Pre-Calculus 0 eview Trigonometry (eference Chapter, Sections. -., pages 74-99) Outcomes: Demonstrate an understanding of angles in standard position, 0 60 Solve problems, using the three primary trigonometric ratios, for angles from 0 to 60 in standard position. For each angle listed below Draw the angle in standard position and indicate in which quadrant the terminal arm is located. Indicate the reference angle on your diagram. c. State two other angles (other than the one in quadrant I) that have the same reference angle. i. 45 ii. 05 iii. 00. Which angle in standard position has a different reference angle than all the others? 05 65 55 85. Determine the measure of the three angles in standard position, (90 60 ) that have a reference angles of 65 0 c. 85 4. Each point listed below is situated on the terminal arm of an angle in standard position. Determine the eact trigonometric values for sin, cos, and tan. P (,-4) Q (-7, -4) c. A (-5, ) 5. Find, 0 60. sin sin cos7 cos c. tan6 tan d. sin98 sin e. cos6 cos f. tan76 tan 6. Determine the eact value of sin0 cos70 c. tan00 d. sin0 e. cos5 f. tan0 7. is an angle in standard position. Find the two other primary trigonometric values if the terminal arm of is in 6 quadrant IV and cos. 7 9 8. Consider in standard position such that tan. 5 Determine the eact values for the other two primary trigonometric ratios. Determine the possible value(s) of, if 0 60 9. Solve for where 0 60 sin cos c. tan d. sin e. cos f. tan 0. Solve each equation, where 0 60. ound to the nearest degree. sin 0.45 cos 0.6 c. tan.4
Quadratic Functions (eference Chapter, Sections.., pages 4-0) Outcomes: Analyze quadratic functions of the form y a( p) q and determine the verte, domain, and range, direction of opening, ais of symmetry, - and y-intercepts. Analyze quadratic functions of the form y a b c to identify characteristics of the corresponding graph, including verte, domain and range, direction of opening, ais of symmetry, - and y-intercepts, and to solve problems.. Indicate the verte, the equation of the ais of symmetry, direction of opening, the maimum or minimum value, the domain and range for each of the following quadratic functions. ( ) f 9 g( ) 8 5 c. y4 d. -intercepts at - & 7, ma value is. Determine the equation, both in verte and standard form, for each of the following quadratic functions. Maimum point is at (-,-5) and graph passes through the point (-,-) -intercepts are at - and 4, with a minimum of -8 c. Graph of function is shown to the right. By inspection, determine the number of -intercepts for each quadratic function. f( ) ( 5) 7 g( ) 4 y 8 4. Sketch the graph of the quadratic function. Identify the verte, domain, range, ais of symmetry, - and y-intercepts, ma/min value. 5. f ( ) 0 Sketch the graph of the function. Identify the verte, domain, range, ais of symmetry, - and y- intercepts. 6. Determine the verte form of the following quadratic functions by completing the square. y 5 9 y 5 0 6 7. A lifeguard at a public beach has 400 m of rope available to lay out a restricted swimming area using the straight shoreline as one side of the rectangle. What dimensions will maimize the swimming area? 8. Calculators are sold to students for $0 each. Three hundred students are willing to buy them at that price. For every $5 increase in price, there are 0 fewer students willing to buy the calculator. What selling price will produce the maimum revenue and what will the maimum revenue be? 9. A ligting fiture manufacturer has a daily costs of C( ) 800 0 0.5, where C is the total daily cost in dollars and is the number of light fitures produced. What is the minimum cost for daily production for this company? How many fitures are produced to yield a minimum cost?
Quadratic Equations (eference Chapter 4, Sections 4.-4.4, pages 06-6) Outcomes: Factor polynomial epressions of the form:, 0 a b a a b y, a 0, b 0 5 6y a( f( )) b( f( )) c, a 0 ( 5) ( 5) 9 a f b g y a b ( ( )) ( ( )), 0, 0 9( ) 4( y ) Solve problems that involve quadratic equations.. Factor. 5 0 c. 4 5. Factor each of the following. Use substitution as a strategy to aid in the factoring. 6 5 5 9 4 c. 5 y 7 49. Solve each quadratic equation by factoring. 4p 4p c. 0..4 0 e. 5 6 0 d. t 4t 6 f. 0q 6 0 6 7 7 q 4. A park has an area of484 m. The width is 8 m shorter than the length. Determine the dimensions of the park. 5. A mural is to be painted on a wall that is 5 m long and m high. A border of uniform width is to surround the mural. If the mural is to cover 75% of the area of the wall, how wide must the border be? 6. Determine the number and nature of the roots. 6 5 6 0 4 9 c. 0 7. Solve. 7 6 5 a 4a 0 8. A field is in the shape of a right triangle. The fence around the perimeter of the field measures 40 m. If the length of the hypotenuse is 7m, find the lengths of the other two sides. 9. A diver jumps upward from a platform. The height, h(t), in metres, of the diver above the water is modeled by h( t) 5t 0t 40, where t is the time in seconds after she leaves the board. Find the diver s maimum height above the water. How long does it take the diver to reach this height? How long is it before the diver enters the water? c. At what time(s) is the diver at a height of 4m? d. How high is the platform above the water?
Systems of Equations (eference Chapter 8, Sections 8.-8., pages 44-460) Outcomes: Solve, algebraically and graphically, problems that involve systems of linear-quadratic and quadratic-quadratic equations in two variables.. Solve each of the following systems of equations using the corresponding graphs. y4 y y y 4. Determine the values of m and n if (, 4) is a solution to the following system of equations: m y m 5y n. Solve each of the following linear-quadratic systems of equations. 9y4 y y 7 y 5 c. 4 y 4 y 7 4. The sum of two whole numbers is 7. Thirteen less than three times the square of the smaller yields the larger number. Determine the two numbers. 5. Solve each system of equations algebraically. y 7 y 4 5 y 4 y 6. Jack is skeet shooting. The height of the skeet is modelled by the equation h 5t t, where h represents the height in metres t seconds after the skeet is released. The path of Jack s bullet is modelled by the equation h.5t with the same units. How long will it take for the bullet to hit the skeet? How high off the ground will the skeet be when it is hit?
Linear and Quadratic Inequalities (eference Chapter 9, Sections 9.-9., pages 464-505) Outcomes: Solve problems that involve linear and quadratic inequalities in two variables. Solve problems that involve quadratic inequalities in one variable.. Determine the inequality that corresponds to the graph illustrated to the right.. Sketch the graph of each of the following linear inequalities. y y6 c. 4y. Solve. 7 0 9 6 c. 0 6 d. 7 4. A firework is shot straight up into the air with an initial velocity of 500 feet per second from 5 feet off the ground. The equation h( t) 6t 500t 5 models the path of the firework, where t is the time in seconds and h(t) is the height of the firework in feet. Determine when the firework will be higher than 000 ft. Etension: When will the firework be below 000 ft? 5. State the inequality represented by the graph to the right. Epress your final answer in standard form. 6. Sketch the graphs of the following inequalities. y 4 y 4 5 c. y 4 4 7. A garden is to be in the shape of a rectangle. The length of the garden must be 4 m longer than the width. The gardener wants the area of the garden to be no greater than 480 square meters. Find all the possible dimensions.
adical Epressions and Equations (eference Chapter 5, Sections 5.-5., pages 7-07) Outcomes: Solve problems that involve operations on radicals and radical epressions with numerical and variable radicands. Solve problems that involve radical equations (limited to square roots).. Write each radical as an entire radical. 5 b b c. 5 k k d. 4 r r. Write each radical in simplest form. 00 450 c. 60 d. 4 4. Simplify. 6 80 0 5 7 c. 5 75 5 d. 6 8 75 648 4. State the side length of a square with an area of 57cm in simplified radical form. 5. Write number. 6. Simplify. 4 9 5 y in simplest form. Identify the values of the variable for which the radical represents a real 5 6 4 4 7 54 5 7 c. 4 5 4 7. Simplify. 4 4 8 9 5 4 c. 7 5 4 d. 5 8. State the restrictions on in each of the following: 9 9. Square the epression 5. 0. Solve each of the following. Check for etraneous roots. 4 9 c. p p 6 d. z8 4z 7
ational Epressions and Equations (eference Chapter 6, Sections 6.- 6.4, pages 0-55) Outcomes: Determine equivalent forms of rational epressions (limited to numerators and denominators that are monomials, binomials or trinomials) Perform operations on rational epressions (limited to numerators and denominators that are mononials, binomials or trinomials) Solve problems that involve rational equations (limited to numerators and denominators that are monomials, binomials or trinomials). State the non-permissible value(s) in each rational epression. 5w 4p c. gj 4 5 p p0. Simplify each rational epression. State the non-permissible values. 8 r r r r c. 4 y 5y y. Simplify. State the non-permissible values. c. 6 4 5 9 0 4 7 4 4. Add or subtract. Give answers in simplest form. Identify all non-permissible values. 8 6 4 4 c. 5 4 5. Simplify each epression. Identify all non-permissible values. 4 4 4 r 7 7 r r r 49 7 6. Simplify each of the following epressions. Identify the non-permissible values. 8 4 4 5 5 4 5 5 6 7 7. When it was raining, Elliott drove for 0 miles. When the rain stopped, he drove 0 mph faster than he did while it was raining. He drove for 00 miles after the rain stopped. If Elliott drove for a total of 0 hours, how fast did he drive while it was raining?
8. Solve. 5 7 5 6 m m b 4 5 c. b b b 8b b 8 b 9. A ski resort can manufacture enough machine-made snow in hours to open its steepest run, whereas it would take naturally falling snow 6 hours to provide enough snow. If the resort makes snow at the same time that it is snowing naturally, how long will it take until the run can open? 0. A positive integer is 4 less than another. The sum of the reciprocals of the two integers is 0. Find the integers.. Mark is kayaking in a river which flows downstream at a rate of mph. He paddles 5 miles downstream and then turns around and paddles 6 miles upstream. The trip takes hours. How fast can Mark paddle in still water?
Absolute Value Functions (eference Chapter 7, Sections 7.- 7., pages 58-9) Outcomes: Demonstrate an understanding of the absolute value of real numbers. Graph and analyze absolute value functions (limited to linear and quadratic functions) to solve problems. Short esponse:. Evaluate c. 5 6 8 9 5 4 7 0.4 5 7 7 6. Write each of the following as a piecewise function. y 5 y 4 7 c. y 0 d. y 6. Sketch the graph of each of the following absolute value functions. State the domain and range of each. y 4 y c. y 4 d. y 4 4. Solve. 8 9 0 48 8 c. 7 d.
SOLUTIONS : Chapter Trigonometry. 80 45 80 65 other angles are : 80 80 65 5 60 60 65 95 60 60 05 55 other angles are : 80 80 55 5 80 80 55 5 c. 80 80 00 80 other angles are : 80 80 80 60 60 60 80 80. If 05 ; 80 80 05 75 If 65 ; 80 80 65 5 If 55 ; 80 55 80 75 If 85 ; 60 60 85 75 Conclusion : 65has a reference angle that is different from the others.
. 4. If 65 80 80 65 5 ; 80 80 65 45 ; 60 60 65 95 If 0 80 80 0 70 ; 80 80 0 90 ; 60 60 0 50 c. If 85 80 80 85 95 ; 80 80 85 65 ; 60 60 85 75 P, 4 ; y 4; r 4 5 5. 4 4 Thensin ;cos ; tan 5 5 P y r 7, 4 7; 4; 7 4 65 5. 4 7 4 Thensin ;cos ; tan 5 5 7 P 5, 5; y ; r 5 69. c. 5 Thensin ;cos ; tan 5 5. 80 5 and 0 in quadrants III & IV 60 60 5 07 so sin sin 07 7 60 7 4 and cos 0 in quadrants I & IV 4 so cos7 cos 4 c. 6 80 6 44 and tan 0 in quadrants II & IV 60 60 44 6 so tan6 tan 6 d. 98 80 98 8 and sin 0 in quadrants I & II 8 so sin 98 sin8 e. 6 6 80 8 and cos 0 in quadrants II & III 80 80 8 99 so cos6 cos 99 f. 76 76 and tan 0 in quadrants I & III 80 80 76 56 so tan 76 tan 56 6. 0 is in quadrant II where sin 0 and 80 0 60.
Since sin 60, then sin0. 0 cos 70 0 r r c. 00 is in quadrant IV where tan 0 and 60 00 60 Since tan 60, then tan00.. d. 0 is in quadrant IV where sin 0 and 60 0 0. Since sin 0, then sin 0. e. 5 is in quadrant IV so cos 0 and 60 5 45. Since cos 45, then cos5. f. 0 is in quadrant III where cos 0 et 0 80 0 Since tan 0, then tan 0.. 7. r 6 7 Since cos and is in quadrant IV 6; r 7; y 7 6 y y Thus sin and tan. r 7 6 y 9 Since tan and could be in quadrant II 5; y 9; r 5 5 9 06 y 9 5 Thus sin and cos. r 06 r 06 y 9 5 could also be in quadrant IV, thus sin and cos r 06 r 06 6. Therefore =80-6 =9 or =60-6 =99 8. 9. sin 60 and since sin 0, is in quadrants III or IV. Thus 80 80 60 40 and 60 60 60 00. cos 45 and since cos 0, is in quadrants I or IV. Thus 45 and 60 60 45 5. tan 45 and since tan 0, is in quadrants II or IV c.. Thus 80 80 45 5 and 60 60 45 5.
d. Since sin 90, 90 is the only solution since 90 is a quadrantal angle. e. f. cos 60 and since cos 0, is in quadrants II or III. Thus 80 80 60 0 and 80 80 60 40. tan 0 and since tan 0, is in quadrants I or III. Thus 0 and 80 80 0 0. 0. sin 0,45 7 and since sin 0, is in quadrants I or II. thus 7 and 80 80 7 5. cos 0,6 69 and since cos 0, is in quadrants II or III. Thus 80 80 69 and 80 80 69 49 c.. tan, 4 67 and since tan 0, is in quadrants II or IV. Thus 80 80 67 and 60 60 67 9. Chapter Quadratic Functions. a ; p ; q 9 so the verte is at, 9 ;A.S.: ; opens down maimum @ f 9; D: ; : f f 9 8 8 4 g g g c. a p q maimum @ y0 4; D: ; : y y 4 a ; p ; q 5 so the verte is at, ;A.S.: ;opens up minimum @ ; D : ; : ; 0; 4 so the verte is at 0,4 ;A.S.: 0; opens down 7 a 0; p ; q ; so the verte is at, ;A.S.: ; opens down maimum @ y when ; D: ; : y y d.., y, ; p, q, 5
. a 5 6 4 a so a 4 y 4 5 (verte form) standard form: y 4 5 4 6 9 5 4 4 6 5 4 4 4 4, y,0 ; p ; q 8 0 a 8 8 9 a so a y 8 (verte form) standard form: y 8 8 4 8 4 6 c., y 0, ; p, q,8 a0 8 6 9 a so a y 8 (verte form) standard form: y 8 6 9 8 4 6 8 4 Since a > 0 & q < 0, f() has -intercepts Since a < 0 & q < 0, g() has 0 -intercepts 4. Verte (,8), domain { }, range { y y 8, y }, AS: =, -intercepts - and 7, y- intercept 7, ma value y = 8 5. Verte (,-8), domain { }, range { y y 8, y }, AS: =, -intercepts and 5, y- intercept 0
6. f ( ) 5 9 f ( ) 5 9 5 5 f ( ) 5 9 4 4 5 9-5 9 f ( ) ; Verte, 4 4 f ( ) 5 0 6 f ( ) 5 4 6 f ( ) 5 4 4 4 6 f ( ) 5 4 4 4-6 f ( ) -5-0 6 f ( ) 5( ) 4 Verte(,4) 7. width y400 length A = (400-) A 400 A 400 A ( 00 ) A ( 00 0000 0000) A ( 00) 0000 Dimensions are 00 m by 00 m. Maimum enclosed area is 0000m. 8. 50 900 6000 b 900 Ma. is @ ; = 750 a 00 Maimum revenue is $750 and it occurs when = which would bring the price of the calculators up to $5. There would be price increases of $5. They would have 0 sales instead of 00. 9. C( ) 800 0 0.5 b 0 Minimum cost @ 0 a 0.5 C(0) $700 0 light fitures can be produced for a minimum cost of $700.
Chapter 4 Quadratic Equations. 5 0 5 4 4 8 5 5 c. 4 5 5. Let a 5 a a 6 5 5 9 a a 9 aa a a 5 5 5 0 8 7 Let b 4 b b 4 b b 8 a a b b 8 8 8 4 c. Let c and d y 7 6 6 4 4 4 4 5 y 7 5c d 5c d 5c d 5 y 75 y 7 49 49 7 7 7 7 4 4 4 4 5 5 y 45 5 y 4 5 y 95 y 7 7 7 7. c. d. 4 p 4 p 4p 4 p0 4p p6 0 p0 or p6 5 6 0 9 4 0 9 or 4 0..4 0 0. 70 0 0. 5 0 5 or t 4t 6 t 4t6 0 t 7t8 0 t t 8 0 t 8 or t e. 0q 6 q 0q 6q 0 5q q 6 0 5q55q 0 5 5 q5q 5 q q 5 0 q or q 5 0 f. 0 6 7 7 6 0 0 7 7 6 7 0 0 7 6568 0 7 6 7 4 5 6 5 4 0 7 5 4 or 0
4. Let be the length and 8 be the width of the park. The area of the park is given by A 8 8 484 8 484 0 54 46 0 54 or 46 The length of the park is 54 m and the width is 54 8 = 46 m 5. The width of the mural is approimately 0.89 m. 6. a b c D b ac 6; 5; 6; 4 5 4 6 6 5 44 69 0 real and distinct roots a b c D b ac 4; ; 9; 4 4 4 9 44 44 0 real and equal roots c. a b c D b ac ; ; 0; 4 4 0 4 0 6 0 non-real roots
7. a 7; b 6; c 5 b b 4ac 6 6 4 7 5 6 76 6 4 a 7 4 4 7 a ; b 4; c 4 4 4 b b 4ac 4 6 (No real solutions) a 6 8. 9. b 0. The diver reaches the maimum height after second. The maimum height is 45 m. a 0 5t 0t 40 0 5( t t 8) 0 It takes the diver 4 seconds to reach the water. 5( t 4)( t ) 0 t 4 and t = - c. a 5; b 7; c 4 5t 0t 40 4 5t 0t 0 0 00 40 t 0.5 and.77 0 The diver reaches a height of 4m at 0.5 s & again at.77 s. d. h 0 40 m
Chapter 8 Systems of Equations., y 0, 4 and, y,5, y,6 and, y,. m y 9m 4 m m 6y n (9) 5(4) n 5 n. 9 y 4 y 9 4 y y 9 4 7 7 7 0 0 If y 9 4 5 Solution:,5 y y y 5 y 5 5 6 7 0 7 0 7 or If 7 y 7 If y Solutions: 7, et,
c. 4 y 4y 7 ( 4 y ) 4y 7 4 8 y 6 4y 7 4 4 4 4 0 ( )( ) 0 4y 7 4 y 7 9 y 9 Solution:, 4. let = smaller number and let y= larger number y 7 y 7 0 0 0 0 0 or If y 4 The two numbers are and 4. 5. y 7 y 7 4 y y 4 7 4 6 4 0 0 or If y 7 6 If y 7 7 Solutions:, 6 and, 7
5 y 4 y (5 y 0) ( y 4 0) 0 6y 6 0 6 6y 0 4 5 6 0 ( )(4 ) 0 4 5 Solutions to the system: (, 7) and, 4 6 6. 5t t.5t 5t 0.5t 0 0.5 0.5 0 t 0.4 and 0.5 0 The skeet was shot at 0.5 seconds after release at a height of 6.75m Chapter 9 Linear and Quadratic Inequalities. points on the graph are (4,) and (0,5) 5 m 0 4 4 y 5 4
. y y6 c. 4y. 7 0 intercepts @ and 7 ; 7 opens up. 7 7 0 so or 9 6 9 6 0 0 roots @ and. y y opens down, so 9 6 0 when. 5 0 6 6 0 0 5 0 roots @ and. 5 y 6 0 opens up, so 6 0 0 when or. c. d. 7 7 4 b b ac 4 7 7 7 0 a 4 7 7 roots @.9 and 0. 4 4 7 7 y 7 opens up, so 7 0 when 4 4 6t 500t 5 000 4. b b ac 500 500 46995 6t 500t 995 0 t a 4 500 0 6 4.70 and 6.6 y 6 500 995 opens down, so 6t 500t 995 0 when 4.70 t 6.6 The firework is above 000 ft between 4.70 and 6.6 seconds. Etension : The firework is below 000 ft at times less than 4.7 seconds and between 6.6 seconds and.6 sec. (Firework hits the ground at.6 seconds)
5. 6. y 4 a ; p ; q 4 y 0 5 0 4 0 6 9 4 0 6 9 4 0 6 5 0 5 5 or y 4 5 a 4 p 4 q 5 9 4 y 0 5 0 4 5 0 5 5 or
c. y a 4 4 4 4 p 4 4 44 q 4 y 0 0 4 4 0 or 7.
Chapter 5 adical Epressions and Equations. 5 5 75 b b b b b b b 6 7 k 5k k 5k k 5k 9k 5k 45k c. 5 r r r r r r 8r r 4r d. 4 4 4 4 7. c. 00 00 0 0 60 80 0 0 d. 450 5 5 5 4 4 4 4 4 84 4 4 4. 6 80 0 64 5 5 4 5 4 5 0 5 5 7 5 0 6 4 c. d. 5 75 5 5 40 5 45 5 6 7 5 7 5 7 7 7 7 7 6 8 6 8 8 8 8 8 75 648 5 6 0 8 8 4. 57 cm (side length of the square) 5. y 6 y y y y, 0, y 0 4 9 5 4 8 4 4 6. 5 6 4 4 5 8 4 7 5 46 5 4 9 7 54 5 7 8 5 5 7 5 5 5 60 4 5 5 44 c. 4 5 4 6 8 5 4 6 5 8 6 5 4 5 5 4 4 0 5 4 4 7. c. 4 4 4 4 7 8 9 5 9 5 9 5 5 4 4 8 4 7 5 4 8 5 4 5 8 5 4 5 8 5 4 5 4 5 7 5 4 4 6 4 6 4
d. 5 05 6 6 0 7 6 7 6 4 0 8. 0 9 9 0 9 9 9. 5 5 9 6 5 6 5 4( 5) 9 5 40 4 5 0. 4 9 6 if 6 4 6 4 4 9 4 6 is a solution 9 9 9 4 88 4 4 88 85 0 5 7 0 5 or 7 if 5 5 5 5 4 5 5 4 9 5 is a solution if 7 7 7 7 6 7 4 7 8 5 9 7 is an etraneous root
c. p p 6 p p 6 p p 6 p 6 p 6 6 p 6 p 6 p 69 p if p Left side 4 ight side 6 9 p is a solution d. 6 z 5 4z 6 z 5 4z 6 z z 6 z z z z z 6 z 6 z 5 4z 6 z z z 6 z z 4z 5 0 5 0 z 5 or z if z 5 Left side 6 5 6 0 6 4 5 ight side 5 4 5 5 0 5 5 z 5 is a solution if z Left side 6 6 4 ight side 5 4 5 4 z is an etraneous root
Chapter 6 ational Epressions and Equations. g 0; j 0 0; 5 p p 0 p 5 p p 5; p c.. c. 4 r r r r r r ; 4 8 8 4 8 r r 4 y 5y y r ; r ; r r y y y y y ; y ; y y. c. 6 5 9 0 4 7 4 5 4 or ; ; ; 4 5 4 7 or ; 0; 4; 5; 4 4 4 or ; 0; ; 4. 5 5 6 4 4 6 6 6 6 6 6 6 8 6 8 4 5 4 6 4 7 6 ; ; ; 6;
5. c. 5 5 4 5 5 5 6 0 ; ; 4 4 r r r r( r ) r r( r ) r ( r 4)( r ) ( r 4) r r r 0,,, 4 4r r ( r ) 4( r ) r r 4( r ) r ( r ) 4( r ) 4 7 7 4 7 4 7 7 7 7 7 7 7 7 7 7 7 49 7 4 4 8 4 4 7 7 4 4 ; 7, 7 7 7 7 7 7 4 4 6. 8 4 4 4 7 5 5 7 5 4 5 5 6 7 5 7 7 5 8 7 0 7 5 7 7 5 5 7 5 7 4 5 5 4 8 5 4 8 4 ; ; 7; 5; ; 5 5 5 5 5,,7,0,, 5 7
7. 0 00 0 0 0( 0) 00 0 ( 0) 0 400 00 0 00 0 0 0 400 0 40 0 ( 0)( 8) 0 8 Elliott drove 0 mph while it was raining. 8. 5 7 5 7 5 7 5 5 7 7 0 4 0 or 4 So is a solution and 4 is a solution. 5 6 m m 5m m 6 m m m m m m m m 5m m 6m m 5m 5 m 6m 6 6m 7m 0 m m 0 m or m m So m is a solution and m is a solution.
b 4 5 c. b b b 8b b 8 b b 4 5 b b b b 8 b 8 b b 4 b 5 b b b 8 b b 8 b b 8 b b 8 b b b b b b b b b b b 8 8 8 4 5 4 b b b b b b 4 5 4 9b0 bb9 0 b0 or b9 b0; b8 so b 0 is etraneous and b 9 is a solution.
9. Time to open hill Fraction completed in hour Fraction of work completed in hours Machine-made snow Natural snow 6 6 6 Together 6 6 6 6 6 4 6 9 hours 0. The two positive numbers are 7 and.
. Distance (miles) Speed (mph) Time (h) With the current 5 + Against the current 6 5 6 Total 4 Mark paddles at 4mph.
Chapter 7 Absolute Value. 5 6 8 9 5 4 7 4 7 4 7 0.4 5 7 4.6 8 7 4.6 8. 8 40. c. 7 7 ( 6) 7 49 6 7 4 7 4 50 5. 5 5 if 5 5 0 invariant point @ ; y 5 5 if 7 4 7 if 7 4 7 0 invariant point @ ; y 4 4 7 47 if 4 0 5 0 invariant points @ 5 and c. y 0 opens up, y 0 if 5 d. 0 if 5 or 6 6 6 0 invariant points @ 6 y 6 if 6 6 6 opens down, y 6 if 6 or 6. Sketch the graph of y 4 and reflect the piece that is negative. y y D : ; : 0
Sketch the graph of y and reflect the piece that is negative. y y D : ; : 0 c. Sketch the graph of negative. y 4 and reflect the piece that is y y D : ; : 0 d. Sketch the graph of negative. y 4 and reflect the parts that are y y D : ; : 0
4. 8 9 0 8 9 No solution 48 8 c. 48 if (case ) 4 8 0 invariant point @ ; 4 8 4 8 if (case ) Case 4 8 8 Case 5 4 8 8 5 4 8 8 4 5 is a solution 4 5 is an etraneous root 4 7 7 7 if (case ) 7 7 0 invariant point @ ; 7 7 7 if (case ) 7 7 Case Case 7 7 40 0 is a solution is a solution 7 60 0 D b 4ac 6 4 0 4 0 since D 0, no real solutions for Case
d. 4 b b 4ac 8 0 a if ( ) or ( ) (case ) if ( ) ( ) (case ) invariant points @ 0.8 and 4.8; y opens up
Case 0.8 or 4.8 4 4 4 8 0 4 0 4 4 is an etraneous root is a solution Case ; 0 0.8 4.8 4 4 4 60 0 6 0 0 is a solution 6 6 is an etraneous root