the task in the time interval between.7 minutes and 7. minutes. t 09. h 0 0 0 h 0 0. False. If c is negative, then ac bc.. a a 0 h 80 a or The minimum relative humidity is 0 and the maimum is 80. a a Matches (b).
Section.8 Other Types of Inequalities 87., and, d 0 Midpoint:,, 7 7., 6 and, 8 d 8 6 8 60 6 Midpoint: 6 8,, 9. 6 6. 0 6 6 7 0 6 6 7 0 or 0 6 60 7 0 7., 0. Answers will vary. Section.8 Other Types of Inequalities You should be able to solve inequalities. (a) Find the critical number.. Values that make the epression zero. Values that make the epression undefined (b) Test one value in each test interval on the real number line resulting from the critical numbers. (c) Determine the solution intervals. Vocabulary Check. critical; test intervals. zeroes; undefined values. P R C. < 0 (a) <? 0 (b) 0 0 <? 0 (c) <? 0 (d) <? 0 6 < 0 < 0 < 0 < 0 No, is not a solution. Yes, 0 is a solution. Yes, is a solution. No, is not a solution.
88 Chapter Equations, Inequalities, and Mathematical Modeling. (a) (b) (c) 9 (d) 9? 7 Yes, is a solution.? 6 is undefined. 0 No, is not a solution. 9 9? 7 No, 9 is not a solution. 9 9? Yes, 9 is a solution.. 6 7. 0 0 Critical numbers:, 7 0 0 7 7 Critical numbers: 7, 9. 9. 9 0 0 Critical numbers: ± Test intervals:,,,,, Test: Is 0? Interval,,, -Value 0 Value of 9 6 9 7 0 9 9 6 9 7 Conclusion Positive Negative Positive Solution set:, 0 < < < 0 7 < 0 Critical numbers: 7, Test intervals:, 7, 7,,, Test: Is 7 < 0? Interval -Value Value of 7 Conclusion, 7 7,, 0 0 9 7 Positive Negative Positive Solution set: 7, 8 7 6 0
Section.8 Other Types of Inequalities 89. 9. 0 0 Critical numbers:, Test intervals:,,,,, Test: Is 0? Interval -Value Value of Conclusion,,, 6 0 7 7 7 7 Positive Negative Positive Solution set:,, < 6 6 < 0 < 0 Critical numbers:, Test intervals:,,,,, Test: Is < 0? Interval -Value Value of Conclusion,,, 0 6 6 6 6 6 Positive Negative Positive Solution set:, 6 0 0 7. < 0 < 0 Critical numbers:, Test intervals:,,,,, Test: Is < 0? Interval -Value Value of Conclusion,,, 0 Positive Negative Positive Solution set:, 0 9. 8 0 8 0 Complete the square. 8 6 6 ± ± Critical numbers: ± Test intervals:,,,,, Test: Is 8 0? Interval,,, -Value 0 0 Value of 8 00 80 0 0 6 Conclusion Positive Negative Positive + Solution set: <, 0 8 6 0
90 Chapter Equations, Inequalities, and Mathematical Modeling. > 0 > 0 > 0 > 0 Critical numbers: ±, Test intervals:,,,,,,, Test: Is > 0? Interval -Value Value of Conclusion, Negative,, 0 Positive Negative 0, Positive Solution set:,,. 9 0 9 8 0 9 0 9 0 0 Critical numbers:, ± Test intervals:,,,,,,, Test: Is 0? Interval -Value Value of Conclusion, 67 Negative,, 0. 8 0..0..7 Positive Negative 0, 7 Positive Solution set:,,. 0 7. Critical number: Test intervals: Test: Is 0? Interval -Value Value of Conclusion, 0 Positive Positive, 0 Solution set:,,, 0 6 < 0 < 0 Critical numbers: Test intervals: 0,, 0, 0,,, Test: Is < 0? By testing an -value in each test interval in the inequality, we see that the solution set is:, 0 0,
Section.8 Other Types of Inequalities 9 9. 0 Critical numbers: Test intervals: 0. 0, ±,,, 0, 0,,, Test: Is 0? By testing an -value in each test interval in the inequality, we see that the solution set is:, 0, 0 Critical numbers:, Test intervals:,,,,, ) Test: Is 0? By testing an -value in each test interval in the inequality, we see that the solution set is:,. y. y 8 6 8 7 (a) y 0 when or. (b) y when 0. 8 (a) y 0 when 0, <. (b) y 6 when. 7. > 0 6 9. < 0 6 > 0 < 0 Critical numbers: 0, ± Test intervals:,,, 0, 0,,, < 0 Critical numbers:, Test: Is > 0? Test intervals:,,,,, By testing an -value in each test interval in the inequality, we see that the solution set is:, 0, Test: Is < 0? 0 By testing an -value in each test interval in the inequality, we see that the solution set is:,, 0. > > 0 > 0 > 0 6 9 8 Critical numbers:, Test intervals:,,,,, Test: Is > 0? By testing an -value in each test interval in the inequality, we see that the solution set is:,
9 Chapter Equations, Inequalities, and Mathematical Modeling. > 0 > > 0 7 7 > 0 Critical numbers:,, Test intervals:,,,, 0,,, 7 Test: Is > 0? By testing an -value in each test interval in the inequality, we see that the solution set is:,,. 9 0 9 9 0 0 0 Critical numbers:,, 6 Test intervals:,,,,, 6, 6, 0 6 8 0 Test: Is 0? By testing an -value in each test interval in the inequality, we see that the solution set is:, 6, 7. 9 0 0 Critical numbers: 0,, ± Test intervals:,,,,, 0, 0,,, Test: Is 0? By testing an -value in each test interval in the inequality, we see that the solution set is:, 0, 0
Section.8 Other Types of Inequalities 9 9. < < 0 < 0 < 0 7 6 < 0 < 0 Critical numbers:,, ± Test intervals:,,,,,,,,, Test: Is < 0? By testing an -value in each test interval in the inequality, we see that the solution set is:,,, 0. y. y 8 6 6 6 6 (a) y 0 when 0 <. (b) y 6 when <. (a) y when or (b) This can also be epressed as y for all real numbers... This can also be epressed as < <.. Critical numbers: Test intervals: 0 7. 0 Test: Is 0? ±,,,,, By testing an -value in each test interval in the inequality, we see that the domain set is:, 7 0 0 Critical numbers: Test intervals:,,,,,, Test: Is 0? By testing an -value in each test interval in the inequality, we see that the domain set is:,,
9 Chapter Equations, Inequalities, and Mathematical Modeling 9. 0 6. 7 0 Critical numbers: 0,, 7 Test intervals:,,, 0, 0, 7, 7, Test: Is 7 0? By testing an -value in each test interval in the inequality, we see that the domain set is:, 0 7, 0..6 < 0. 0..9 < 0 0.. < 0 Critical numbers: ±. Test intervals:,.,.,.,., By testing an -value in each test interval in the inequality, we see that the solution set is:.,. 6. 0...6 > 0 6. The zeros are. ±. 0..6. 0. Critical numbers: 0.,. Test intervals:, 0., 0.,.,., By testing an -value in each test interval in the inequality, we see that the solution set is: 0.,... >.. > 0....... 7.8 8.68.. > 0 > 0 Critical numbers:.9,.6 Test intervals:,.6,.6,.9,.9, By testing an -value in each test interval in the inequality, we see that the solution set is:.6,.9 67. s 6t v 0 t s 0 6t 60t 69. (a) 6t 60t 0 6tt 0 0 t 0, t 0 It will be back on the ground in 0 seconds. (b) 6t 60t > 8 6t 60t 8 > 0 6t 0t > 0 t 0t < 0 t t 6 < 0 < t < 6 seconds L W 00 W 0 L LW 00 L0 L 00 L 0L 00 0 By the Quadratic Formula we have: Critical numbers: L ± Test: Is L 0L 00 0? Solution set: L.8 meters L 6. meters
Section.8 Other Types of Inequalities 9 7. R 7 0.000 and C 0 0,000 P R C 7 0.000 0 0,000 0.000 0,000 0.000 0,000 70,000 0.000,000,000 0 Critical numbers: 0,000, 0,000 (These were obtained by using the Quadratic Formula.) Test intervals: 0, 0,000, 0,000, 0,000, 0,000, By testing -values in each test interval in the inequality, we see that the solution set is 0,000, 0,000 or 0,000 0,000. The price per unit is p R 7 0.000. P 70,000 For 0,000, p $. For 0,000, p $0. Therefore, for 0,000 0,000, $0.00 p $.00. 7. C 0.00t 0.6t.t 9., 0 t (a) 80 (d) t C 6 8. (b) 0 0 t C 70. 6 7.6 8 7.9 0 7.6 76.8 79.6 (c) C 7 when t 0.. C will be greater than 7% when t, which corresponds to 0. 7 8. 8 87.8 9 90. 0 9. 96.8 00. 0. C will be between 8% and 00% when t is between 7 and. These values correspond to the years 07 to 0. (e) 8 C 00 when 6.8 t.89 or 7 t. (f) The model is a third-degree polynomial and as t, C.
96 Chapter Equations, Inequalities, and Mathematical Modeling 7. R R R R RR R R R R R R Since R, we have R R 77. True The test intervals are,,,,,, and,. R R 0 R R 0. Since R > 0, the only critical number is R. The inequality is satisfied when R ohms. 79. b 0 8. To have at least one real solution, b 6 0. This occurs when b or b. This can be written as,,. b 0 0 To have at least one real solution, b 0 0. b 0 0 b 0b 0 0 Critical numbers: b ±0 ±0 Test intervals:, 0, 0, 0, 0, Test: Is b 0 0? Solution set:, 0 0, 8. (a) If a > 0 and c 0, then b can be any real number. If a > 0 and c > 0, then for b ac to be greater than or equal to zero, b is restricted to b < ac or b > ac. (b) The center of the interval for b in Eercises 79 8 is 0. 8. 0 87. ) 89. Area lengthwidth
Review Eercises for Chapter 97 Review Eercises for Chapter. y. y 0 y 8 0 y 0 0 y y. y 0 y 7. y y y Line with -intercept, 0 and y-intercept 0, 6 Domain:, 6 0 y y 0 6 9. y 0 0 y is a parabola. ± ± y 0 8 y. y -intercepts: 0 ± ± or y-intercept: y 0 y 9 y The -intercepts are, 0 and, 0. The y-intercept is 0,.. y y Intercepts:, 0, 0, y y No y-ais symmetry y y No -ais symmetry y y No origin symmetry
98 Chapter Equations, Inequalities, and Mathematical Modeling. y 7. Intercepts: ±, 0, 0, y y y-ais symmetry y y No -ais symmetry y y No origin symmetry y y Intercepts:, 0, 0, y y No y-ais symmetry y y No -ais symmetry y y No origin symmetry y 6 7 6 9. y y Domain:, Intercepts:, 0, 0, y No y-ais symmetry 7 6 y y No -ais symmetry y y No origin symmetry 6. y 9 y. y 6 Center: 0, 0 y 0 Radius: Center:, 0 (0, 0) Radius: y 6 (, 0) 8 6. y 6 7. Endpoints of a diameter: 0, 0 and y 6 Center:, Radius: 6 y Center: Radius: Standard form:, 6 0 0 6,, r 0 0 9 y y 8 6 (, ( 8 8
Review Eercises for Chapter 99 9. F, 0 0 (a) (c) When 0 8 6 0 F 0 0 0 0, F 0. pounds. (b) Force (in pounds) F 0 0 0 8 6 0 Length (in inches). 6. 6 0 0 Identity All real numbers are solutions. 7 7 7 7 7 7 7 0 0 Identity All real numbers are solutions.. 0 7. 0 0 0 9. 8 6 9 6 0 0. 8 0. y 8 0 -intercept: 0 8 7 0 8 7 9 y-intercept: The -intercept is y 0 y, 0 and the y-intercept is 0,.. y 7. -intercept: 0 y-intercept: y 0 y 8 The -intercept is, 0 and the y-intercept is 0, 8. y -intercept: y-intercept: The -intercept is 0 y 0 y, 0 and the y-intercept is 0,. 9..8y 0. 0..9..h -intercept:.80 0. 0 0. y-intercept:.8y 0.0 0 y.8 9 The -intercept is, 0 and the y-intercept is 0, 9..9 6. 8.8h 88.0 8.8h 0 h The height is 0 inches.
00 Chapter Equations, Inequalities, and Mathematical Modeling. Verbal Model: September s profit October s profit 689,000 Labels: Let September s profit. Then 0. October s profit. Equation: 0. 689,000. 689,000,000 0. 6,000 Answer: September profit: $,000, October profit: $6,000. Let height of the streetlight. 7. Let the number of original investors. By similar triangles we have: 90,000 Each person s share is. 0 6 90,000 If three more people invest, each person s share is. 0 Since this is $00 less than the original cost, we have: 6 ft 90,000 The streetlight is feet tall. ft 00 90,000 ft 90,000 00 90,000 90,000 70,000 00 700 90,000 00 700 70,000 0 00 08 0 00 9 0, etraneous or 9 There are currently nine investors. 9. Let the number of liters of pure antifreeze. 6. 0% of 0 00% of 0% of 0 0.00.00 0.00 0.0.00 0.70 0 0.70 7 6 7 liters V r h V r h V r h 6. t 0t 0 t 0 rate time distance st car 0 mph t 0t nd car mph t t t hour or 0 minutes 6. 0 0 0 0 0
Review Eercises for Chapter 0 67. 6 69. 8 7. 0 0 ±8 0 ± ± 6 0 6 6 6 6 ±6 6 ± 6 7. 7 0 7. 7 0 ± 7 ± M 000 (a) 000 0 when 0 feet and 0 feet. (b),000 0 0 (c) The bending moment is greatest when 0 feet. 77. 6 6 i 79. i i i 8. 7 i i 7 i i 7i 8. i 8i 6i 0i 0 6i 8. 0 8i i 0 0i 6i i 87. 6i 6 i 6 i i i i i 0i i 6 0i 7 0 7 7 i 89. i i i i i i i i 8 i 9 8 i i 8 i i i i 9. 0 ± ± i ± i 9. 0 0 9. 0 9 ±9 ± i 0 0 0 or 0 0 or