Formulae v2

8/2/2019 Formulae v2

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[ Baroka ] [2010] [ MATHEMATICS FORMULAE ] 1

(1) NUMBERS

(a) Types of numbers

Whole numbers Eg. : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,

Integer numbers Eg. : , 5, 4, 3, 2, 1, 0, 1, 2, 3, 4,

Odd numbers Eg. : 1, 3, 5, 7, 9, 11, 13, 15, 17,

Even numbers Eg. : 2, 4, 6, 8, 10, 12, 14, 16, 18,

Prime numbers Eg. : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, ...

Perfect square numbers Eg. : 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, ...

Cube numbers Eg. : 1, 8, 27, 64, 125, 216, 343, 512, ...

(2) POLYGONS / ANGLES

(a) Triangles

Equilateral Triangles Isosceles Triangles Right-angled Triangles Scalene Triangles

a = 180 2b

b =2

180 a

a + b= 90

a = 90 b

b = 90 a

c = a + b

(b) Rhombus

All sides are equal in length.

Opposite sides are parallel.

Opposite angles are equal in size.

Diagonal bisect each other in right angle.

a + b = 180

(c) Type of Polygon

Polygons / sidesum of interior angles

(n2) 180Regular Polygons

exterior angle + interior angle = 180

sum of exterior angle = 360

Triangle / 3 180

Quadrilateral / 4 360Pentagon / 5 540

Hexagon / 6 720

Heptagon / 7 900

Octagon / 8 1080

Nonagon / 9 1260

Decagon / 10 1440

(d) Angles

Acute angle Right angle Obtuse angle Reflex angle

60 60

60

a

b b

a

b

cb

a

a

ab

b

Angle at centre =

n

360

Interior angle = 180 n

360

Exterior angle =n

360

x < 90 x = 9090


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(e) Properties of angle

a = c and b = d a + b + c = 180 a + b + c = 360

a = c and b = da = c and b = d a + d = 180 and b + c = 180

a + b = c a + b = c a + b = c

b

cd

a

cb

ac

ba

b

a

c

d

bad c

bad c

b

a

cb

a

c

a

c b


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(3) INDICES

(a) Table for Numbers Power ofn

n 3 2 1 0 1 2 3 4 5 6 7

2 81

21

1 2 4 8 16 32 64 128

3 271

91

31

1 3 9 27 81 243 729

5 1251 251 51 1 5 25 125 625 3125

6 2161

361

61

1 6 36 216 1296

2 = 21

4 5 = 21

25

3 = 21

9 6 = 21

36

4 = 21

16 7 = 21

49

(b) Indices and Law of Indices

a n = aa a ( n times of a )

a0 = 1

an

=n

a

1, a n =

na1

,

n

b

a

=

n

a

b

na

1

= n a

nm

a = mn a = n ma am an = am + n

am an =n

m

a

a= am

n

(am)n = (an)m = amn

an bn = (ab)n,n

n

n

b

a

b

a

Example 1 :

2

4

k

kk= ???

= k4 + 1

(2)

= k4 + 1

+ 2

= k7

Example 2 :

(3f5g)

2 (f4)3f2g7 = ???

= 9f10

g2 f12 f2g7

= 9f10 + (12) (2)

g2 7

= 9f10 12 + 2

g5

= 9f0 g5

= 9g5

Example 3 :

3x 1 = 81

3x 1 = 34

x 1 = 4

x = 4 + 1

x = 5

41


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(4) PYTHAGORAS THEOREMc = 22 ba c

2 = a2 + b

2

b = 22 ac b2 = c2 a2

a = 22 bc a2 = c

2 b2

Group 1 Group 2 Group 3 Group 4

a /b b /a c/c a /b b /a c/c a /b b /a c/c a /b b /a c/c

3 4 5 5 12 13 7 24 25 8 15 17

6 8 10 10 24 26 14 48 50 16 30 34

9 12 15 15 36 39

12 16 20

(5) SPEED, CHANGE OF RATES

(a) Speed

time

cedisspeed

tan

speed

cedistime

tan distance = speed time

(b) Change of Rates

Example 1 :

90 kmh1 = ?? ms1

h

km

1

90=

s

m

60601

100090

= 25 ms1

Example 2 :

900 m min1 = ?? kmh1

min1

900 m=

h

km

60

11000

900

= 54 kmh1

Example 3 :

50 ms1 = ?? kmh1

s

m

1

50=

h

km

6060

11000

50

= 180 kmh1

(6) LINEAR EQUATIONS I, II

(a) Linear Equation

Example 1 :

2k 3 = k 2k+ 2

2kk + 2k= 2 + 3

3k = 5

k =3

5

Example 2 :

10k= 3 7k

10k+ 7k = 3

17k = 3

k =17

3

Example 3 :

5 8k= 12 + 4k

8k 4k = 12 5

12k = 17

k = 12

17

k =12

17

ac

b

2k 3 = k 2 (k 1) 2k =5

73 k

)3(242

5kk

5 8k= 4 (3 k)


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(b) Simultaneous Linear Equation

Example 1 :

2k 3m = 9, 7k+ 3m =9 k= ?, m = ?

Example 2 :

x + 2y = 6 (i),2

3xy = 7 x = ?, y = ?

2k + 7k = 9 + (9)

9k = 18

k=9

18

k = 2

2k 3m = 9

2(2) 3m = 9

4 3m = 9

3m = 9 + 4

3m = 5

m =3

5

m =3

5

x + 3x = 6 + (14)

4x = 8

x =4

8

x = 2

x + 2y = 6

2 2y = 6

2y = 6 + 2

2y = 8

y =2

8

y = 4

Example 3 :

2k3w = 10 (i), 4k+ w = 1 k= ?, w = ?

Example 4 :

7x 5y = 45 (i), 2x + 3y = 4 (ii) x = ?, y = ?

(i) 2,

6w (+w) = 20 (1)

7w = 21

w =7

21

w = 3

4k+ w = 1

4k + (3) = 1

4k 3 = 1

4k= 1 + 3

4k= 2

k=4

2

k= 2

1

(i) 2, (ii) 7

10y (+ 21y) = 90 28

31y = 62

y =31

62

y = 2

2x + 3y = 4

2x + 3(2) = 4

2x + 6= 4

2x = 4 6

2x = 2

x =2

2

x = 1

2k 3m =

97k + 3m = 9+ )

x + 2y = 63x 2y = 14+ )

4k 6w = 204k + w = 1 )

14x 10y = 9014x + 21y = 28 )


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(7) LINEAR INEQUALITIES

(a) Linear Inequalities in One Unknown

x < 3 x= , 6, 5, 4 x3 x = , 5, 4, 3.

x > 2 x= 3, 4, 5, x 2 x= 2, 3, 4,

3 2

5

Example 8 :

2x 5

x2

5

x2

5

Example 9 :

x < 10

Example 10 :

x 10

Example 11 :

x > 10

Example 12 :

x10

x

3 2(a) x

3 2(b)

x

3 2(c)

3 2(d)

x

3 2(e) x

3 2(f)

x3 2

(g) x3 2

(h)

2

x< 5

2

x 5

2

x< 5

2

x 5


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(8) ALGEBRAIC FRACTIONS

Example 1 :

=mv

vv

15

25(3

=mv

vv

15

253

=mv

v

3

1

Example 2 :

= 26

)2(3

m

mm

=2

6

23

m

mm

=2

3

1

m

m

Example 3 :

=mp

pp

2

)1(2321

=mp

pp

2

23

=mp

p 12

Example 4 :

3

2

x

9

52

x

x

=9

)5()3(22

x

xx

=9

5622

x

xx

=9

112

x

x

(9) ALGEBRAIC FORMULAE

Example 1 :

5h 20 = 7g

5h = 7g + 20

5

207

gh

Example 2 :

v

1+

u

1=

f

1, v = ???

v

1=

f

1

u

1

v

1

= fu

fu

v =fu

fu

Example 3 :

5m p 3m = 21

5m 3m = 21 +p

2m = 21 +p

m =2

21 p

Example 4 :

LT 3L = TLT

LT+LTT= 3L

2LTT= 3L

T(2LT 1) = 3L

T =12

3

LL

Example 5 :

32

h

g, g = ???

h

g2= 32

2 + g = 9h

g = 9h 2

Example 6 :

m = 5 3n2, n = ???

3n2 = 5 m

n

2

= 3

5 m

3

5 mn

m5

1

mv

v

15

25

(3v)

(3v)

=mv

v

15

55 5

5

(3m)m2

1

26

2

m

m

(3m)

= 26

22

m

m 2

2

2(p)m2

3

mp

p2

11

(p) 2

=mp

p

2

24

2

2

(x3)

3

2

x

9

5

2

x

x

(x3)

=

g

h )4(5 = 7, h = ???

73

5

m

pm, m = ?

L =3

)1(

T

LT, T = ???

h

g2 = 9


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(10) CIRCUMFERENCE / PERIMETER AND AREA OF A CIRCLE

(a) the circumference / perimeter of circle = 2r

(b) the length of arc AB =360

2r

(a) the area of circle = r2

(b) the area of sector AOB =360

r2

Example :

(a) the length of arc NR

=360

30 2 7

22 7

= 33

2

the length of arc QP

=360

60 2 7

22 14

= 143

2

the perimeter of the whole diagram

= ON + NR + RQ + QP + PO

= 7 + 33

2+ 7 + 14

3

2+ 14

= 463

1

(b) the area of sector ONR

=360

30 7

22 72

= 126

5

the area of sector ORM

=360

60 7

22 72

= 253

2

the area of sector OQP

=360

60 7

22 142

= 1023

2

the area of the shaded region

= 1023

2 25

3

2+ 12

6

5

= 896

5

O

r

B

A

O

r

O

r

B

A

O

r

(a) the perimeter, in cm, of the whole diagram,

(b) the area, in cm2, of the shaded region.

60

N R

Q

PMO 7 cm7 cm


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(11) AREA / TOTAL SURFACE AREA / VOLUME

(a) Area

(i) Square (ii) Rectangle (iii) Parallelogram

(iv) Triangle

Area = ab2

1

(v) Trapezium

Area = ))((2

1hba

(b) Total Surface Area of a Solid

(i) Cylinder

Total surface area = 2r2 + 2rh

(ii) Cone

Total surface area = r2 + rl

(iii) Sphere (iv) Hemisphere

(c) Volume of a Solid

(i) Cube (ii) Cuboid

(iii) Cylinder (iv) Semi-Cylinder

aArea = a a

= a2 b

a

Area = ab

b

a

Area = ab

b

aa

bb

ah

h b

b

a

a

h

b

a

r

h h

2r

r

l

rl

r

r Total surface area = 4 r2 Total surface area = 3r2r

a = a3

Volume = a a a c

ab

Volume = abc

h

r

Volume = base area h

= r2hh

r

Volume =2

2hr


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(v) Cone (vi) Pyramid

(vii) Sphere (viii) Hemispheres

(ix) Prism

Volume = base area h

=2

1abh

(x) Prism

Volume = base area t

=2

1(a + b)(h) t

Example 1 :

the volume of cone

=3

1(

7

22) (9)2 (14)

= 1188

the volume of cylinder

=7

22(3)2 (7)

= 198

the volume of the remaining solid

= 1188 198

= 990

Example 2 :

the volume of pyramid

=3

1(14 6) (8)

= 224

the volume of prism

= [2

1(10 + 14) (FG) ] (6)

= 72FG

224 + 72FG = 584

72FG = 584 224

72FG = 360

FG =72

360

FG = 5

Volume =3

1 base area h

=3

1r2h

hr

Volume =3

1 base area h

=3

1abh

h

a

b

Volume =3

4r3r Volume =

3

2r3

r

h

b

a

b

a

b

ah

h b

a

a

t

h

bh

t

The diagram shows a solid cone with radius 9 cm and

height 14 cm. A cylinder with radius 3 cm and height 7

cm is taken out of the solid. Calculate the volume, in

cm3, of the remaining solid.

V

E

D

A B

G

H

F C

6 cm10 cm

The height of the pyramid is 8 cm and FG = 14 cm.

It is given that the volume of the combined solid is

584 cm3. Calculate the length, in cm, of AF.


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(12) STANDARD FORM

(a) Significant figures

Type 1 ~ whole number Type 2 ~ decimal number < 1 Type 3 ~ decimal number > 1

Example :

63864 (3 s.f.) 63900 304638 (4 s.f.) 304600

Example :

0.00368 (2 s.f.) 0.0037 0.08195 (3 s.f.) 0.0820

Example :

1.75692 (3 s.f.) 1.76 1.00746 (4 s.f.) 1.007

(b) Standard form ~ A number in the form of A 10n, where 1 A 10 and n is an integerType 1 ~ number > 1 Type 2 ~ number < 1

Example :

522.6 5.226 102

5140000 5.14 106

Example :

0.565 5.65 101

0.00042 4.2 104

(c) Perform operation involving any two number and express the answer in the standard form

a 10n + b 10n = (a + b) 10n a 10n b 10n = (a b) 10n

Example :

1.8 105 + 88000 = 1.8 105 + 0.88 105

= (1.8 + 0.88) 105

= 2.68 105

Example :

0.0000025 1.3 107 = 2.5 106 0.13 106

= (2.5 0.13) 106

= 2.37 106

a 10m b 10n = (a b) 10m + n a 10m b 10n =

b

a 10mn

Example :

2.7 106 8.2 102 = (2.7 8.2) 106 + (2)

= 22.14 104

= 2.14 105

Example :

23

2

)103(

1086.4

=6

2

109

1086.4

=9

86.4 102 (6)

= 0.54 104

= 5.4 103

(13) QUADRATIC EXPRESSIONS AND EQUATIONS

(a) Expanding Brackets

= ac + adbcbd = acad bc + bd = a2 + ac + ab + bc = a2ac + abbc

Example :

(2x 1) (x + 1)

= 2x2+ 2xx 1

= 2x2 + x 7

Example :

(3x 1) (2x 7)

= 6x2 21x 2x + 7

= 6x2 23x + 20

Example :

(x + 2) (x + 3)

= x2+ 3x + 2x + 6

= x2+ 5x + 6

Example :

(x + 5) (x 4)

= x2 4x + 5x 20

= x2 +x20

(a + b)2

= a2 + 2ab + b

2

(a b)2

= a2 2ab + b2

(a + b) (a b)

= a2b2

a (b + c d)

= ab + acad

Example :

(m + 3)2

= m2+ 6m + 9

Example :

(3x 2y)2

= 9x2 12xy + 4y2

Example :

(2x + 1) (2x 1)

= 4x2 1

Example :

3p (4q p + 4)

= 12pq 3p2 + 12p

(a b) (c + d) (a b) (cd) (a + b) (a + c) (a + b) (ac)


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(b) Factorization

ax + ay

= a (x +y )

a2b2

= (a + b) (a b)

= ab + ac + bd+ cd= a (b + c) + d(b + c)

= (b + c) (a + d)

a (b c) + d(c b)= a (b c) d(b c)= (b c) (a d)

Example 1 :

5 30k= 5 (1 6k)

Example 2 :

18pq 15q= 3q (6p 5)

Example 1 :

81 64d2

.= 92 82d2

= (9 + 8d) (9 8d)

Example 2 :

2x2 72 .

= 2 (x2 36)

= 2(x + 6) (x 6)

Example 1 :

2m + 2n + mn + n2

= 2 (m + n) + n (m + n)

= (m + n) (2 + n)

Example 2 :

cd+ 5c d 5= c (d+ 5) (d+ 5)= (d+ 5) (c 1)

Example 1:

8eu 2ew +fw 4fu= 2e (4uw) +f(w 4u)= 2e (4uw) f(4u w)= (4u w) (2ef)

Example 2 :

pq q2 4q + 4p= q (p q) 4(q p)= q (p q) + 4(p q)= (p q) (q +p)

(c) Solving quadratic equations

Example 1 :

x2 + 9x + 20 = 0

(x + 4) (x + 5) = 0

x = 4, x = 5

Example 2 :

q2 18q + 45 = 0

(q 3) (q 15) = 0

q = 3, q = 15

Example 3 :

2n2

+ 9n 5 = 0

(2n 1) (n + 5) = 0

n = 2

1

, n = 5

Example 4 :

12k2 5k 3 = 0

(3k+ 1) (4k 3) = 0

k= 31

, k= 4

3

9x20

4+x

x 4x

5 5x

x2

q2 18q45

3+

q

q 3q1515q

2n2 9n5

1+

n

2n n5 10n

12k2 5k3

1+

4k

3k 4k

3 9k


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(14) SETS

(a) Universal sets (), elements (), subsets (), empty set ({ }, ), intersection (), union (),complements of sets ()

Example :

= { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }

A = { 3, 4, 5 } , B = { 4, 5, 6, 7, 8 } , C = { numbers that are greater than 10 }

3 A, 4 A, 5 A

0 B, 1 B, 2 B, 3 B, 9 B

C = { } or C =

A B

n (B) = 5

A B

{ } A, B

the number of subsets of A = 23 = 8

subset of A = { }, {3}. {4}, {5}, {3, 4},{3, 5}, {4, 5}, {3, 4, 5}

A = { 0, 1, 2, 6, 7, 8, 9 }

A B = { 4, 5 }

(A B) = {0, 1, 2, 3, 6, 7, 8, 9 }

A B = { 3, 4, 5, 6, 7, 8 }

(A B) = { 0, 1, 2, 9 }

** the number of subsets of a given set with n elements = 2n

** all sets have an empty set as its subset.

(b) Venn Diagrams

set A set B set C

set A set B set C

set A B set A C set A B

set ( A B ) set ( A C ) set A B

4 3 5

A BC

6

9

7 8

0

1 2

4 3 5

A BC

6

9

7 8

0

1 2

4 3 5

A BC

6

9

7 8

0

1 2

A BC

9

0

1 2

4 5

6 7

8 3 3

A BC

9

0

1 2

4 5

6 7

8

A BC

9

0

1 2

4 5

6 7

8 3

3

A BC

9

0

1 2

4 5 7

6

8

4 3 5

A BC

6

9

7 8

0

1 2

3

A BC

6

9

7 8

0

1 2

4 5

3

A BC

9

0

1 2

7 6

8 4 5

A BC

9

0

1 2

4 5

6 7

8 3 3

A BC

9

0

1 2

7 6 8

4 5


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set A B set ( A B ) set ( B C )

(c) Various Examples :

set ( P Q ) R set G H F set A B C

set A (B C ) set B C A set A B C

A BC

9

0

1 2

6 7

8 3

4 5

A BC

9

0

1 2

3 4 5

6

8 7

A BC

9

0

1 2

3 4 5

6

8 7

P

Q R

H

G F A

B C

A

B C

A

B C

A B

C


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(15) MATHEMATICAL REASONING

(a) Statement

Statement~ the sentences that is either true of false

Non-statement~ the sentences that is neither true nor false

Example :

3 is a prime number true statement

32 + 22 = (3 + 2)2 false statement

x +x = 2x true statement

7 < 6 false statement

Example :

3m 2 = 6,

3k+ 7, 3 + 7, R (P Q)

** All questions, commands, exclaimations,

algebraic expressions are non statement.

(b) Quantifier all, someExample 1 :

Object : odd numbers

Property : multiple of 5

Some odd numbers are multiple of 5.

Example 2 :

Object : cuboids.

Property : cross section in the shape of rectangular.

All cuboids has cross section in the shape ofrectangular.

Example 3 :

(i) of the prime numbers are odd numbers. some

(ii) pentagons have five sides. all

(c) Operations on Statements ~ Negation not or no ~P, not PExample :

All equilateral triangles are isosceles triangles (true) Notall equilateral triangles are isosceles triangles (false)

5 is a odd number (true) 5 is not a odd number (false)

P = A quadrilateral has five sides (false) ~P = No quadrilateral has five sides (true)

(d) Operations on Statements and or orp Q p and q

true True true

true False false

false True false

false False false.

p q p or q

true true true

true false true

false true true

false false false

Example 1 :

8 2 = 4 and 82= 62 the truth = ???

true and false = false

Example 2 :

8 > 7 or 32= 6 the truth = ???

true or false = true

Example 3 :

(i) 42 = 8, (ii) 0.4 =5

2, (iii) 6 < 5 (i) = false, (ii) = true, (iii) = true

true = 42 = 8 or 0.4 =5

2 , 42 = 8 or 6 < 5, 0.4 =5

2 or 6 < 5, 0.4 =5

2 and 6 < 5

false = 42 = 8 and 0.4 =5

2, 4

2= 8 and 6 < 5


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(e) Implication

Form 1

antecedent : p

consequent : q

Implication : Ifp, then q

Example :

antecedent : n < 3consequent : n

2< 9

Implication : If n < 3, then n2 < 9 [ false ]

Form 2

p if and only if q

Implication I : Ifp, then q Implication II : If q, then p

Example :

3m > 15 if and only if m > 5

Implication I : If 3m > 15, then m > 5 Implication II : If m > 5, then 3m > 15

Form 3

Implication : If p, then q

Converse : If q, then p

Example :

Ifx > 9, then x > 5

Converse : Ifx > 5, then x > 9 [ false ]

(f) Arguments

Form 1

Premise 1 : All A are B

Premise 2 : C is A

Conclusion : C is B

Example :

Premise 1 : All hexagons have six sides

Premise 2 : PQRSTU is a hexagon

Conclusion : PQRSTU has six sides

Form 2

Premise 1 : Ifp, then q

Premise 2 : p is true.

Conclusion : q is true.

Example :

Premises 1 : Ifx is greater than 0, thenx is a positive number

Premises 2 : 6 is greater than 0

Conclusion : 6 is a positive number

Form 3

Premise 1 : Ifp, then q.

Premise 2 : Not q is true

Conclusion : Not p is true.

Example :

Premises 1 : If set K is a subset of set L, then KL = LPremises 2 : KL LConclusion : Set K is not a subset of set L

(g) Induction the process of making a general conclusion from specific cases.Example 1 :

1 = 3 1 2

4 = 3 2 27 = 3 3 2

Conclusion : 3 n 2, n= 1, 2, 3, .

Example 2 :

2 = 3 1 1

11 = 3 4 126 = 3 9 1

Conclusion : 3 n2 1, n= 1, 2, 3, .

Example 3 :

1 = 2 = + 3 = + 24 = + 3

Conclusion : + n, n= 0, 1, 2, 3 .

+ (n 1), n= 1, 2, 3, 4 .

Example 4 :

1 =2

21

1 + 2 =2

32

1 + 2 + 3 =2

43

Conclusion :2

)1( nn, n= 1, 2, 3, 4 .


17/56


Example 5 :

The number of subsets in a set with 2 elements is 22



Conclusion : The number of subsets in a set with n elements is 2n, n = 2, 3, 4, .

(h) Deduction the process of making a specific conclusion based on a given general statementExample :

The sum of the interior angles of a n- sided polygon is (n 2) 180

Specific case : PQRSTU ialah sebuah poligon.

Conclusion : The sum of the interior angles of PQRSTU is 720


18/56


(16) THE STRAIGHT LINE

(a) Type of straight line and their respective gradient, m

(b) y-intercept (c), and y-intercept of a straight line

find x-intercept, sub. y = 0

findy-intercept, sub. x = 0

Example 1 :

3x 4y = 24, y-intercept = ???

x = 0,

3(0) 4y = 24

4y = 24

y =4

24

= 6

Example 2 :

3y 2x = 6, x-intercept = ???

y = 0,

3(0) 2x = 6

2x = 6

x =2

6

= 3

(c) Gradient of a straight line, m

m =distancehorizontal

distancevertical m =

12

12

xx

yy

=

21

21

xx

yy

m =intercept-x

intercept-y

Example 1 :

m =

2

4= 2

Example 2 :

P (0, 2),

m =04

23

=4

1

Example 3 :

m =

4

8= 2

Example 4 :

k= ???

)3(3

6

k

=3

1

6

6k=

3

1

k6 = 2k = 2 + 6

k= 4

Example 5 :

x-intercept = ???

x

3=

4

1

x

3=

4

1

x = 12

x

y

O

m > 0 (m +if)

x

y

O

m < 0 (mif)

x

y

O

m = 0

x

y

O

m

x

y

y-intercept

x-intercept(0, y)

(x, 0)O

4 Q

y

xO

2

P

y

xO

R (4, 3)

P

2 8

Qy

xO4

P

Q (3, 6)

y

xO

R (3, k)

m =3

1 m =

4

1

y

xO

3 F

E


19/56


(d) Equation of a straight line

x = 5 y = 3 x = 5 y = 3

x +y = 3 y = x + 3 y x = 3 y =x + 3 y =x x y = 0 y = x x + y = 0

(e) Form equation of a straight line, y =mx + c, where erceptyc

gradientm

int

Example 1 :

y = mx + c

y = 2x + 4

Example 2 :

m =03

06

= 2

y = mx + c

y = 2x

Example 3 :

y = mx + c8 = 2(3) + c

8 = 6 + c8 + 6 = c

14 = c

y = 2x + 14

Example 4 :

y = mx + c3 = m(1) + 63 = m + 6m = 6 3

m = 3

y = 3x + 6

Example 5 :m =

)2(4

26

=2

1

y = mx + c

6 = 2

1

(4) + c

6 = 2 + c

4 = c

y =2

1x + 4

Example 6 :

equation of QR = ???

y = 0,

2x + 0 = 5

2x = 5

x2

5

x

y

5O

y

x

3

Ox

y

(5, 3)

Ox

y

(5, 3)

O

y

x

3

O 3x

3

3

y

O

( 3, 3)

y

xO

x

y

O

( 3, 3)

x

y

O

4 m = 2y

xO

P (3, 6)

m = 2

P

y

xO

Q (3, 8)

(1, 3)

y

xO

6

Q (4, 2)

y

xO

R (4, 6)

2x +y = 5

O Q

y

x

P

R


20/56


(f) Solve problems involving the equation of a straight line

Example 1 :

a = ???

y = 2x + 3

a = 2(2) + 3

a = 7

Example 2 :

2x 7y = 14, m = ???

7y = 2x + 14

7y = 2x 14

y = 7

2

x + 2

m =7

2

(g) Parallel lines, m1 = m2

Example 1 :

2y =x + 6 is parallel to 4y =px + 9, p = ???

Example 2 :

h = ???

2

1 =

42

3

h

2

1 =

2

3

h

2h 6 = 2

2h = 4

h = 2

2y =x + 6

y =2

1x + 3

m1 =2

1

4y =px + 9

y =4px +

49

m2 =4

p

2

1=

4

p

2p = 4

p =2

4

p = 2

(h) Form equation of a parallel line

Example 1 :

equation of PQ = ???

m =

2

4= 2

y = mx + c5 = 2 (3) + c

5 = 6 + c

5 + 6 = c

11 = c

y = 2x + 11

Example 2 :

equation of ST = ???

x + 2y = 14

2y = x + 14

y = x2

1 + 7

y = mx + c

5 =2

1 (2) + c

5 = 1 + c4= c

y =2

1 x 4

(i) Distance, Midpoint

distance = 2122

12 )()( yyxx midpoint, (x, y) =

2,

2

2121 yyxx

Example :

distance PQ = 22 ])2(4[)19(

= 10

Example :

291 = x

5 = x

2

2 y

= 8

2 +y = 16

y = 14

x

y

O

(2, a)

y = 2x + 3

y

xO

F (4, 3)

C

D

y =2

1x + 1

E (2, h)

O

R (0, 4)

S (2, 0)

y

x

P (3, 5)

O

y

x

P

R

x + 2y = 14

T (2, 5)

S

y

x

(1, 2)

P ( 9, 4)

O x

y

Q (x, 8)

P (1, 2)

R (9, y)

x = ???, y = ???


21/56


(17) STATISTICS I, II, III

(a) Mode of a ungrounded data

Mode = the value of data with the highest frequency

Example 1 :

6, 7, 7, 11, 5, 6, 11, 13, 14, 11, 8

5, 6, 6, 7, 7, 8, 11, 11, 11, 13, 14

mode = 11

Example 2 :

Score 2 4 6 8 10Frequency 3 15 7 12 9

mode = 4

Example 3 :

Score 0 1 2 3 4

Frequency 1 3 7 x 5

mode = 2, the maximum value ofx = ???

x < 7 x = 6

Example 4 :

Score 0 1 2 3 4

Frequency 1 7 0 x 2

mode = 3, the minimum value ofx = ???

x > 7 x = 8

(b) Median of a ungrounded data

Median = the middle value when a set of data is arranged in ascending order

Example 1 :

5, 3, 3, 5, 7, 7, 1

median = 5

Example 2 :

24, 23, 12, 19, 16, 17

median =2

1917 = 18

Example 3 :

Number of books 1 2 3 4 5

Number of pupils 3 0 1 5 6

median = 4

Example 4 :

Saiz of shoes 1 2 3 4 5

Number of students 8 14 12 x 3

median = 3, range of x = ???

8 14 1 11 x 3

8 14 11 1 x 3.

8 + 14 = 11 +x + 322 =x + 14

8 =x

8 + 14 + 11 =x + 333 =x + 3

30 =x

8 x 30

(c) Mean of a ungrounded data

mean =dataofnumberthe

dataofvaluestheallofsum mean =

frequencytotal

frequencyvalueofsum )(

Example 1 :

68, 62, 84, 75, 78, 89

mean =6

797875846268

= 76

Example 2 :

Mark 74 78 82 86

Frequency 5 10 2 3

mean =32105

)3(86)2(82)10(78)5(74

= 78.6

1, 3, 3, 5, 5, 7, 7 12, 16, 17, 19, 23, 24

1, 1, 1, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5


22/56


(d) Measure of Dispersion ~ range, first / lower quartile (Q1), third / upper quartile (Q3), interquartile range

range = ( largest smallest ) value of data

Q1 = the value that divides the values of data that are less than median into 2 equal parts

Q3 = the value that divides the values of data that are greater than median into 2 equal parts

Interquatile range = Q3 Q1

Example 1 :

5, 30, 45, 29, 25, 6, 21, 8, 28, 4

4, 5, 6, 8, 21, 25, 28, 29, 30, 45

range = 45 4 = 41

Q1 = 6

Q3 = 29

Interquartile range = 29 6 = 23

Example 2 :

8, 12, 6, 10, 6, 7, 13, 3, 8, 10, 13, 19

3, 6, 6, 7, 8, 8, 10, 10, 12, 13, 13, 19

range = 19 3 = 16

Q1 =2

76 = 6.5

Q3 =2

1312 = 12.5

Interquartile range = 12.5 6.5 = 6

(e) Solve problem involving ungorounded data

Example 1 :

Score 1 3 6 x 12 14

Frequency 1 1 2 3 1 1.

1, 3, 6, 6, x, x, x, 12, 14

2

12x= 11

x + 12 = 22

x = 22 12x = 10

Example 2 :

3, 3, 6, x, x, 3

mode = 3, median = 4. Two new pieces of data, 4 and 7 put into the set, mean = ???

3, 3, 3, x, x, 62

3 x= 4

3 +x = 8

x = 8 3x = 5

mean

=8

74655333

= 4.5

(f) Class interval, lower / upper limit, lower / upper boundary, size of class interval, midpoint

Example :

Class interval 1115 1620 2125 2630 3135 3640 4145.

lower limit = 16, upper limit= 20

size of class interval = 5

= upper. B lower. B

= midpoint2 midpoint1

lower boundary = 15.5, upper boundary = 20.5

midpoint = 2limlim itupperitlower

=2

boundaryupperboundarylower

= 18

third quartile = 11, x = ???


23/56


(g) Frequency table, Cumulative Frequency, Modal class, Mean, Range

Example :

Donation, x 1115 1620 2125 2630 3135 3640 4145

Frequency, f 1 3 6 10 11 7 2

`

Cumulative F. 1 4 10 20 31 38 40.

modal class = the class interval with the highest frequency = 3135

mean =

f

fx=

frequencyofsum

frequencymidpoofsum )int( =40

)2(43)7(38)11(33)10(28)6(23)3(18)1(13 = 30

range = midpoint of (hightest lowest ) class = 43 13 = 30

(h) Histogram, frequency polygons, ogive, first quartile, third quartile, interquartile range

Histogram

lower / upper boundary.

frequency

**

the frequency polygon can beconstructed based on a histogram

Example : [ base on frequency table in (g) ]

Frequency Polygons

midpoint

frequency

**

the frequency polygon should add aclass with zero frequency before the

first class and after the last class


Ogive

upper boundary

cumulative frequency

**

add a class with zero frequency

before the first class


Histogram

2

4

6

10

8

Frequency

Donation

10.

5

0

15.

5

20.

5

25.

5

30.

5

35.

5

40.

5

45.

5

Frequency

polygon

2

4

6

10

8

Frequency

Donation

8

0

13

18

23

28

33

38

43

48

34.5

Cumulative Frequency

Donation0

10.

5

15.

5

20.

5

25.

5

30.

5

35.

5

40.

5

45.

5

10

20

30

40

30.525.5

21 N

43 N

41 N

Q1 Q3med


24/56


(18) PROBABILITY I / II

(a) Probability of an event, P (A)

P (A) =)(

)(

Sn

An, 0 P (A) 1

Example 1 :

P (B) =10

3

P (R) =10

7

Example 2 :

20

x=

5

1

5x = 20

x =

5

20

x = 4

Example 3 :

12xx

=3

2

3x = 2x + 24

3x 2x = 24

x = 24

(b) Probability of the complement of an event, P(A)P (A) = 1 P (A) P (A) + P (A) = 1

Example 1 :

P (G) = 1 7

2

=7

5

Example 2 :

P (D) = 1 3

1

9

2

=9

4

Example 3 :

x

4=

6

1

x = 24

(c) Probability of combined event

P(A and B) = P(A B) =)(

)(

sn

BAn = P(A) P(B) P(A or B) = P (A B) =

)(

)(

sn

BAn = P(A) + P(B)

Example 1 :

(i) P (RR)

=95

95

=81

25

(ii) P (RB)

=95

94

=81

20

(iii) P (only one R)

= P (RB or BR)

=9

5

9

4+

9

4

9

5

=81

40

B P(B) = ??3

7

10

R P(R) = ??

?? (5

1)W

B

20

?? (3

2)Y

G12

( ?? )

(

7

2)B

G

(3

1)W

L

D

(92 )

( ?? )

(21 )

R

G

B (3

1)

4

??

( )

9

R

B

5

4

2 marbles are chosen at random,

with replacement


25/56


Example 2 :

(i) P (first is F and second is M)

= P (FM)

=12

7

11

5

=132

35

(ii) P (both are M)

= P (MM)

=12

5

11

4

=33

5

(iii) P (both are same gender)

= P (FF or MM)

=12

7

11

6+

12

5

11

4

=66

31

(iv) P (a F and a M)

= P (FM or MF)

=12

7

11

5+

12

5

11

7

=

66

35

(v) P (at least a F)

= P (FM or MF or FF)

=12

7 11

5+

12

5 11

7+

12

7 11

6

=

33

28

= 1 P (MM)

or = 1 12

5

11

4

=

33

28

12

F

M

7

5

2 workers are selected at random,


26/56


(19) CIRCLES II / III

(a) Properties of angle in a circle, Cyclic Quadrilaterals

(b) Properties of the tangle to circle

O

aO

b

a = b = 90 Oa

2a

O

a

2a

Oa

2a

bb

aa aa

a

aO

aa

O2a

a

c

bd

a a + b = 180

c + d = 180 e

a

a = e

tangent

ab

a + b = 180

tangent

a + b = 90

b

tangent

ba

a

b

b

a + b = 90

tangent

aa

b

tangentb

aa

b

b

a

a

tangent


27/56


(20) TRIGONOMETRY I / II

(a) Trigonometrical ratios

sin =H

O

cos =H

A

tan =A

O

sin =H

O

cos =H

A

tan =A

O

(b) The values of trigonometric ratios of 30, 45, and 60 (Special angles) sin cos tan

3021

2

3

452

1 2

1 1

60 23 21 3

(c) The value of sine, cosine and tangent, of an angle

In a unit circle,

sin = the value of coordinate-y

cos = the value of coordinate-x

tan =xcoordinateofvaluethe

ycoordinateofvaluethe

=

cos

sin

(d) The values of the angles in quadrant I which correspond to the value in other quadrants

in other quadrant, > 90 Corresponding angle in quadrant I

II 180

III 180

IV 360

(e) Finding the angles, given the value of sine, cosine and tangent

Quadrant Angle

I , from calculator

II 180

III 180 +

IV 360

Example 1 :

sinx = 0.5299, 0x 360 x = ???sinx +if, x I, IIsin 32 = 0.5299 (from the scientific calculator)

x = 32, 148

Example 2 :

cosx = 0.7721, 0x 360 x = ???cosx if, x II, IIIcos 39.46 = 0.7721 (from the sc. calculator)

x = 140.54, 219.46

(f) Solve problem involving sine, cosine and tangent

OH

A

23

60

302

11

1

1

2 45

45

3

1

Quadrant 1

0 < < 90 sin +if cos +if tan +if

Quadrant 2

90 < < 180 sin +if cos if tan if

Quadrant 3

180 < < 270 sin if cos if tan +if

Quadrant 4

270 < < 360 sin if cos +if tan if


28/56


Example 1 :

answer

Example 2 :

answer

tan BCA =4

3sin ECD =

5

4

BC

6=

4

3

CD

4=

5

4

BC = 8 CD = 5

EC = 3, BE = 8 3 = 5

(g) Compare and differentiate the graph of sine, cosine and tangent for angle between 0x 360

y = sinx y = cosx y = tanx

y = sin 2x y = cos 2x y = tan 2x

PQ R

S

8 cm

9 cm

17 cm

y cosy = ???

PQ R

S

8 cm

9 cm

17 cm

y

cosy = cos

=10

6

= 5

3

15 cm

10 cm

6 cm

tan BCA =4

3,

sin ECD =5

4

BE = ???

D

C

B

A

6 cm

4 cm

E

x

y

O 90 180 270360

1

1 1

x

y

O 90 180 270360

1

y

x

O 90 180

270

360

x

y

O 90 180 270360

1

1 1

x

y

O 90 180 270360

1

x

y

O 90 180 270 360


29/56


(21) ANGLES OF ELEVATION AND DEPRESSION

(a) Angle of elevation and angle of depression

Example 1 :

answer

Example 2 :

answer

(b) Solve problems involving angle of elevation and angle of depression

Example 1 : Example 2 :

answer answer

tan =60

45

= 3652

tan 42 = 10t

10 tan 42 = t

9.004 = t

h = 9.004 + 3

= 12.004

horizontal line

= angle of elevation

= angle of depression

M

RN

P

QThe angle of elevation

of P from M is ???

M

RN

P

Q PMQ

P

Q

T

S

R

The angle of depression

of P from T is ???

P

Q

T

S

R

TPS

P

Q

95 m50 m

S

R60 m

the angle of

depression of

S from Q is

???T U

V

h m

10 m

S

3 m

the angle of

elevation of

V from S is

42, h = ???

P

Q

95 m 50 m

S

R60 m

45 m

60 m

T U

V

h m

10 m

S

3 m

t42

10 m


30/56


(22) LINES AND PLANES IN 3-DIMENSIONS

(a) Normal to a plane, Orthogonal Projection

(b) The line of intersection between two planes, the point of intersection between two planes

(c) Steps of determine the angle between a line and a plane / base

determine and shade the plane (base).

determine and draw the line mark lower point (point on the base), mark upper point (point on the other side).

from the upper point draw the normal to the plane (base)

then, connect it to the lower point (orthogonal projection to the line).

the angle is at the lower point.

by using trigonometrical ratios, calculate the angle.

Example :

answer

UXV

plane (base)

normal plane (base)

normal

lineorthogonal

projectionline

orthogonal

projection

line of intersection line of intersection

point of intersection

point of intersection

angle between line XU and the plane WXYZ ???

RQ

U

S

Z

W

X

V

Y

U

RQ

S

Z

W

XY

V

RQ

U

S

Z

W

X

V

YLw.

Up.

RQ

U

S

Z

W

XY

Nor.V

RQ

U

S

Z

W

XY

V


31/56


(d) Step of determine the angle between two planes

determine and shade the first plane (base).

determine the second plane.

determine the line of intersection of the two planes / the point of intersection of the two planes.

from the second plane, determine the line that will be chosen.

after choosing the line mark lower point (point on the base), mark upper point (point on the other side).

from the upper point draw the normal to the plane (base)

then, connect it to the lower point (orthogonal projection to the line). the angle is at the lower point.

by using trigonometrical ratios, calculate the angle.

Example :

answer

PRS

line of intersection / point of intersection

the line that will be chosen

the angle between plane CRP and plane CDRS ???A

B C

D

R

SP

A

B C

D

R

SP

A

B C

D

R

SP

QA

B C

D

R

SP

Q

A

B C

D

R

SP

A

B C

D

R

SP

A

B C

D

R

SP


32/56


(23) NUMBER BASES

(a) Comparision between number in bases ten, two, five and eight

Base 10 0 1 2 3 4 5 6 7 8 9 10

Base 2 02 12 102 112 1002 1012 1102 1112 10002 10012 10102

Base 5 05 15 25 35 45 105 115 125 135 145 205Base 8 08 18 28 38 48 58 68 78 108 118 128

(b) The value of a digit of a number in bases two, eight and five

Base 22

92

82

72

62

52

42

32

22

12

0

512 256 128 64 32 16 8 4 2 1

Base 88

48

38

28

18

0

4096 512 64 8 1

Base 55

55

45

35

25

15

0

3125 625 125 25 5 1

Example 1 :

5 4 3 2 1 0

1100112

the value of the digit 1

= 1 24 = 16

or

= 1 16 = 16

Example 2 :

3 2 1 0

75028


= 5 82 = 320

or

= 5 64 = 320

Example 3 :

2 1 0

2415


= 4 51 = 20

or

= 4 5 = 20

(c) Changing numbers in base 2, base 8, base 5 base 10base 2 base 10 base 8 base 10 base 5 base 10

Example 1 :

101102

= (1 24) + (1 22) + (1 21)

= 22

Example 2 :

10568

= (1 83) + (5 81) + (6 80)

= 558

Example 3 :

3245

= (3 52) + (2 51) + (4 50)

= 89

(d) Changing numbers in base 10 base 2, base 8, base 5 [type 1]Base 10 base 2 ( 2) ~ cal. base 10 base 8 ( 8) ~ cal. base 10 base 5 ( 5)Example 1 :

1210 = ???2

= 11002

Example 2 :

28810 = ???8

= 4078

Example 3 :

14410 = ???5

= 10345

remainder1 22

6 02

3 02

1 12

0 1

remainder2 6 38

3 2 78

4 08

40

remainder1 4 45

2 8 45

5 35

015

0 1


33/56


(e) Changing numbers in base 10 base 2, base 8, base 5 [type 2]base 10 base 2 ~ cal. base 10 base 8 ~ cal. base 10 base 5

Example 1 :

25 + 22 + 1 = ???2

= 1001012

Example 2 :

2 (83) + 5 (8) + 7 = ???8

= 20578

Example 3 :

53 + 3 = ???5

= 10035

(f) Base 2 base 8 (group of three digits)base 2 base 8 ~ cal. base 2 base 8 ~ cal.

Example 1 :

10111012 = ???8

= 1358

Example 2 :

5628 = ???2

= 1011100102

(g) Base 2 base 5, Base 8 base 5base 2 base 5 Base 8 base 5

base 2 base 5 = base 2 base 10 base 5 base 5 base 2 = base 5 base 10 base 2

base 8 base 5 = base 8 base 10 base 5 base 5 base 8 = base 5 base 10 base 8

(h) Addition and subtraction of two number in base two

Example 1 : ~ cal. Example 2 : ~ cal.

Remarks : Step convert a number in base 10, 2 or 8 to a number in any of these bases by using calculator

Step perform addition and subtraction of numbers in base 2

mode mode (3) base 10 DEC (d), base 2 BIN (b), base 8 OCT (o)

0

25 2324 22 21 20

( 2 ) ( 1 )

1 1 1 200

83 82 81 80

( 8 ) ( 1 )

0 7 852

53

52

51

50

( 5 ) ( 1 )

0 3 501

5 831

1 0 1 2

4 2 1 4 2 1 4 2 1

1 0 1 1

0 10 2

4 2 1 4 2 1 4 2 1

2 865

1 0 1 1 1 0

02

+ 02

02

12

+ 02

12

02

+ 12

12

12

+ 12

1 02

1

112

+ 12

1002

11

02

02

02

12

02

12

12

12

02

1 02 12

12

0 2

1 0 02 12

1 12

10 2

2

1 1 1 0 12

+ 1 1 1 02

1 0 1 0 1 12

111 0 2

1 1 0 0 0 12

1 1 0 12

1 0 0 1 0 0 2

1 2


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(24) GRAPH OF FUNCTIONS

(a) Graphs of linear functions

y = mx, m > 0 y = mx, m < 0 y = mx + c, m > 0 y = mx + c, m < 0

(b) Graphs of quadratic functions

y = ax2, a > 0 y = ax

2+ c, a > 0 y = ax2 + bx, a > 0 y = ax2 + bx + c, a > 0

y = ax2, a < 0 y = ax

2+ c, a < 0 y = ax

2+ bx, a < 0 y = ax

2+ bx + c, a < 0

(c) Graphs of cubic functions

y = ax3, a > 0 y = ax

3, a < 0 y = ax

3+ c, a > 0 y = ax

3+ c, a < 0

(d) Graphs of reciprocal functions

y =x

a, y = ax

1, xy = a, a > 0 y =x

a, y = ax

1, xy = a, a < 0

x

y

O

m +ify

Ox

mif

x

y

O

c

m +if

x

y

O

c mif

y

xO

a +if

y

xO

ca +if

a +ify

xO

y

xO

c

a +if

y

xO

aifaif

y

xO

cy

xO

aify

xOc

aif

y

xO

a +ify

xO

aif y

xO

c

a +if

c

y

xO

aif

y

xO

a +if

y

xO

aif


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(e) Solve problems involving functiuons

Example 1 :

y = 2x2

+px , (4, 0)

0 = 2 (4)2

+p (4)

0 = 32 + 4p

4p = 32

p =4

32

p = 8

Example 2 :

y = 2x3 + 2 , (h, 0)

0 = 2h3 + 2

2h3 = 2

h3=

2

2

h3

= 1

h = 3 1

h = 1

(f) Region representing inequalities in two variables

Example 1 :

Example 2 :

4

y = 2xn

+px,

y

p = ???

O

y = 2x3 + k

(h, 0)

y

xO

2 h = ???

y =xy

x

x +y = 5O

shaded the region which

satisfies the three inequalities

x +y 5, y x, and x < 5

y =xy

x

x +y = 5O

5

5

x = 5

y

xO

y =x

y = 2x + 8

state the threeinequalities which satisfy

the shaded region

y = 88

y

xO

y =x

y = 2x + 8

0, c < 0

y = ax3 + cx + d,

a < 0, c > 0y =

2x

a, a > 0 y =

2x

a, a < 0

x 4 16

y 6 3

relation between y

and x = ???

p 8 w

q 3 10

r 4 12 r pq, w = ???

w x y

40 4 2

m 6 4

w varies directly as

the square ofx and

inversely asy,

m = ???

y

xO

d

a +if, c ify

xO

d

aif, c +if

y

xO

a +if

y

xO

aif


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(28) GRADIENT AND AREA UNDER A GRAPH

(a) Distance-time graph

speed =in timechange

distanceinchange

average speed =takentimetotal

travelleddistancetotal

Example :

distance travelled in the first 6 s = 8 m

speed, in the first 6 s =s

m

6

8=

3

4ms1

distance travelled in the last 5 s = 12 m

speed, in kmh1, in the last 3 s

= speed in the last 5 s =h

km

36005

100012

= 8.64 kmh1

length of time, the particle is stationary = 4 s

average speed, in ms1

, for the period of 15 s = s

m

15

20

distance travelled in the first 4 s = ???

speed in the first 4 s = speed in the first 6 s

6

8

4

d d= )4(

6

8 d=

3

15 m

(b) Speed-time graph

distance = area under the graph

rate of change in speed = gradient of graph rate of change in speed +if acceleration

rate of change in speed if deceleration

Example :

distance travelled in the first 3 s=

21 (3)(6) = 9 m

distance travelled, with uniform speed= (5)(6) = 30 m

distance travelled in the last 2 s=

21 ( 6 + 12 ) (2) = 18 m

distance travelled in the first 6 s.=

21 ( 3 + 6 ) (6) = 27 m

uniform speed = 6 ms1

the length of time, moves with uniform speed= 5 s

average speed for the period of 10 seconds

=

s

m

10

57= 5.7 ms

1

the rate of change in speed in the first 3 s

=03

06

= 2 ms2

acceleration in the last 2 s

=810

612

= 3 ms2

speed, in the last second = ???

acceleration ~ in the last second = in the last 2 s

89

6

v

= 810

612

v 6 = 3

v = 9

stationary / stopDistance (m)

Time (s)

constant / uniform speed

d

Time (s)

Distance (m)

106

O

4

12

154

Speed (ms1)

Time (s)

constant / uniform speed

acceleration /increasing speed

deceleration /decreasing speed

(10, 12)

O Time (s)

Speed (ms1

)

83

6

12

106

(8, 6)

9

(3, 6) (9, v)


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(c) Solve problem involving distance-time graph and

Distance-time graph Speed-time graph

Example 1 :

speed of AB and BC are the same, T = ???

4

8=

4

12

T

2 =4

12

T

2T 8 = 12

2T = 12 + 8

2T = 20

T =2

20

T = 10

Example 1 :

total distance travelled for the period of ts is 148,

t = ???

25 + 63 + 15t 180= 14815t= 148 25 63 + 180

15t= 240

t= 16

Example 2 :

deceleration is 5m s2, u = ??

117

0

u

= 5

4u

= 5

u = 5 (4)

u = 20

Example 2 :

average speed for the period tmin. is 30 km h1

,t = ???

st

m18= 30 km h1

h

km

t

60

18= 30 km h1

18 = 30 (60

t)

18 (30

60) = t

36 = t

Distance (m)

Time (s)4O

4

12

T

B

A C

Speed (ms1)

Time (s)O

1

9

21

5 12 t

t 12

9

9

19

21

7

5

d =21 (1+ 9) (5) = 25

d= 9(7) = 63

d =21 (9+ 21) (t 12)

= 15 (t 12)

= 15t 180

Speed (m s1

)

Time (s)7O

6

u

11

(7, u)

(11, 0)

Time (min.)

Distance (km)

171

O

10

18

t


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(29) BEARING

(a) Bearing

the bearing of a point B from a point A is the angle at a measured clockwise from the north tothe line joining A and B.

written in a three-digit form, from 000 to 360

Example 1 :

The bearing of M from N is 060. Diagrams shows the positions of M and N = ???

Example 2 :

P is due south of Q. The bearing of R from Q is 150 and the bearing of P from R is 300

Diagrams shows the posititon of P, Q and R = ???

Example 3 :

bearing of M from N = ???

answer

Example 4 :

bearing of G from F = ???

answer

= 360 70= 290

= 360 105 60= 195

Example 5 :

E lies to the north of G,

bearing of G from F = ???

answer

Example 6 :

Q lies to the west of P,

bearing of Q from R = ???

answer

North

60

N

North

60

N

M

North

N

North

Q

P

300

North

P

R

150

North

Q

R

North

Q

P

answer

combine

R

NorthM

N

7075

E

N

G

F

NorthM

N

70North

70

75

E

N

G

F

N

60

105

E

50

20GF

P

R70

Q

50


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= 360 50 20

= 290

= 360 40

= 320

F

E

50

20G

North

North

50

Q 40

50

P

60

R

N

90

(W)

30


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Example 7 :

bearing of L from K is 110,bearing of K from J = ???

answer

Example 8 :

bearing of H from K is 080,bearing of G from H = ???

answer

= 180 142

= 038

= 360 20 100

= 240

Example 9 :

bearing of P from R is 215,bearing of P from Q = ???

answer

Example 10 :

bearing of P and R from Q is 050, and 290,bearing of R from P = ???

answer

= 360 25 90 = 245 = 360 102 = 258

N

42

30

K

L

J

140

K

G H

N

42

30

K

L

J

N

110

108

142

140

K

G H

N

80

20

100N

Q

R

P

60

R

32

Q

P

215

N

P

R35

Q

R

P

60 25

35

NN

30

R

32Q

P

50

290

70

N

N

28

102


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(30) EARTH AS A SPHERE

(a) Longitudes [ xE or yW ]

(b) Latitudes [ xN or yS ]

(c) Difference between two longitudes, Difference between two latitudes

Difference between two longitudes, Difference between two latitudes,

same direction = find the difference

Example :

20E, 50E = 50 20 = 30

30W, 120W = 120 30 = 90

same direction = find the difference

Example :

20N, 50N = 50 20 = 30

30S, 120S = 120 30 = 90

opposite direction = find the sum

Example :

10E, 70W = 10 + 70 = 80

if the sum of two angles of longituded> 180, = 360 (the sum)

Example :

120E, 80W = 360 (120 + 80)= 160

opposite direction = find the sum

Example :

10N, 70S = 10 + 70 = 80

N

S

Great circle

meridian

( half of great circle )

10E

N

S

30

10

Greenwich Meridian(Longitude 0)

30W

N

S

equator

parallel of latitude

parallel of latitude35S

N

3520

equator

(latitude 0)

20N


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(d) Find longitude, latitude

Find longitude Find latitude

Example :

20W (50) 30E (50) 80E(diference if +ifchange direction)

Example :

160E (50) 150W (50) 80W(sum > 180 (360 sum)change direction)

dif. if,

+ifch. dir.

Example :

(50N)

(40)

10N

(40)

(30S)

sum > 90,

(180sum)

Example :

(30S)

(40)

70S

(40)

(70S)

(e) Diameter of the earth, Diameter of the parallel of latitude

a + b = 180

Example :

P (30N, 130E), Q (30S, 130E)

(f) Location of a place

Example 1 :

R (40N, 110E)

Example 2 :

latitude of P = 30N 45 = 15S

longitude of P = 80E + 60 = 140E P (15S, 140E)

E EW+

W EW+

S

N

N

+

S

S

N

+

N

S

aU

aS

diameter of

the earth

S

bW aE

N diameter of theparallel of latitude

diameter of

the earth

N

S

P = ??(30N, 50W)

Q = ??

O

N

S

P R

40

70

R = ???

Greenwich

Meridian

O

N

S

P

R

80E60

30 N

45

P = ???


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(g) Distance on the surface of the earth

along a meridian along the equator along the parallel of latitude, yN / S

distance = 60, = different in latitude

=60

tan cedis

distance = 60, = different in longitude

=60

tan cedis

distance = 60 cosy, = different in longitude

=y

cedis

cos60

tan

(h) Shortest distance ( distance along a great circle : ------------------- )

shortest distance

= (180 2a) 60shortest distance

= (180 ab) 60shortest distance

= (180 a + b) 60

Remark :

knot = unit of speed nautical mile = unit of distance

speed =time

cedis tan time =

speed

cedis tan distance = speed time

N

S

distance 0distance

N

S

yN

distance

N

S

aNaN

bNaN

bS


53/56


(i) Problem solving

Example 1 :

The diagram shows four points, P, Q, R and X, on the surface of the earth. P lies on longitude of

80W. QR is the diameter of the parallel of latitude of 50N. X lies 5820 nautical miles due southof P.

Solution :

(a) longitude of Q = 80W 35 = 45W R (50N, 135E)

(b) different in longitude of QR = 180 QR = 180 60 cos 50 = 10421.63

(c) different in latitude of PX =60

5820= 97

latitude of X = 50N 97 = 47S

(d) different in longitude of RR = 145PR = 145 60 = 8700

time =600

8700= 14.5 hours

Example 2 :

P (60S, 70E), Q, and R are three points on the surface of the earth. PQ is the diameter of theparallel of latitude 60S. R lies 4800 nautical miles due north of P.(a) State the longitude of Q.

(b) Find the latitude of R.

(c) Calculate the distance, in nautical miles, from P to Q measured along the parallel of latitude.

(d) An aeroplane took off from Q and flew towards P using the shortest distance, as measured along

the surface of the earth, and then flew due north to R. Given that its average speed for thewhole flight was 560 knots, calculate the total time taken for the flight.

Solution :

(a) longitude of Q = 110W

(b) different in latitude of PR =60

4800= 80

latitude of R = 60S 80 = 20N

(c) different in longitude of PQ = 180 QR = 180 60 cos 60 = 5400

(d) different in between PQ = 180 60 60 = 60PQ = 60 60 = 3600

Total distance = 3600 + 4800 =8400 time =

560

8400= 15 hours

N

S

O

3550

X

PQ

R

(a) Find the position of R.(b) Calculate the shortest distance, in nautical miles,

from Q to R, measured along the surface of the earth.

(c) Find the latitude of X.

(d) An aeroplane took off from P and flew due west to R

along the parallel of latitude with an average speed of

600 knots. Calculate the time, in hours, taken for

the flight.

80W

N

S

O

3550

X

PQ

R

45W 100E

135E

50N

145

N

S

O

P60S

4800

70E110W

Q

R


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(31) PLANS AND ELEVATIONS

(a) Types of lines are used when drawing the plans and elevations of solids

thick solid lines ( ) visible edges

thick dashed lines ( ----------------------- ) hidden or unvisible edges

(b) Plans and elevations

(c) Examples

Solid Plan Viewed from X Viewed from Y

object

side elevation

plan

front elevation

X

Y

10 cm

8 cm

4 cm

10 cm

8 cm

10 cm

8 cm

YX

8 cm

10 cm4 cm

10 cm

8 cm

10 cm

4 cm

Y

X

9 cm

5 cm12 cm

5 cm

12 cm

9 cm

12 cm

9 cm

5 cm

X

Y

3 cm

2 cm

3 cm

3 cm

1 cm

3 cm

2 cm

1 cm

3 cm

1 cm

2 cm

2 cm

3 cm

1 cm

2 cm

7 cm

6 cm5 cm

5 cm

X

Y 6 cm

2.5 cm

2.5 cm

5 cm

5 cm

7 cm

2 cm

5 cm

6 cm


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Solid Plan Viewed from X Viewed from Y

5 cm

2 cm

4 cm

7 cm 2 cm

3 cm

X

Y 5 cm

4 cm

2 cm3 cm

5 cm

7 cm2 cm

2 cm

3 cm

2 cm

2 cm

4 cm

5 cm

5 cm

2 cm

2 cm

7 cm 5 cm

X

Y 5 cm 2 cm

5 cm

7 cm

2 cm

2 cm

5 cm

5 cm

5 cm

2 cm

2 cm

3 cm

XY

2 cm6 cm

3 cm

4 cm7 cm

4 cm

8 cm4 cm

3 cm

3 cm

4 cm 8 cm

4 cm2 cm

5 cm

4 cm

6 cm

3 cm

4 cm2 cm

3 cm

3 cm 4 cm

4 cm

6 cm 2 cm

X

Y3 cm 4 cm

3 cm

3 cm 4 cm

2 cm

6 cm

2 cm

4 cm

4 cm

6 cm

5 cm

X6 cm

6 cm

2 cm3 cm

7 cm

3 cm

Y

6 cm

5 cm

1 cm

2 cm

3 cm6 cm

5 cm

1 cm

2 cm

6 cm7 cm

3 cm6 cm

6 cm

Y

5 cm

3 cm

4 cm6 cm

3 cm

2 cm

2 cm

X

4 cm

5 cm

3 cm

3 cm

2 cm

2 cm

3 cm

2 cm6 cm

5 cm3 cm

4 cm 2 cm

2 cm

3 cm


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Formulae v2

Documents

Transcript of Formulae v2