# ssm00 - Chapter 0 Preliminaries Chapter 0 Section 0.1 5. 3...

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Chapter 0 Preliminaries 1 Chapter 0 Section 0.1 5. 32 1 1 39 3 x x x +< < < (, 3 ) −∞ 7. 237 24 2 x x x −< <− () ,2 −∞ − 9. 43 6 2 3 x x x >− 2 , 3  −∞   11. 41 7 36 x x ≤+< ≤< [3, 6) 13. 222 3 421 1 2 2 1 2 2 x x x x −< − < −<− < >> −<< 1 2 15. 2 340 (4 ) (1 ) 0 xx +−> +− > –4 (x+4) 0+ 1 (x-1) 1 –4 (x+4)(x-1) 0 +– x < –4 or x > 1 (,4 )( 1 ,) −∞ − 17. 2 60 (3 ) (2 )0 −−< −+ < 3 (x-3) –2 (x+2) –2 3 (x-3)(x+2) 00 + (–2, 3) 19. 2 2 2 34 0 4 3 x x x +> This is always true. (,) −∞ ∞ 21. x 43 4 17 x x −< < (–1, 7) 23. 31 x 13 1 42 x x x x −<−<− << (2, 4) 25. 212 x 21 2o r212 23 2 1 or 22 +<− > > , , −∞ − 27. 2 0 2 x x + > –2 (x+2) 2 (x-2) 2 –2 x+2 x-2 x x < –2 or x > 2 (,2 2 −∞ −

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Chapter 0 Preliminaries 2 29. 2 2 2 2 0 (4 ) (2 ) (1 ) 0 ) xx x x −− > + −+ > + 2 (x-2) 0+ –1 (x+1) –4 (x+4) 2 + 2 –4 –1 (x-2)(x+1) (x+4) 2 00 x + - ++ x <–4or–4< x <–1or x >2 (,4 )( 4 ,1 2 ,) ∞− ∪−−∪ ∞ 31. 3 8 0 ) x x < + 0 (–8x) 0 +– (x+1) 3 0 –+ –1 0 –1 0 x -+ –8x (x+1) 3 x x >0 (,1 0 −∞ − 33. 22 {(2, 1), (4, 4)} 2) (4 1) 49 13 d =− + =+ = 35. {( 1, 2), (3, 2)} [ 3 (1 ) ] [2 (2 ) ] 16 0 4 d = 37. {(0 ,2) ,( 2 ,6)} ( 2 0) (6 2) 41 6 20 d −= + = 39. {(1, 1), (3, 4)} (3 1) (4 1) 13 d + = {(1, 1), (0, 6)} (0 1) (6 1) 12 5 2 6 d + =+= 2 2 {(3, 4), (0, 6)} (0 3) (6 4) 94 1 3 d + = 2 3 ) (1 3 ) 6 ) += This statement is true, so yes, this is a right triangle. 41. {( 2, 3), (2, 9)} [2 ( 2)] (9 3) 16 36 52 d + = {( 2, 3), ( 4, 13)} [ 4 ( 2)] (13 3) 0 0 1 0 4 d = + = {(2, 9), ( 4, 13)} ( 4 2) 9) 36 16 52 d + = 2 (5 2 ) (5 2 ) 0 4 ) This statement is true, so yes, this is a right triangle. 43. x y 01 (3, 3200) (2, 2450) (1, 1800) (0, 1250) 23 5 4 0 500 1000 1500 2000 2500 3000 3500 The y -values are increasing by 550, 650, 750, … The next population is 3200 + 850 = 4050. 45. x y 0123 (3, 3910) (2, 3960) (1, 3990) (0, 4000) 5 4 3900 3910 3920 3930 3940 3950 3960 3970 3980 3990 4000 4010 The y -values are decreasing by 10, 30, 50, … The next population is 3910 – 70 = 3840. 47. When a calculation is done using a finite number of digits, the result can only contain a finite number of digits. Therefore, the result can be written as a quotient of two integers (where the denominator is a power of 10), so the result must be rational. 49. () 71 2 7 1 2 .013 or 1.3%
Chapter 0 Preliminaries 3 51. P win % .551 .568 .587 .593 .404 .414 .538 .556 .605 .615 Section 0.2 5. (1, 2) and (3, 6): 62 4 2 31 2 m == = (3, 6) and (0, 0): 06 6 2 03 3 m −− = Since the slopes are the same, the points must be colinear. 7. (2, 1) and (0, 2): 21 1 1 02 2 2 m = (0, 2) and (4, 0): 1 40 4 2 m = Since the slopes are the same, the points must be colinear. 9. (3, 1) and (4, 4): 41 3 3 43 1 m = (4, 4) and (5, 8): 84 4 4 54 1 m = Since the slopes are not equal, the points are not colinear. 11. (4, 1) and (3, 2): 1 34 1 m = (3, 2) and (1, 3): 32 1 1 13 2 2 m = Since the slopes are not equal, the points are not colinear. 13. 2 m = 15. m 17. 1(6 ) 55 2 2 m −−− = 19. 0.4 ( 1.4) 1.0 5 1.1 0.3 1.4 7 m = 21. 2 (2, 5) P = 0 y x 2( 1) 3 yx =− + 23. 2 (2, 2) P = 0 y x 2( 4 − + 25. 2 (0, 1) P = 0 y x 1 1 y = 27. 2 (4, 2) P = y x 1 1 1 (2 ) 1 2 +

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Chapter 0 Preliminaries 4 29.
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## This note was uploaded on 04/19/2008 for the course MATH 105 taught by Professor Schroder during the Spring '08 term at Millersville.

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ssm00 - Chapter 0 Preliminaries Chapter 0 Section 0.1 5. 3...

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