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# 11 x1 t01 01 algebra & indices (2014)

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### 11 x1 t01 01 algebra & indices (2014)

1. 1. Methods In Algebra Like terms can be added or subtracted, unlike terms cannot.
2. 2. Index Laws a m  a n  a mn
3. 3. Index Laws a m  a n  a mn a m  a n  a mn
4. 4. Index Laws a m  a n  a mn a m  a n  a mn a  m n  a mn
5. 5. Index Laws a m  a n  a mn a m  a n  a mn a  m n  a mn a0  1
6. 6. Index Meaning  : top of the fraction
7. 7. Index Meaning  : top of the fraction  : bottom of the fraction
8. 8. Index Meaning  : top of the fraction  : bottom of the fraction x a b power
9. 9. Index Meaning  : top of the fraction  : bottom of the fraction x a b power root
10. 10. Index Meaning  : top of the fraction  : bottom of the fraction x a b power  b xa root OR   x b a
11. 11. Index Meaning  : top of the fraction  : bottom of the fraction x a b power  b xa root OR   x b e.g. (i ) x 3  a
12. 12. Index Meaning  : top of the fraction  : bottom of the fraction x a b power  b xa root OR   x b e.g. (i ) x 3 1  3 x a
13. 13. Index Meaning  : top of the fraction  : bottom of the fraction x a b power  b xa root OR   x b e.g. (i ) x 3 1  3 x a (ii ) a 5b 7 
14. 14. Index Meaning  : top of the fraction  : bottom of the fraction x a b power  b xa root OR   x b e.g. (i ) x 3 1  3 x a (ii ) a 5b 7 a5  7 b
15. 15. 3 (iii ) x  4 a 9b  2  4
16. 16. 3 (iii ) x  4 a 9b  2 4 3a 9  4 2 4x b
17. 17. 3 (iii ) x  4 a 9b  2 4 1 4 (iv) x  3a 9  4 2 4x b
18. 18. 3 (iii ) x  4 a 9b  2 4 1 4 (iv) x  4 x 3a 9  4 2 4x b
19. 19. 3 (iii ) x  4 a 9b  2 4 1 4 (iv) x  2 3 (v ) y  4 x 3a 9  4 2 4x b
20. 20. 3 (iii ) x  4 a 9b  2 4 1 4 4 x 2 3 3 y2 (iv) x  (v ) y  3a 9  4 2 4x b
21. 21. 3 (iii ) x  4 a 9b  2 4 1 4 4 x 2 3 3 y2 (iv) x  (v ) y  3 2 (vi ) x  3a 9  4 2 4x b
22. 22. 3 (iii ) x  4 a 9b  2 4 1 4 4 x 2 3 3 y2 (iv) x  (v ) y  3 2 (vi ) x  x3 3a 9  4 2 4x b
23. 23. 3 (iii ) x  4 a 9b  2 4 1 4 4 x 2 3 3 3a 9  4 2 4x b y2 (iv) x  (v ) y  3 2 (vi ) x  x3  x2 x
24. 24. 3 (iii ) x  4 a 9b  2 4 1 4 (iv) x  2 3 (v ) y  3 2 (vi ) x  4 3 3a 9  4 2 4x b x x2 x3  x2 x x x
25. 25. 3 (iii ) x  4 a 9b  2 4 1 4 4 x 2 3 3 3a 9  4 2 4x b y2 (iv) x  (v ) y  3 2 (vi ) x  x 3  x2 x x x see OR 3 2 x 
26. 26. 3 (iii ) x  4 a 9b  2 4 1 4 4 x 2 3 3 3a 9  4 2 4x b y2 (iv) x  (v ) y  3 2 (vi ) x  x 3  x2 x x x see OR 3 2 x  1 1 2 x think
27. 27. 3 (iii ) x  4 a 9b  2 4 1 4 4 x 2 3 3 3a 9  4 2 4x b y2 (iv) x  (v ) y  3 2 (vi ) x  x 3 see 3 2 x  OR 1 1 2 x x x  x2 x x x think 1 x and x 1 2
28. 28. (vii ) m 27 4 
29. 29. (vii ) m 27 4 64 3  m m
30. 30. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2
31. 31. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2 2
32. 32. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2 2 n6
33. 33. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2 p 500 2 n6
34. 34. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2 p 500 2 n 6 28 q
35. 35. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2 p 500 c 6 c 2 n 6 28 q
36. 36. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2 p 500 c 6 c r 69 2 n 6 28 q
37. 37. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2  2 (ix)    3 2  p 500 c 6 c r 69 2 n 6 28 q
38. 38. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2  2 (ix)    3 2  3     2 2 p 500 c 6 c r 69 2 n 6 28 q
39. 39. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2  2 (ix)    3 2  3     2 9  4 2 p 500 c 6 c r 69 2 n 6 28 q
40. 40. (vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2  2 (ix)    3 2  3     2 9  4 p 500 c 6 c r 69 2 n 6 28 q 2 Exercise 1A; 1c, 2d, 3b, 4d, 5b, 6ad, 7bc, 8a, 9b, 10d, 11cf, 12ac, 13bd, 15, 17, 18* Exercise 6A; 1adgi, 2behj, 3ace, 4ace, 5bdfh, 6ace, 7adgj, 8behj, 9bd