# 11 x1 t01 01 algebra & indices (2014)

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Information about 11 x1 t01 01 algebra & indices (2014)
Education

Published on January 29, 2014

Author: nsimmons

Source: slideshare.net

Methods In Algebra Like terms can be added or subtracted, unlike terms cannot.

Index Laws a m  a n  a mn

Index Laws a m  a n  a mn a m  a n  a mn

Index Laws a m  a n  a mn a m  a n  a mn a  m n  a mn

Index Laws a m  a n  a mn a m  a n  a mn a  m n  a mn a0  1

Index Meaning  : top of the fraction

Index Meaning  : top of the fraction  : bottom of the fraction

Index Meaning  : top of the fraction  : bottom of the fraction x a b power

Index Meaning  : top of the fraction  : bottom of the fraction x a b power root

Index Meaning  : top of the fraction  : bottom of the fraction x a b power  b xa root OR   x b a

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

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

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 

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

3 (iii ) x  4 a 9b  2  4

3 (iii ) x  4 a 9b  2 4 3a 9  4 2 4x b

3 (iii ) x  4 a 9b  2 4 1 4 (iv) x  3a 9  4 2 4x b

3 (iii ) x  4 a 9b  2 4 1 4 (iv) x  4 x 3a 9  4 2 4x b

3 (iii ) x  4 a 9b  2 4 1 4 (iv) x  2 3 (v ) y  4 x 3a 9  4 2 4x b

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

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

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

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

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

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 

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

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

(vii ) m 27 4 

(vii ) m 27 4 64 3  m m

(vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2

(vii ) m 27 4 64 3  m m 1 7 1 6 500  28 6 69 (viii ) n p q c r  2 2

(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

(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

(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

(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

(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

(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

(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

(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

(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

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