7 Algebra. 7.1 Manipulation of rational expressions. 5x x x x 2 y x xy y. x +1. 2xy. 13x

Size: px

Start display at page:

Download "7 Algebra. 7.1 Manipulation of rational expressions. 5x x x x 2 y x xy y. x +1. 2xy. 13x"

Doreen Morgan
5 years ago
Views:

1 7 Algera 7.1 Manipulation of rational expressions Exercise 7A 1 a x y + 8 x 7x + c 1x x -10 e xy - 8 y f x + 1 g -7x - 5 h - x i xy j x - x 10 k 1 6 l 1 x m 1 n o 1x p x + a 7x + 9 (x +1)(x + ) 11x -10 (x )(x +1) c -1x (x +1)(x + ) 8-10x (x -1)(x +1) e x +1 (x -1)(x -1) x + x - x - 9 a 9x +1 (x +1)(x + ) 1x +19 (x -1)(x +1) c x + x -1 (x +1) x +1 (x -1)(x -1) 5 a x - 1 x + 1 x + 1 x + c x -1 x - x + 1 x - 1 e x + 5 x a x(x -1) x + x(x - 5) x - c x(x - ) x - 5 x(x - 6) x - 7 e x(x - 5) x + f x(x + ) x a x + x + c x + 5 x 7 e (x + 1) f x + 8 a x(x +1) x - x(9x - 8) x - 7 c x(x + ) x -

2 6(x -1) 5x + 6 e 6x(x +1) 8x - x(x - 5) x + 7. Use an manipulation of formulae an expressions Exercise 7B 1 m = gv m = t r = C π = A h 5 l = P - w 6 p = m - 7 a = v - u t 8 = A π 9 a n = W - t t = W n 10 p = k 11 a t = u v u = v + t 1 a m = k n n = k - m 1 r = T 5 1 a w = K 5n n = K - w 5 Exercise 7C 1 a.5 a = c - a 60 a = (s - ut) t a = ac c = + a t = r p + 5 e = ( 1-1) 6 a 5 u = v - as c s = v - u a

3 æ t ö 7 a L = è ç π ø G Check stuents proof 8 a R = D + πr π r = πr - D π c π = D R - r 9 a x = 5 or 5 x = 11+ y c y = x -11 æ 10 a a = T ö è ç ø 11 T = + c - a c (c + ) c = a æ ö è ç T ø - 1 a 1 f = uv u + v 1 x = yz y + z 1 s = r t h = gi 5i -1 c u = fv v - f v = fu u - f 16 k = j t 17 c = 81 a 18 h = 9gi( 1e f ) 7. The factor theorem Exercise 7D a (x + 1)(x )(x + 5) (x + )(x 1)(x + ) c (x + 1)(x + )(x + ) (x )(x )(x 5) e (x + 1)(x )(x ) f (x 1)(x + )(x + ) a x = 1, x =, x = 5 x = 1 c x =, x =, x = x =, x = 1, x = 5 e x =, x = 6, x = 7 f x =, x = 10, x = 5

4 5 1 an 6 a =, = 1, c = 1 7 x, x +, x + 8 x + an x Exam-style questions 1 (x +1)(x + ) (x -1)(x + ) x +1 x (x + ) x = y v = uf u - f 6 C = (πr) = π r Þ C π = π r π = πr = A 7 a f( ) = = 0 (x )(x )(x + ) 8 a f() = = 0 (x ) 9 (x )(x + 5)(x 5) 10 x = 1, or 11 a x = 1 or x = 0, 1 or c x =, or 8 Sequences 8.1 Numer sequences Exercise 8A 1 a 8,, 0: a 6, 8, :a 5 c 0 000, , : multiply y 10 19,, 5: a e 6, 55, 6: a 9 f 05, 115, 65: multiply y g 18,, 6: a

5 h 65, 15, 15 65: multiply y 5 a 16, 6. 7 c 1, 6, 6 e 11, 169 f 11, 1 g, 65 h 78, 108 a 8, 96, 19, 9, 5 c,, 1 8, 5, e 6,, 1 f 19,, 5 g 8, 6, 5 h 0.065, , a 1, : a previous terms 9, 6: next square numer c 7, 76: a previous terms 16, : cue numers 8. The nth term of a sequence Exercise 8B 1 a, 5, 6, 7, 8, 5, 8, 11, 1 c, 8, 1, 18, 9, 1, 17, 1, 5 a c 5 8. The nth term of a linear sequence Exercise 8C 1 a 1, 15, n + 1 5, 9, n + 1 c, 8, 5n +, 8, 6n e 0,, n + f 7,, 7n 5 g 1, 5, n h, 7, n 1 i 17, 0, n 1 a n + 1, 151 n + 5, 105 c 5n, 8 n, 197 e 8n 6, 9 f n +, 5 a r 0th c 100th = 99 a i n + 1 ii 01 iii 101, 5th 5 a i n + 1 ii 01 iii 99 or 101, 9th an 50th c i n + 1 ii 01 iii 100, r i n + 6 ii 06 iii 100, 7th e i n + 5 ii 05 iii 101, th f i 5n + 1 ii 501 iii 101, 0th n + 1 n + 1 Getting closer to ( 0.6) c i (6.p.) ii (6.p.) 6 a (6.p.), (6.p.),5 7, 7 10 i ii c For n, n - 1 n - 1 n n = 7 a Sequence goes up in s; first term is + 9

6 n c Because it ens up as n n 79th 8. The nth term of a quaratic sequence Exercise 8D 1 a 6, 9 8, 51 c, 56 7, 100 e 78, 105 f 109, 18 g, 57 h 178, i 11, 15 j 66, 91 a n² n² + c n² + n n² + e n² + n f n²+ 1 g n² + n + 1 h 5n² i n² + n j n² n a 0, 1, 8 1 6th term n² is positive since n is always positive an n² is positive. n is always positive since when n = 1, n = 1, so first term = 1 an terms are increasing y each time, so n is always positive. Positive ivie y positive is always positive. 5 n n 1 6 a n n c (n )( n ) or 1n² 1n The limiting value of a sequence as n Exercise 8E 1 a 1 c 1 e 0 f a 1 0 c 0 5 e 1 5 f 0 Exam-style questions 1 9, 1; the numers go up in 1s. No, one sequence the numers are always a multiple of three (n); in the other the numers are always one less that a multiple of three (n 1). Hence no term in common. a, 5 1 an 6 The eighth c 8

n + n or n(n + ) 5 Four 9 Pythagoras theorem an trigonometry 9.1 Pythagoras theorem Exercise 9A 1 10. cm 8.5 cm 0.

7 n + n or n(n + ) 5 Four 9 Pythagoras theorem an trigonometry 9.1 Pythagoras theorem Exercise 9A cm 8.5 cm 0.6 cm The square in the first iagram an the two squares in the secon have the same area. 5 a 15 cm 1.7 cm c 6. cm 18. cm 6 a 5 m 6 m c 50 cm 7 There are infinite possiilities, e.g. any multiple of,, 5 such as 6, 8, 10; 9, 1, 15; 1, 16, 0; multiples of 5, 1, 1 an of 8, 15, cm Exercise 9B 1 No. The foot of the laer is aout 6.6 m from the wall. Aout 17 minutes, assuming it travels at the same spee. 1 units a.85 m.8 m (There is only a small ifference.) 5 Yes, ecause + 7 = cm Exercise 9C 1 a. cm.8 cm c 50.0 cm. cm 15.6 cm a

8 The areas are 1 cm an 1.6 cm respectively, so triangle with 6 cm, 6 cm, 5 cm sies has the greater area cm or 0 m 7 a 10 cm 6 cm c 9.6 cm Exercise 9D 1 a i 1. cm ii 1 cm iii 9. cm 15. cm No, 6.6 m is longest length a 0.6 cm 15.0 cm 1. cm 5 a 8.9 m 9 m cm cm 8 a 11. cm 7 cm c 8.06 cm 9 a 50.0 cm 5.8 cm c 8. cm 7.0 cm 9. Trigonometry in right-angle triangles Exercise 9E 1 a c a i 0.57 ii 0.57 i 0.08 ii 0.08 c i 0.91 ii 0.91 Same e i sin 15 is the same as cos 75. ii cos 8 is the same as sin 8 iii sin x is the same as cos (90 x) a c.8 infinite Has values > 1 5 a c a 5, 5, 5 1, 1 1, 5 1 c 7 5, 5, 7 7 a c e 87. f 5.0

9 8 a c e 78.5 f a c e 69.5 f Error message, largest value 1, smallest value 1 Exercise 9F 1 a c. a 5.9 cm 5.75 cm c 1. cm a.57 cm 6.86 cm c 100 cm a 5.1 cm 9.77 cm c 11.7 cm 15.5 cm 5 a c a 5.5 cm 1.8 cm c 1.0 cm 8.6 cm 7 a 5.59 cm 6.6 c 9.91 cm a c a 9.0 cm 7.51 cm c 7.1 cm 8.90 cm 10 a 1.7 cm 8. c 7.0 cm 1. Exercise 9G 1 a c e 67.9 f 0.1 a. 9.8 c. 9.5 e 58.7 f 8.7 a c e 9.7 f 6 g 50. h 51. i 18 j.8 a Sies of right-han triangle are sine an cosine Pythagoras theorem c Stuents shoul check the formulae. Exercise 9H 1 65 The safe limits are etween 1.0 m an.05 m. The laer will reach etween 5.6 m an 5.90 m up the wall. 1 a 8 km 75 km 5 km

10 6 170 km 7 One way is stan opposite a feature, such as a tree, on the opposite ank, move a measure istance, x, along your ank an measure the angle, θ, etween your ank an the feature. With of river is x tan θ. This of course requires measuring equipment! An alternative is to walk along the ank until the angle is 5 (if that is possile). This angle is easily foun y foling a sheet of paper. This way an angle measurer is not require. Exercise 9I km 9 m a 156 m 00 No. The new angle of epression is tan -1 = an half of 5 is a m 6 a 1.5 m 17.8 m 7 1. m The angle is 16 so Cara is not quite correct. Exercise 9J a 5 cm 58.6 c 0.5 cm a.6 m 75.5 c 7. a c 1.5 cm a It is.6 ; use triangle XDM where M is the mipoint of BD; triangle DXB is isosceles, as X is over the point where the iagonals of the ase cross; the length of DB is cosine of the require angle is The sine rule an the cosine rule Exercise 9K a.6 m 8.05 cm c 19. cm a c an the

11 .7 m a i 0 ii m m 6. m km 8 1 Exercise 9L 1 a 7.71 m 9.1 cm c 7. cm a i 76. ii 15.1 iii 90 Right-angle triangle a 10.7 cm 1.7 c cm 58. km at km ; the largest angle is opposite the longest sie. Exercise 9M 1 a 8.60 m 90 c 7. cm 6.9 e 7.5 f 6. cm 7 cm 11.1 km a A = 90 ; this is Pythagoras theorem A is acute c A is otuse 5 1 m Exam-style questions 1 8 cm 8 cm a cm or 11. cm cm

12 cm 8 a 1. cm 9 c km m cm

sin A cos A Georgia Milestones Geometry EOC Study/Resource Guide for Students and Parents Page 73 of 182

sin A cos A Georgia Milestones Geometry EOC Study/Resource Guide for Students and Parents Page 73 of 182 UNIT 3: RIGHT TRIANGLE TRIGONOMETRY This unit investigates the properties of right triangles. The trigonometric ratios sine, cosine, an tangent along with the Pythagorean theorem are use to solve right