Chapter 6 indices
jasimahpuari
718 views
13 slides
Aug 12, 2014
Slide 1 of 13
1
2
3
4
5
6
7
8
9
10
11
12
13
About This Presentation
indices
Slide Content
Module PMR
CHAPTER 6 : INDICES
NUMBERS IN INDEX FORM
Example:
Express in the form
of repeated
multiplication.
aaaaa ´´´=
4
1)
5
h 2) ()
3
h-
3)
2
÷
ø
ö
ç
è
æ
e
d
4)
4
5 5) ()
3
1-
6)
4
3
2
÷
ø
ö
ç
è
æ
7)
2
4
3
÷
ø
ö
ç
è
æ
-
Example:
Write in index
notation.
h x h x h x h x h x h
=
8) g x g x g x g9).
()()()
()()dd
ddd
-´-
´-´-´-
10) ÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
n
e
n
e
n
e
11) 7 x 7 x7 x 7 x 712)
(-11)(-11)(-11)
13) ÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
7
2
7
2
7
2
7
2
14) ÷
ø
ö
ç
è
æ
-÷
ø
ö
ç
è
æ
-
9
7
9
7
Example:
Evaluate
222222
5
´´´´=
= 32
15)
3
10 16) ()
4
3-
17)
2
4
3
÷
ø
ö
ç
è
æ
18)
3
4
1
÷
ø
ö
ç
è
æ-
19)
3
2
1
3÷
ø
ö
ç
è
æ 20) ()
3
4.0 21) ( )
2
05.0-
LAWS OF INDICES
Indices 64
CHAPTER 6 : INDICES
NUMBERS IN INDEX FORM
Example:
Express in the form
of repeated
multiplication.
aaaaa ´´´=
4
1)
5
h 2) ()
3
h-
3)
2
÷
ø
ö
ç
è
æ
e
d
4)
4
5 5) ()
3
1-
6)
4
3
2
÷
ø
ö
ç
è
æ
7)
2
4
3
÷
ø
ö
ç
è
æ
-
Example:
Write in index
notation.
h x h x h x h x h x h
=
8) g x g x g x g9).
()()()
()()dd
ddd
-´-
´-´-´-
10) ÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
n
e
n
e
n
e
11) 7 x 7 x7 x 7 x 712)
(-11)(-11)(-11)
13) ÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
7
2
7
2
7
2
7
2
14) ÷
ø
ö
ç
è
æ
-÷
ø
ö
ç
è
æ
-
9
7
9
7
Example:
Evaluate
222222
5
´´´´=
= 32
15)
3
10 16) ()
4
3-
17)
2
4
3
÷
ø
ö
ç
è
æ
18)
3
4
1
÷
ø
ö
ç
è
æ-
19)
3
2
1
3÷
ø
ö
ç
è
æ 20) ()
3
4.0 21) ( )
2
05.0-
LAWS OF INDICES
Indices 64
Module PMR
LAW 1
nmnm
aaa
+
=´
Example:
Simplify
35
bb´
35+
=b
8
b=
1)
62
gg´ 2)
693
hhh ´´ 3) ()()
32
ee-´-
Example:
Simplify
43
22´
43
2
+
=
7
2=
4)
62
55´ 5)
683
666 ´´ 6)()()
45
33-´-
Example:
Simplify
32
43 bb´
( )
32
43
+
´= b
5
12b=
7)
62
54 yy´ 8)
43
310 ss´
9)
45
8
3
2 kk´
Example:
Simplify
7265
94 shsh ´´
( )
7675
94
++
´= sh
1312
36sh=
10)
232
37 mbm ´´
11)
234
9
3
2
vvv ´´
12) papa ´´-
443
36
LAW 2
nmnm
aaa
-
=¸
Example:
Simplify
47
jj¸
47-
=j
3
j=
1)
49
nn¸
2)
2
5
a
a 3)
34
4tt¸
Example:
Simplify 4
5
¸4
2
= 4
5 - 2
= 4
3
4) 5
4
¸5
3
5)
2
4
7
7 6)
57
1010
-
¸
Example:
Simplify
35
412 aa¸
35
4
12
-
=a
2
3a=
7)
45
540 gg¸
8)
2
8
4
20
y
y 9)
510
2718 nn¸
Example:
Simplify
10)
85
210
-
¸uu
11)
2
4
2
18
-
w
w
12)
48
9
2
3
-
¸mm
Indices 65
LAW 1
nmnm
aaa
+
=´
Example:
Simplify
35
bb´
35+
=b
8
b=
1)
62
gg´ 2)
693
hhh ´´ 3) ()()
32
ee-´-
Example:
Simplify
43
22´
43
2
+
=
7
2=
4)
62
55´ 5)
683
666 ´´ 6)()()
45
33-´-
Example:
Simplify
32
43 bb´
( )
32
43
+
´= b
5
12b=
7)
62
54 yy´ 8)
43
310 ss´
9)
45
8
3
2 kk´
Example:
Simplify
7265
94 shsh ´´
( )
7675
94
++
´= sh
1312
36sh=
10)
232
37 mbm ´´
11)
234
9
3
2
vvv ´´
12) papa ´´-
443
36
LAW 2
nmnm
aaa
-
=¸
Example:
Simplify
47
jj¸
47-
=j
3
j=
1)
49
nn¸
2)
2
5
a
a 3)
34
4tt¸
Example:
Simplify 4
5
¸4
2
= 4
5 - 2
= 4
3
4) 5
4
¸5
3
5)
2
4
7
7 6)
57
1010
-
¸
Example:
Simplify
35
412 aa¸
35
4
12
-
=a
2
3a=
7)
45
540 gg¸
8)
2
8
4
20
y
y 9)
510
2718 nn¸
Example:
Simplify
10)
85
210
-
¸uu
11)
2
4
2
18
-
w
w
12)
48
9
2
3
-
¸mm
Indices 65
Module PMR
()
9
36
36
36
2
2
6
12
612
z
z
z
zz
=
=
=
¸
+
--
-
LAW 3 ()
nm
n
m
aa
´
=
Example:
Simplify
()
15
53
5
3
h
h
h
=
=
´
1) ()
2
4
k 2) ()
5
2
-
g 3) ()
2
7-
v
Example:
Simplify
()
12
42
4
2
3
3
3
=
=
´
4) ()
3
5
4 5) ()
5
2
6
-
6) ()
3
5
7
-
Example:
Simplify
( )
2015
5453
5
43
2
2
2
m
m
m
=
=
´´
7) ( )
2
34
ed 8) ( )
4
23
52
-
9) ( )
2
64
3
-
g
LAW 4
n
n
a
a
1
=
-
Example:
Simplify
r
l
-
r
l
1
=
1)
k
c
-
2)
p
m
-
3 3)
t
dv
-
Example:
Simplify
3
2
-
=
3
2
1
4)
3
4
-
5)
1
7
-
6)
2
3
-
Indices 66
()
9
36
36
36
2
2
6
12
612
z
z
z
zz
=
=
=
¸
+
--
-
LAW 3 ()
nm
n
m
aa
´
=
Example:
Simplify
()
15
53
5
3
h
h
h
=
=
´
1) ()
2
4
k 2) ()
5
2
-
g 3) ()
2
7-
v
Example:
Simplify
()
12
42
4
2
3
3
3
=
=
´
4) ()
3
5
4 5) ()
5
2
6
-
6) ()
3
5
7
-
Example:
Simplify
( )
2015
5453
5
43
2
2
2
m
m
m
=
=
´´
7) ( )
2
34
ed 8) ( )
4
23
52
-
9) ( )
2
64
3
-
g
LAW 4
n
n
a
a
1
=
-
Example:
Simplify
r
l
-
r
l
1
=
1)
k
c
-
2)
p
m
-
3 3)
t
dv
-
Example:
Simplify
3
2
-
=
3
2
1
4)
3
4
-
5)
1
7
-
6)
2
3
-
Indices 66
Module PMR
8
1
=
Example:
Simplify
5
8
4
1
4´
=
58
44
-
´
=
()58
4
-+
=
58
4
-
=
3
4
= 64
7)
2
3
8
1
8´ 8) ( )
2
2
3
10
10
1
´ 9)
()
3
2
81
m
m´
LAW 5 1
0
=a
Example:
12
0
=
1)
0
3
2
÷
ø
ö
ç
è
æ
=
2) ()
0
3- = 3) ()
0
5.4=
LAW 6 FRACTIONAL INDICES
Example:
Rewrite by using
the root and power
symbol
9
3
2
3
9=
1) 3
1
64
2) 5
1
32
3) 3
2
8
Example:
Write in the form of
index
5
3
2
2
3
5=
1)
3
8 2)
3
5
10 3) 16
Example:
Evaluate
1) 3
1
27
2) 5
2
32
3) 4
3
81
Indices 67
8
1
=
Example:
Simplify
5
8
4
1
4´
=
58
44
-
´
=
()58
4
-+
=
58
4
-
=
3
4
= 64
7)
2
3
8
1
8´ 8) ( )
2
2
3
10
10
1
´ 9)
()
3
2
81
m
m´
LAW 5 1
0
=a
Example:
12
0
=
1)
0
3
2
÷
ø
ö
ç
è
æ
=
2) ()
0
3- = 3) ()
0
5.4=
LAW 6 FRACTIONAL INDICES
Example:
Rewrite by using
the root and power
symbol
9
3
2
3
9=
1) 3
1
64
2) 5
1
32
3) 3
2
8
Example:
Write in the form of
index
5
3
2
2
3
5=
1)
3
8 2)
3
5
10 3) 16
Example:
Evaluate
1) 3
1
27
2) 5
2
32
3) 4
3
81
Indices 67
Module PMR
()
2
2
4
2
1
2
2
1
=
=
Example:
Simplify
2
3
6
3
2
3
4
3
2
3
4
9
9
9
99
=
=
=
´
+
1) 3
5
3
2
1010¸
2) ()
4
2
1
6
gg´
3) ( )
3
2
1
4 --
¸pp
COMBINATION OF LAWS
Example:
Simplify
2
2
2
2
0
936
936
=
=
=
¸´
-+
b
b
bbb
1)
317 -
´¸ggg
2)
4
59
m
mm´ 3)
435
232 ggg ¸´
Simplify
( )
33
3612
31622
31
2
3
4
4
2
2
yx
yx
yxyx
yxxy
=
=
¸=
¸
-+
-
-
4) ( )
1
3
2
2
-
¸xyyx 5) ( )
24
2
4
3
--
´yxxy 6) ( ) yxxy
3
2
2
¸
Evaluate
2
1
2
1
8
1
3
1
3
3
1
=
ú
ú
û
ù
ê
ê
ë
é
÷
ø
ö
ç
è
æ
=
÷
ø
ö
ç
è
æ
7)
2
1
25
1
÷
ø
ö
ç
è
æ
8)
2
1
9
4
÷
ø
ö
ç
è
æ
9)
3
1
27
8
÷
ø
ö
ç
è
æ
Evaluate 10)
322
284 ¸´ 11) ()
3
27
93´
-
12) ()
5
1
2
1
4
3242 ¸´
Indices 68
()
2
2
4
2
1
2
2
1
=
=
Example:
Simplify
2
3
6
3
2
3
4
3
2
3
4
9
9
9
99
=
=
=
´
+
1) 3
5
3
2
1010¸
2) ()
4
2
1
6
gg´
3) ( )
3
2
1
4 --
¸pp
COMBINATION OF LAWS
Example:
Simplify
2
2
2
2
0
936
936
=
=
=
¸´
-+
b
b
bbb
1)
317 -
´¸ggg
2)
4
59
m
mm´ 3)
435
232 ggg ¸´
Simplify
( )
33
3612
31622
31
2
3
4
4
2
2
yx
yx
yxyx
yxxy
=
=
¸=
¸
-+
-
-
4) ( )
1
3
2
2
-
¸xyyx 5) ( )
24
2
4
3
--
´yxxy 6) ( ) yxxy
3
2
2
¸
Evaluate
2
1
2
1
8
1
3
1
3
3
1
=
ú
ú
û
ù
ê
ê
ë
é
÷
ø
ö
ç
è
æ
=
÷
ø
ö
ç
è
æ
7)
2
1
25
1
÷
ø
ö
ç
è
æ
8)
2
1
9
4
÷
ø
ö
ç
è
æ
9)
3
1
27
8
÷
ø
ö
ç
è
æ
Evaluate 10)
322
284 ¸´ 11) ()
3
27
93´
-
12) ()
5
1
2
1
4
3242 ¸´
Indices 68
Module PMR
()
()
32
1
2
22
22
42
5
49
2
29
2
3
3
=
=
´=
´=
´
-
-
-
-
13) 5
3
4
1
3
1
32168 ´´
14) 23
2
327
-
¸
15) ( )
2
2
3
999
-
´¸ 16) ()
5
0
6
rr¸
17)
( )
2
2
52
4
4
ab
ba 18) ()
2
3
24
666 ¸´
-
19) ()
5
2
3
xx¸
-
-
20) Given .2166
2
=
-x
Find x.
21) Given that .1255
3
=
-x
Find x.
22) If
64
1
3
=
-
y . Find y.
Common Errors
Indices 69
()
()
32
1
2
22
22
42
5
49
2
29
2
3
3
=
=
´=
´=
´
-
-
-
-
13) 5
3
4
1
3
1
32168 ´´
14) 23
2
327
-
¸
15) ( )
2
2
3
999
-
´¸ 16) ()
5
0
6
rr¸
17)
( )
2
2
52
4
4
ab
ba 18) ()
2
3
24
666 ¸´
-
19) ()
5
2
3
xx¸
-
-
20) Given .2166
2
=
-x
Find x.
21) Given that .1255
3
=
-x
Find x.
22) If
64
1
3
=
-
y . Find y.
Common Errors
Indices 69
Module PMR
Errors Correct Steps
1.
22
22
-
´
=
()22
2
-+
=
0
2
= 2
1.
22
22
-
´
=
()22
2
-+
=
0
2
= 1
2.
24
55
-
¸
=
24
5
-
=
2
5
2.
24
55
-
¸
=
()24
5
--
=
24
5
+
=
6
5
3.
4
52
-
´
k
kk
=
452-+
k
=
3
k
3.
4
52
-
´
k
kk
=
)4(52--+
k
=
47+
k
=
11
k
Questions based on PMR Format
1. Evaluate
3
1
2
3
16
-
÷
÷
ø
ö
ç
ç
è
æ
2. Evaluate ( )
2
3
284 -
cba
3. Find the value of 3
2
5
3
832
-
¸
4. Given that 2
1
3
4
2
8127¸=k
.Find the value of .k
Indices 70
Errors Correct Steps
1.
22
22
-
´
=
()22
2
-+
=
0
2
= 2
1.
22
22
-
´
=
()22
2
-+
=
0
2
= 1
2.
24
55
-
¸
=
24
5
-
=
2
5
2.
24
55
-
¸
=
()24
5
--
=
24
5
+
=
6
5
3.
4
52
-
´
k
kk
=
452-+
k
=
3
k
3.
4
52
-
´
k
kk
=
)4(52--+
k
=
47+
k
=
11
k
Questions based on PMR Format
1. Evaluate
3
1
2
3
16
-
÷
÷
ø
ö
ç
ç
è
æ
2. Evaluate ( )
2
3
284 -
cba
3. Find the value of 3
2
5
3
832
-
¸
4. Given that 2
1
3
4
2
8127¸=k
.Find the value of .k
Indices 70
Module PMR
5. If 343
47
=´
-
aa , find the value of .a
6. Simplify 6
5
2
1
23 pp´
7. Simplify
27
5
33
3
-
-
´
8. Simplify
372
4
-
´´ aab
a
9. Simplify
()
5
2
1
432
prq
rqp
´
´´
-
-
10. Simplify
34
25
-
-
´
´´
km
mkk
PMR past year questions
2004
1.Given that 162
2
=
-x
, calculate the value of x.
( 2 marks )
Indices 71
5. If 343
47
=´
-
aa , find the value of .a
6. Simplify 6
5
2
1
23 pp´
7. Simplify
27
5
33
3
-
-
´
8. Simplify
372
4
-
´´ aab
a
9. Simplify
()
5
2
1
432
prq
rqp
´
´´
-
-
10. Simplify
34
25
-
-
´
´´
km
mkk
PMR past year questions
2004
1.Given that 162
2
=
-x
, calculate the value of x.
( 2 marks )
Indices 71
Module PMR
2.Simplify ( )()
95
3
2
4
2
2 kmkmk ¸´ .
( 3 marks )
2005
1.Evaluate
3
2
2
1
2
1
8
123´
.
( 3 marks )
2.Given ()(),333
212 xx
=
-
calculate the value of x.
( 2 marks )
2006
1.Simplify
2
4
-
´
k
kk
.
( 2 marks )
2.Find the value of 2
3
2
1
2
2183 ´´
.
( 3 marks )
Indices 72
2.Simplify ( )()
95
3
2
4
2
2 kmkmk ¸´ .
( 3 marks )
2005
1.Evaluate
3
2
2
1
2
1
8
123´
.
( 3 marks )
2.Given ()(),333
212 xx
=
-
calculate the value of x.
( 2 marks )
2006
1.Simplify
2
4
-
´
k
kk
.
( 2 marks )
2.Find the value of 2
3
2
1
2
2183 ´´
.
( 3 marks )
Indices 72
Module PMR
2007
1.a). Find the value of 2
1
2
3
55¸
.
b). Simplify ( )
2
3
4
hhg´.
( 3 marks )
2008
1.Simplify
4
52
m
mm´
-
.
( 2 marks )
2.Find the value of
a).
13
22
-
¸
b). ( )
3
1
63
32´
-
( 3 marks )
CHAPTER 6 : INDICES
ANSWERS
1. hhhhh ´´´´ 2. ()()()hhh -´-´-
3. ÷
ø
ö
ç
è
æ
´÷
ø
ö
ç
è
æ
e
d
e
d 4. 5555 ´´´
Indices 73
2007
1.a). Find the value of 2
1
2
3
55¸
.
b). Simplify ( )
2
3
4
hhg´.
( 3 marks )
2008
1.Simplify
4
52
m
mm´
-
.
( 2 marks )
2.Find the value of
a).
13
22
-
¸
b). ( )
3
1
63
32´
-
( 3 marks )
CHAPTER 6 : INDICES
ANSWERS
1. hhhhh ´´´´ 2. ()()()hhh -´-´-
3. ÷
ø
ö
ç
è
æ
´÷
ø
ö
ç
è
æ
e
d
e
d 4. 5555 ´´´
Indices 73
Module PMR
5. ()()()111 -´-´-
6. ÷
ø
ö
ç
è
æ
´÷
ø
ö
ç
è
æ
´÷
ø
ö
ç
è
æ
´÷
ø
ö
ç
è
æ
3
2
3
2
3
2
3
2
7. ÷
ø
ö
ç
è
æ
-´÷
ø
ö
ç
è
æ
-
4
3
4
3 8.
4
g
9. ()
5
d-
10.
3
÷
ø
ö
ç
è
æ
n
e 11.
5
7 12. ( )
3
11-
13.
4
7
2
÷
ø
ö
ç
è
æ
14.
2
9
7
÷
ø
ö
ç
è
æ
-
15. 1000 16. 81
17.
16
9
18.
64
1
- 19.
8
343
or
8
7
42
20. 0.064
21. 0.0025
LAW 1
1.
8
g 2.
18
h 3. ()
5
e- 4.
8
5
5.
17
6 6. ()
9
3- 7.
8
20y 8.
7
30s
9.
9
4
3
k
10.
34
21bm 11. ()
3
5
7
-
12.
57
18pa-
LAW 2
1.
5
n 2.
3
a 3. t4 4. 5
5. 7
2
6.
12
10 7. g8 8.
6
5y
9.
5
3
2
n
10. 5u
13
11.
6
9w
12.
12
2
27
m
LAW 3
1.
8
k 2.
10-
g 3.
14-
v
4.
15
4 5.
10
6
-
6.
15
7
-
7.
68
ed 8.
812
52
--
9.
128
3
-
g
LAW 4
1.
k
c
1
2.
p
m
3
3.
t
v
d
4.
3
4
1
5.
7
1
6.
2
3
1
7. 8 8. 10 9.
2
m
LAW 5
1. 1 2. 1 3. 1
LAW 6
1.
3
64 2.
5
32 3.
2
3
8
1. 3
1
8
2. 5
3
10
3. 2
1
16
1. 3 2. 4 3. 27
4.
10
1 5.
7
g 6. p
COMBINATION OF LAWS
Indices 74
5. ()()()111 -´-´-
6. ÷
ø
ö
ç
è
æ
´÷
ø
ö
ç
è
æ
´÷
ø
ö
ç
è
æ
´÷
ø
ö
ç
è
æ
3
2
3
2
3
2
3
2
7. ÷
ø
ö
ç
è
æ
-´÷
ø
ö
ç
è
æ
-
4
3
4
3 8.
4
g
9. ()
5
d-
10.
3
÷
ø
ö
ç
è
æ
n
e 11.
5
7 12. ( )
3
11-
13.
4
7
2
÷
ø
ö
ç
è
æ
14.
2
9
7
÷
ø
ö
ç
è
æ
-
15. 1000 16. 81
17.
16
9
18.
64
1
- 19.
8
343
or
8
7
42
20. 0.064
21. 0.0025
LAW 1
1.
8
g 2.
18
h 3. ()
5
e- 4.
8
5
5.
17
6 6. ()
9
3- 7.
8
20y 8.
7
30s
9.
9
4
3
k
10.
34
21bm 11. ()
3
5
7
-
12.
57
18pa-
LAW 2
1.
5
n 2.
3
a 3. t4 4. 5
5. 7
2
6.
12
10 7. g8 8.
6
5y
9.
5
3
2
n
10. 5u
13
11.
6
9w
12.
12
2
27
m
LAW 3
1.
8
k 2.
10-
g 3.
14-
v
4.
15
4 5.
10
6
-
6.
15
7
-
7.
68
ed 8.
812
52
--
9.
128
3
-
g
LAW 4
1.
k
c
1
2.
p
m
3
3.
t
v
d
4.
3
4
1
5.
7
1
6.
2
3
1
7. 8 8. 10 9.
2
m
LAW 5
1. 1 2. 1 3. 1
LAW 6
1.
3
64 2.
5
32 3.
2
3
8
1. 3
1
8
2. 5
3
10
3. 2
1
16
1. 3 2. 4 3. 27
4.
10
1 5.
7
g 6. p
COMBINATION OF LAWS
Indices 74
Module PMR
1.
3
g 2.
10
m 3.
4
3g 4.
45
8yx
5.
62
9yx
-
6.
31
yx
-
7.
5
1
5
1
=
-
8.
3
2
9.
3
2 10. 128 11. 243 12. 8
13. 32 14. 81 15. 81 16.
5-
r
17.
83
4ba 18. 1 19. x 20. 5=x
21. 6 22. 4
Questions Based on PMR Format
1.
4
1 2.
3126 -
cba 3. 32 4. 3
5. 7
6. 3
4
6p 7.
10
3
1 8.
2-
b
9.
qr
p
10.
2
1
km
PMR QUESTIONS
YEAR SOLUTION AND MARK SCHEME
SUB
MARK
FULL
MARK
2004
42
22 =
-x
x or 42=-x
6=x
1
1 2
2004
956844
2 kmkkm ¸´
96854
16
-+-
km
51
16km
-
1
1
1 3
2005
()
()
2
3
4
6
2
6
2
6
8
36
2
1
3
2
3
2
1
2
3
2
2
1
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
1
1
1
3
2005 212 +=-xx
122 +=-xx
3=x
1
1 2
Indices 75
1.
3
g 2.
10
m 3.
4
3g 4.
45
8yx
5.
62
9yx
-
6.
31
yx
-
7.
5
1
5
1
=
-
8.
3
2
9.
3
2 10. 128 11. 243 12. 8
13. 32 14. 81 15. 81 16.
5-
r
17.
83
4ba 18. 1 19. x 20. 5=x
21. 6 22. 4
Questions Based on PMR Format
1.
4
1 2.
3126 -
cba 3. 32 4. 3
5. 7
6. 3
4
6p 7.
10
3
1 8.
2-
b
9.
qr
p
10.
2
1
km
PMR QUESTIONS
YEAR SOLUTION AND MARK SCHEME
SUB
MARK
FULL
MARK
2004
42
22 =
-x
x or 42=-x
6=x
1
1 2
2004
956844
2 kmkkm ¸´
96854
16
-+-
km
51
16km
-
1
1
1 3
2005
()
()
2
3
4
6
2
6
2
6
8
36
2
1
3
2
3
2
1
2
3
2
2
1
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
1
1
1
3
2005 212 +=-xx
122 +=-xx
3=x
1
1 2
Indices 75
Module PMR
2006
7
)2(14
k
k
--+
1
1 2
2006
( )
()
2
1
2
1
2
1
22
2
3
2
1
2
2233
2293
´´´
´´´
108
427
23
23
´
´
1
1
1
3
2007 a)
1
55or
b)
123
3123
gh
horgh
1
1
1
3
2008
m
orm
mormormorm
1
2
45263
-
-+--
1
1
2
2008 a) 162
4
=
b) =
´´-
´
3
1
6
3
1
3
32
4
9
9
2
1
32
21
=´
´
-
1
1
1
3
Indices 76
2006
7
)2(14
k
k
--+
1
1 2
2006
( )
()
2
1
2
1
2
1
22
2
3
2
1
2
2233
2293
´´´
´´´
108
427
23
23
´
´
1
1
1
3
2007 a)
1
55or
b)
123
3123
gh
horgh
1
1
1
3
2008
m
orm
mormormorm
1
2
45263
-
-+--
1
1
2
2008 a) 162
4
=
b) =
´´-
´
3
1
6
3
1
3
32
4
9
9
2
1
32
21
=´
´
-
1
1
1
3
Indices 76
Tags
Categories
GeneralDownload
Download Slideshow Get the original presentation fileQuick Actions
Statistics
- Views 718
- Slides 13
- Favorites 1
- Age 4132 days
Related Slideshows
22
Pray For The Peace Of Jerusalem and You Will Prosper
RodolfoMoralesMarcuc
33 views
26
Don_t_Waste_Your_Life_God.....powerpoint
chalobrido8
36 views
31
VILLASUR_FACTORS_TO_CONSIDER_IN_PLATING_SALAD_10-13.pdf
JaiJai148317
33 views
14
Fertility awareness methods for women in the society
Isaiah47
30 views
35
Chapter 5 Arithmetic Functions Computer Organisation and Architecture
RitikSharma297999
29 views
5
syakira bhasa inggris (1) (1).pptx.......
ourcommunity56
30 views