express ulangkaji mathe pt3

8/20/2019 Express Ulangkaji Mathe Pt3

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[ SM. ST. PETER TELIPOK ] [ LEE CHIONG TEE ] [ EXPRESS ULANGKAJI MATEMATIK PT3 / 50 ]

EXPRESS ULANGKAJI MATEMATIK PT3 / 50

(A) NOMBOR BULAT

→ nilai tempat, nilai digit, pembundaran

(B) NOMBOR PERPULUHAN

→ nilai tempat, nilai digit, pembundaran

(C) PECAHAN ~ kalkula!"

→ pecahan wajar, pecahan tak wajar, pecahan bercampur ; pecahan setara

Pecahan wajar, Pecahan setara Pecahan bercampur ↔ Pecahan tak wajar

4

32 #

4

11

($) INTEGER

• 50 m d !"#"$ "%"& '"() * −50 • &($( m+,,-") 10°C * 10

(E) URUTAN % POLA NOMBOR

→ nombor berpola

G", 1 3 5 11 13 15 1 1 21 . . .

G+," 2 4 6 10 12 14 1 16 20 22 . . .

+%d"," 2 3 5 11 13 1 1 23 2 31 3 41 43 4 53 5 1 1 3 63 6 7

G",d"",-",d"", 3 ⇒ 3 12 15 16 21 24 2 30 . . .

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-",d"", 5 ⇒ 5 10 15 20 25 30 35 40 45 50 . . .

("&" d(" 1 4 1 25 3 4 4 61 100 121 144 1 1 125 . . .

→ faktor, faktor perdana ~ kalkula!"

Faktor Faktor perdana

45 * 1 × 45

* 3 × 15

* 5 ×

∴ 8")9% !"- 45 * 1 3 5 15 45 →

8")9% +%d"," !"- 15 * 5 3 13

→ GSTK (gandaan sepuna terkecil!, FST" (faktor sepuna terbesar!, faktor sepuna, gandan sepuna

∴ -",d"", &+(,:" !"- 1 d", 40

* -",d"", 60* 60 10 . . .

∴ 8")9% &+(,:" !"- 1 d", 40

* 8")9% !"- 6* 1 2 4 6

∴ -",d"", &+(,:" !- 16 24 d", 40 * -",d"", 2

∴ 8")9% &+(,:" !- 16 24 d", 40 * 8")9% !-

(&) NOMBOR BERARAH ~ kalkula!"

NOTA ; peraturan melakukan operasi bercampur

→ +%)"m" '"(", 9+%"& d"'"m (%(,-",

→ +m(d", '"(", 9+%"& d"%"! ")"( !"$"- d"% % + ",",

→ "$% '"(", 9+%"& )"m!"$ ")"( )9'" d"% % + ",",

* 1.06 − 2.

* − 1.2

* 12 − 1

* −4

* 3.406 1.5

* 4.06

* 13 24

* 3

* 16 − 21

* −3

#15

2 ÷

5

13

#3

2 ÷

5

13

(G) KUASA $UA ' PUNCA KUASA $UA % KUASA TIGA ' PUNCA KUASA TIGA ~ kalkula!"

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(H) UKURAN ASAS ~ kalkula!"

Sistem #$ jam

1.00 ".m. * J"m 0100

2.00 ".m. * J"m 02003.00 ".m. * J"m 0300

4.00 ".m. * J"m 0400

5.00 ".m. * J"m 0500

.00 ".m. * J"m 000

.00 ".m. * J"m 000

6.00 ".m. * J"m 0600.00 ".m. * J"m 000

10.00 ".m. * J"m 1000

11.00 ".m. * J"m 1100

12.00 .m. * J"m 1200

1.00 .m. * J"m 1300

2.00 .m. * J"m 14003.00 .m. * J"m 1500

4.00 .m. * J"m 100

5.00 .m. * J"m 100

.00 .m. * J"m 1600

.00 .m. * J"m 100

6.00 .m. * J"m 2000.00 .m. * J"m 2100

10.00 .m. * J"m 2200

11.00 .m. * J"m 2300

12.00 ".m. * 2400 / 0000

(I) SU$UT $AN GARIS

→ jenis sudut

sudut tirus sudut tepat % tegak sudut cakah sudut refleks

θ < 0° θ * 0° 0° < θ < 160° 160° < θ < 30°

3


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→ jenis garis

&" selari dengan ' &" berserenjang dengan ' &" bersilang dengan '

&(d() !+%)+,)",-", !(=( "d"'"$ &"m"

→ sudut pada garis lurus

sudut pelengkap % sudut penggenpa sudut bersebelahan sudut putaran lengkap

• " a° d", b° "d"'"$ &(d() +'+,-"

⇒ a b * 0

• " a° d", b° "d"'"$ &(d() +,--+,"

⇒ a b * 160 a b * 160 a b c * 30

→ ciri)ciri garis selari

sudut sepadan sudut berselang seli hasil tambah sudut pendalaman

a b * 160

NOTA

(J) POLIGON

→ poligon * bukan poligon + rajah tertutup & disempadani oleh tiga atau lebih garisan

9'-9, !(", 9'-9, !(", 9'-9,

→ segi tiga

segi tiga sama sisi segi tiga sama kaki segi tiga bersudut tegak segi tiga tidak sekata

∴ a * 160 − b − b

∴ b *

2

160 a−

∴ a b * 0

∴ a * 0 − b

∴ b * 0 − a

a b c * 160

>'",-", -"%& &m+)% ? !'",-", "& &m+)%

3 1 0 0

4


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→ segi empat

segi empat segi empat sama segi empat tepat segi empat selari

a b c d * 30e * a b c

a b * 160

>'",-", -"%& &m+)% ? !'",-", "& &m+)%

0 4 2 0

rombus lelaang Trapeium

a b * 160

a b * 160

c d * 160

a b * 160

>'",-", -"%& &m+)% ? !'",-", "& &m+)%

2 1 0 1

→ poligon sekata + semua sisinya sama panjang & semua sudut pedalaman adalah sama saiz

P!l!*Bla*a* ++

( n )Bla*a*

,ak+ +-."

Bla*a*

a!*al

Ha+l a-a1

+uu ,.ala-a*

(n 2) × 40

Ha+l a-a1

+uu ,.lua"a*

&+- )-" &"m" && 3 3 0 160

30

&+- +m") &"m" 4 4 2 30

+,)"-9, 5 5 5 540

$+&"-9, 20

$+)"-9, 14 00

9)"-9, 6 6 20 1060

,9,"-9, 2 120

d+"-9, 10 10 35 1440

N!a $"&' )"m!"$ &(d() +,d"'"m", / +'("%", "d"'"$ &"m" (,)( 9'-9, &+")" d", )d" &+")"

• &(d() +'("%", &(d() +d"'"m", * 160°

• &(d() +d"'"m", *

sisibilangan

pedalaman sudut tambahhasil

• &(d() +'("%", *n

°30 ⇒ n *

peluaran sudut

°30

• &(d() "d" (&") *n

°30 ⇒ n *

pusat pada sudut

°30

(K) PERATUSAN

→ penukaran pecahan % perpuluhan kepada peratusan * sebalikna

pecahan ↔ peratusan perpuluhan ↔ peratusan

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→ peratus kenaikan % penurunan ; peratus untung % rugi ; peratus diskaun

• +%")(& +,"", / +,(%(,", * @100"&"' ,'"

)(%(,,'"," /,'"

×

• +%")(& +(,)(,-", / +%(-", * @1009& $"%-"

%(- ,'" /(,)(,- ,'"×

• +%")(& d&"(, * @100 ("'","&"'$"%-"

d&"(, ,'"×

→ faedah ; di-iden ; komisen

• 8"+d"$ * %,&"' × "d"% 8"+d"$ × m"&"

• dd+, * (m'"$ +'"!(%", × "d"% dd+,

• 9m&+, * $"%-" ("'", × "d"% 9m&+,

→ pelbagai contoh

• 6

4 × 100 * 506 ?

4

2 × 100 *

506

B2 RD"' m+m!+' &+!("$ =9m()+% d+,-", $"%-"RM2000. " m+,("' 9m()+% )+%&+!()d+,-", $"%-" RM100. H)(,-", +%")(&

+%(-", $"%-" 9m()+% )( F

• 2000 − 100 * 400

• 2000

400 × 100 * 206

B3 C"% ,'" "$% !"- 50 " " !+%)"m!"$&+!",:" 16@.

B4 H"%-" &+!("$ %"d9 "'"$ RM35.60. I" d("'d+,-", d&"(, 20@ d&=9(,). C"% $"%-" ("'", %"d9 )( F

B5 AD% m+,-"m!' !"$"-", d"'"m &(")( +%)",d,-", (D. " m+,"#"! 5@ d"%"d"&9"'",&9"'", )( d+,-", !+)('. " m+,"#"! 4

&9"'", d+,-", &"'"$. C"% (m'"$ !'",-",&9"'", d"'"m (D )(

B K+(,)(,-", 30@ d+%9'+$ &+'+"& &(")( !"%",-", d("' "d" $"%-" RM50. C"% $"%-"

9& !"%",-", )( F

(L) TEOREM P7THAGORAS 8 PERIMETER % LUAS

→ Teorem Pthagoras

3 9 5 5 2 3 : 29 25 4 5 :

6 10 10 24 2 14 46 50 1 30 34


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12 15 15 3 3

12 1 20 ; 90 9 20 2 2;

H * 22 O A + ? H 2 * A2

O2 A * 22 O H − ? A2 * H 2 −

O2O * 22 A H − ? O2 * H 2 − A2

→ perimeter, P

• P+%m+)+% * (m'"$ ",",- -"%& :",- m+,()( &(")( %""$

segi empat sama segi empat tegat segi tiga sama sisi segi tiga sama kaki

P * a a a a

* 4a

P * a a b b

* 2a 2b

P * a a a

* 3a

P * a b b

* a 2b

→ luas, .

segi empat sama segi empat tegat segi empat selari

• L * a × a

• a * L

• L * a × b

• a *b

L


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• PQ * QR * S * !

→

sudut pada pusat bulatan, sudut pada lilitan bulatan, sudut sisi empat kitaran sudut pada lilitan bulatan sudut pada lilitan (dicangkum oleh semibulatan!

hubungan antara sudut pada pusat dengan sudut pada lilitan (pada suatu lengkok ang sama panjang!

sisi empat kitaran lain)lain

(N) PEPEJAL GEOMETRI

→ ,.,.=al' la*a* ++' la*a* u>u' la*a* ,."-ukaa*

pepejal bilangan sisi bilangan bucu bilangan permukaan

(!(& 12 6

(!9d 12 6

&',d+% 2 0 3

9, 1 1 2

%"md B&+- +m") 6 5 5P%&m" B&+-)-" 5

%&m" B)%"+D(m 12 6

S8+%" 0 0 1

→ ,.,.=al ' .*a*a* ' =u-la1 lua+ ,."-ukaa* ( /P.) ' isipadu (0 )

Pepejal "entangan /umlah luas permukaan 1sipadu

(!(&

a2 a3

(!9d2 [ ab ac bc ]

?

a3

?

6


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2ab c B2a 2b'("& )"" × a

&',d+%

2π j2 2π jhπ × j2 × h

?

'("& +%")", %+,)"& × h

9m

π j2 π jt 3

1 × π × j2 × h

?

31

× '("& )"" × h

%"md B&+- +m")

ab 2 [ bk at 2

1

2

1+

]

?ab at bk

3

1 × a × b × h

?

31 × '("& )"" × h

%&m" B&+- )-"2 [ ab

21

] ah bh th

?

ab h Ba b t

2

1 × a × b × h

?

'("& +%")", %+,)"& × h

Pepejal "entangan /umlah luas permukaan 1sipadu

%&m" B)%"+D(m2 [ CCBB

21 hba +

at bt ht kt

?

Ba bBh Ba b h k Bt

2

1× Ba b × h × t

?

'("& +%")", %+,)"& × h

&8+%"

4 × π × j2

3

4 × π × j3

$+m&8+%"

3 × π × j2

3

2

×

π

×

j

3

→ .*a*a* ,aa "

C!*!1 ; C!*!1 2 ;


10/41


(O) UNGKAPAN ALGEBRA

→ pembolehubah, objek

→

sebutan algebra + hasil darab suatu nombor dengan pembolehubah

→ sebutan serupa, sebutan tak serupa

sebutan serupa sebutan tak serupa

5h −h

h h

2 2 "y

3

2 "y

5

y"

− y"

g −3 g 2 g

5 k

3 p 2 2abc −4bcd

5

2

de#

→ ungkapan algebra

( terdiri daripada satu sebutan algebra $ gabungan sebutan algebra dan nombor dengan operasi % atau dan

− '

ungkapan algebra bilangan sebutan "ilangan pembolehubah pembolehuhah

3 " − 2 2 1 "

5 − 3c ( 3 2 c (

−2 "y 4abc 3 3 5 " y a b c

3 y2 y − 11 4 1 y

→ kembang

→ faktor + 3

* p B p − m

* 4e B1 − 3 #

→ faktor + #

a2 − b2 * Ba b Ba − b

10


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1 * 12 4 * 22 * 32 1 * 42 25 * 52 3 * 24 * 2

4 * 62 61 * 2 100 * 102 121 * 112 144 * 122 1 * 132 1 * 142

→ faktor + 4

11


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→ faktor + $

→ pecahan algebra

12


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(P) RUMUS ALGEBRA ? CATATAN ,."ka"a "u-u+ +.*a+a ,!+@

→ perkara rumus

• m * 5 − 3n2 Bn

3n2 * 5 − m

n2 *

3

5 m−

3

5 mn

−=

• 32

=+h

g B g

h

g +2 * 32

2 g * h

g * h − 2

→ menentukan nilai suatu pembolehubah

• !+% y * 2 p − 4( 3r . C"%

B ,'" y ""!'" p * 5 ( * −1 d", r * 3

∴ y * 2B5 − 4B−1 3B3

* 16

B ,'" ( ""!'" y * 4 p * d", r * 2

∴ 4 * 2B − 4( 3B2

4 * 14 − 4(

4 − 14 − * −4(

4 * (

() PERSAMAAN LINEAR

→ konsep kesamaan 5 ,

→ persamaan linear dalam satu pembolehubah

persamaan linear dalam satu pembolehubah bukan persamaan linear dalam satu pembolehubah

3 − 2

p * 1 10a − 6 * 2a k 15 * 2 " 5 "

2 * 3 " y2 − y 2 * 0 t 3 1 *

→ persamaan linear dalam dua pembolehubah

persamaan linear dalam dua pembolehubah bukan persamaan linear dalam dua pembolehubah

4 " y * 6 5h 4b * 2b " y − 2 * 0m

1 n * 5 ab − a *

(

p * 2 "2 − 1 *

2 y

→ menulis persamaan linear dalam satu pembolehubah bagi maklumat ang diberi

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B1 a° 2a° d", 50° "'"$ )-" &(d() "d" &(")(

-"%& '(%(&.

• a 2a 50 * 160

B2 P",",- &++,- 9&)+% "d"'"$ 3 "' '+!"%,:" L =m. P+%m+)+% 9&)+% )( "'"$ 3=m.

• 2 L 2B L 3 * 3

→ selesaikan persamaan linear

" 3 * 5

" * 5 − 3

" − 3 * 5

" * 5 3

2 " *

" *2

" *

C!*!1

B" 2n * 3n − 4

/awapan

C!*!1 2

/awapan

B" 2n − 3n * −4

−n * −4

n * 4

B! 10k * 3 − k

10k k * 3

1k * 3

k *1

3

B" 12 * 3n 4 * n

B! # − # * −31

# − # * −31 −

−5 # * −40 # * 5

40−−

# * 6

→ menulis persamaan linear dalam dua pembolehubah bagi maklumat ang diberi

B1 "'"m %""$ d !"#"$ &+)" &")( +)" m+#"' &+9%",- +m",. S+)" +m", $",:" d!+,"%", !+%m", &")( d"%"d" +m") +,& +%m",", !9'" &+" BS $9 B H !9'" "%%,- d", =")(%. >9'"&+" d", $9 d)(,(", d"'"m %""$ )(. >'",-", +m", !9'" "%,- "d"'"$ d(" "' d"%"d" +m", =")(%. B ) * !'",-", +m", !9'" "%,- * * !'",-", +m", =")(%

S H H S S + ,

H H -

. /

S H H H S 0

H S 1

H 2 3 S +4++

S +, +-

H H S +. +/ +0 +1 +2

H H

• ) * * 16 • ) * 2*

B2 M"% m+m!+' 0 +,- &+)+m :",- !+%$"%-" 30 &+, d", RM 1. J(m'"$ ,'" &+)+m )( "'"$RM 42.50. B " * !'",-", &+)+m 30 &+, y * !'",-", &+)+m RM 1

• " y * 0 • 30 " 100 y * 4250

→ selesaikan persamaan linear serentak + kaedah gantian

C!*!1 C!*!1 2

→ selesaikan persamaan linear serentak + kaedah penghapusan

NOTA +a-a a*a ( ) 8 ak +a-a a*a ()

C!*!1 C!*!1 2

14


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6( * −24

( * 3 k * −3

k *31

−

C!*!1 3

−( * −2

( * 2

C!*!1 9

(R) KETAKSAMAAN LINEAR

→ konsep ketaksamaan 6 , 7

→ membina ketaksamaan linear daripada maklumat ang diberi

B1 M"%"$ '('(& " !"- (", M")+m") "'"$ 40 m"%"$. ............ " ≥ 40

B2 T,-- m"&m(m " +,d+%"", :",- !9'+$ m+'+"& )+%9#9,- "'"$ 5.5 m+)+%. ............ " ≤ 5.5

B3 G" !('",", " C AD+%%" m+'+!$ RM3500. ............ " 3500

→ mewakilkan ketaksamaan linear pada garis nombor dan begitu juga sebalikna

. . . − −5 −4 * " " * 3 4 5 . . . " * −2 −1 0 1 2

. . . −5 −4 −3 * " " * 2 3 4 . . . " * −3 −2 −1 0 1

" * −2 −1 0 1 " * −3 −2 −1 0 1 2

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→ menentukan % mewakilkan nilai sepuna bagi dua ketaksamaan linear serentak pada garis nombor

→ menelesaikan ketaksamaan linear

" 3 < 5

" < 5 − 3

" − 3 5

" 5 3

2 " ≤

" ≤ 3

2 "

≥ 3

" ≥

−2 " <

" D

2

−

2− "

3

" −

− " ≤ 3

" ≥ −3

− " ≥ −3

" ≤ 3

• S+'+&"", ; 4 − 2 " ≤ 10.

−2 " ≤ 10 − 4

−2 " ≤

" ≥ 2

−

" ≥ −3

• S+'+&"", ; − 5 " − ".

−5 " " < − −4 " < −1

" 4

1

−

−

" 4

1

→ menelesaikan ketaksamaan linear serentak

→ contoh)contoh lain

penambahan penolakan pendaraban pembahagian

[ -* ] * m,. m,. [ -*F ] * m,. − m". [ -*F ] * Bm,. Bm,. [ -*F ] *.

.m,

mak

[ -akF ] * m". m". [ -akF ] * m". − m,. [ -akF ] * Bm". Bm". [ -aF ] *.m,

.mak

B1 !+% 0


17/41


• " * 1 2 • y * 3 4 5

• ,'" m,m(m "

y *

2

3

B2 !+% 2 r < d", 2 − s < 4 d+,-", +"d"", r d", s "'"$ ,)+-+%. C"% ,'" )+%!+&"% !"- r − &.

• r * 2 3 4 5 6 ; • 2

− s < 4

− s ≤ 4 − 2

− s ≤ 2

s ≥ −2

∴ s * 0 1 2 ...

• ,'" )+%!+&"% r − &

* − B−1

* 10

(S) PEMBINAAN GEOMETRI

P+m!"$"- d(" &"m"&+%+,",- A5

>+%&+%+,",- d+,-", A5d", m+'"'( 5

>+%&+%+,",- d+,-", A5d", m+'"'( P

P+m!"$"- d(" &"m" ∠ A5*

S+- +m") &+'"% A5*6

∠ A5* * 0°

∠ A5* * 120° ∠ A5* * 30° ∠ A5* * 0°

∠ A5* * 45° ∠ A5* * 5° ∠ A>C * .5°

(T) PENJELMAAN

→ translasi

y

" ="% )%",&'"& 9!+ ? m+ B-(,"", +%)"& &(%$ (,)( m+m!",)( m+,"#"!

1


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[ SM. ST. PETER TELIPOK ] [ LEE CHIONG TEE ] [ EXPRESS ULANGKAJI MATEMATIK PT3 / 50 ] 16


19/41


→ pantulan ="% "& ",)('", ? m+ B-(,"", +%)"& &(%$ (,)( m+m!",)( m+,"#"!

$(%"", +,+'m"",

⇓

",)('", "d" "& "

$(%"", +,+'m"",

⇓

",)('", "d" "& y

$(%"", +,+'m"",

⇓

",)('", "d" -"%& y * 3

$(%"", +,+'m"",

⇓

",)('", "d" -"%& " * 5

→ putaran ="% (&") ()"%", ? m+ B-(,"", +%)"& &(%$ (,)( m+m!",)( m+,"#"!

$(%"", +,+'m"",

⇓

()"%", 0° () "%"$ "m

"d" (&") O

$(%"", +,+'m"",

⇓

()"%", 0° '"#", "%"$ "m

"d" (&") O

$(%"", +,+'m"",

⇓

()"%", 160° "d" (&") O

→ kekongruenan ="% m+ B-(,"", +%)"& &(%$ (,)( m+m!",)( m+,"#"!

→ bentuk serupa +,:+'+&"", m"&"'"$

B2 !+,)( "d"'"$ &"m" " → B1 sudut sepadan adalah sama B2 sisi sepadan adalah berkadaran

C!*!1

/awapan

1


20/41


B" RS B! ∠SPQ

→ pembesaran ="% (&") +m!+&"%", ? 8")9% &"'"

& → " +m!+&"%", "d" (&") O d+,-", 8")9% &"'" 2

$(%"", +,+'m"", ⇒ & → ' +m!+&"%", "d" (&") O d+,-", 8")9% &"'"21

& → +m!+&"%", "d" (&") O d+,-", 8")9% &"'" −1

• '("& m+ * k 2 × '("& 9!+ &"'" 8")9% k *objek panjang

imej panjang

(U) LOKUS $ALAM $UA $IMENSI

20


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() LUKISAN BERSKALA

→ menulis skala bagi lukisan dan objek ang diberi

• &"'" → &"D '(&", ; &"D 9!+ &+!+,"%

&"'" 1 ;21

? 2 ; 1 ⇒ 9!+ m+,"d !+&"% d+,-", 1 (,) +"d" 2 (,)

&"'" 1 ; 2 ⇒ 9!+ m+,"d +=' d+,-", 2 (,) m+,"d 1 (,)

C!*!1

O=.k

Luk+a*

Skala ( n) 1 ; 1 2 ; 1 → 1 ;21

1 ; 2

.

→ melukis bentuk geometri mengikut skala 3 8 n

C!*!1

C!*!1 2

→ penelesaian masalah

C!*!1

21


22/41


/awapan

22


23/41


() STATISTIK

→ mod

-! * ,'" d")" d+,-", ++%"", )+%),-- Bd")" d+,-", !'",-", )+%!",:"

4 2 6 3 4 3 3 5 3

B1 C"% &9% m9d !"- &+) d")" d ")"&.

• 2 3 3 3 3 4 4 5 6

• m9d * 3

Skor 10 20 30 40 50

Kekerapan 2 5 2 1

B2 C"% &9% m9d !"- d")" d ")"&.

• &9% m9d * 0

Skor 0 1 2 3 4

Kekerapan 1 3 " 5

B3 J" &9% m9d "'"$ 2 ="% ,'" m"&m(m ".

• " <

" *

Skor 1 2 3 4 5

Kekerapan 10 k 11 11

B4 J" &9% m9d "'"$ 2 ="% ,'" m,m(m k .

• k 11

k * 12

→ median

-.a* * ,'" :",- !+%"d" d )+,-"$)+,-"$ &(")( &+) d")" :",- )+'"$ d&(&(, m+,-() )+%)! m+," ? m+,(%(,

5 3 3 5 1

B1 C"% &9% m+d", !"- &+) d")" d ")"&.

∴ m+d", * 5

4 5 2 3 2 1 5 2 4 6

B2 C"% &9% m+d", !"- &+) d")" d ")"&.

∴ m+d", *2

43+ * 3F5

Skor 0 1 2 3 4 5 "ilangan murid 2 1 1 5 1

B3 C"% &9% m+d", !"- d")" d ")"&

∴ m+d", *2

21+ * F5

→ min

m, * databilangan

datanilai semuatambahhasil m, *

apanke jumlah

apanketengahtitik skor jumlah

+%

C+% /B ×

B1 C"% m, &9% !"- &+) d")" d !"#"$.

6 2 64 5 6 6

•

6656426 +++++ * :

B2 H)(,- m, m"%"$ (", !"- d")" d !"#"$.

9arkah 4 6 62 6

Kekerapan 5 10 2 3

•

32105

C3B6C2B62C10B6C5B4

++++++

*

:4F

23


24/41


B3 M, (m(% E,= J('"&% P(", J+,,8+% d",)-" 9%",- "," m+%+" "'"$ 31 )"$(,. M,(m(% !"- )-" "," m+%+" "'"$ 1 )"$(,.H)(,- m, (m(% E,= J('"&% d", P(",J+,,8+%.

• 5 × 31 * 155

• 3 × 1 * 5

• 2

5155 − * 9;

B4 S")( &+) d")" :",- )+%d% d"%"d" ,9m!9%

m+m(,:" m, 42. A"!'" &")( ,9m!9% d)"m!"$ m, m+,"d 36. C"% ,'" ".

• × 42 * 252

• × 36 * 2

• " * 2 − 252 * 9

→ jadual kekerapan

C!*!1

")" d"'"m %""$ d !"#"$ m+,(,(", m"%"$ :",- d+%9'+$ 20 9%",- +&+%)" d"'"m &")( (D.

B" M+,--(,"", d")" )( '+,-"", "d("' ++%"", d %(",- "#"",.

B! N:")"", m9d.

/awapan

B"

Ma"ka1 1 2 3 4 5

K.k."a,a*

⇓

B! 4

→

piktograf

C!*!1

J"d("' d !"#"$ m+,(,(", !'",-", !(( d &(d() !"="", !"- )-" !("$ +'"&.

K.la+ Bla*a* uku

Am", 5

>+&)"% 0

C+%d"& 0

M"'(m") !"- +'"& >+&)"% d)(,(", &++,($,:" d"'"m )9-%"8 d %(",- "#"",. L+,-"", )9-%"8 )( (,)( m+#"' &+m(" m"'(m") d"'"m "d("'.

/awapan

24


25/41



26/41


→ carta bar

C!*!1

J"d("' d !"#"$ m+,(,(", )-" ")) :",- d&+%)" 9'+$ &+(m('", 50 9%",- m(%d.

Ak Bla*a* -u"

C")(% 24

H9 7

K9% 16

B" C"% ,'" 7 .

B! S+)+%(&,:" #"'", &+m(" d")" )( d+,-", m+'(& &")( ="%)" "'",- "d" %(",- "#"",.

/awapan

B" 50 − 24 − 16 * 6

→ carta pai

C!*!1

J"d("' d !"#"$ m+,(,(", !'",-", m(%d :",- !+%m", +m") +,& +%m",",.

M"'(m") !"- +%m",", !"dm,)9, d)(,(", &++,($,:" d"'"m ="%)" " d %(",- "#"",.

L+,-"", ="%)" " )( (,)( m+#"' &+m(" m"'(m") d"'"m "d("'.

/awapan

2


27/41


C!*!1 2 R""$ d !"#$ "'"$ ="%)" " :",- m+,(,(", !'",-", !("$ %"m!()", :",- dm"", 9'+$ 4 9%",-

m(%d.

H)(,-

B" m, !'",-", !("$ %"m!()", :",- dm"", 9'+$ &+9%",- m(%d

B! &(d() &+)9% :",- m+#"' "d.

/awapan

→ graf garis

C!*!1

J"d("' d !"#"$ m+,(,(", !'",-", m(%d d &+!("$ &+9'"$ :",- m+,d"") &9% 6A d"'"m &(")( ++%&"", (,)( )+m9$ 'm" )"$(, .

B" N:")"", m+d",.

B! >+%d"&"%", "d("' '(& &")( -%"8 -"%& d %(",- "#"",.

/awapan

B" 200

B!

2


28/41


(X) GRA& &UNGSI

→ mengira nilai pembolehubah bersandar, apabila nilai pembolehubah tidak bersandar diberi:

B1 J"d("' m+,(,(", ,'",'" +m!9'+$(!"$ " d", y !"- 8(,-& y * 2 "2 − 1.

" −1 0 1 2

y p −1 1 (

H)(,- ,'" p (.

• p * 2B−12 − 1 * 1

• ( * 2B22 − 1 *

• 1 * 4

→ mengenali bentuk graf apabila diberi fungsi dan begitu juga sebalikna

Graf .inear Graf Kuadratik Graf Kubik

→ cari pintasan), pintasan)

,*a+a* % pada paksi) → k!!"*a # 0 ,*a+a* % pada paksi) → k!!"*a # 0

• C"% ,)"&", " !"- 3 " − 4 y 24 * 0.

3 " − 4B0 24 * 0

3 " 24 * 0

3 " * −24

" * −6

• C"% ,)"&", y !"- 3 " − 4 y * 24.

3B0 − 4 y * 24

−4 y * 24

y * −

26


29/41


→ pelbagai contoh

B1 S")( -"%& '(%(& d+,-", 8(,-& y 8 3 " k m+'"'( )) B3 1 ="% ,'" k .

• 1 * 3B3 k

1 * k

1 − * k

* k

B2 T) B−3 11 m+m("&", 8(,-&

B3 J"d("' d !"#"$ m+,(,(", ,'",'" +m!9'+$(!"$ " d", y !"- &(")( 8(,-&.

" −2 1 2

y −5 1 3

(,-& )( "'"$

→ melukis graf fungsi dengan skala ang diberi

C!*!1

" −2 −1 0 1 2 3 4

y 12 5 0 −3 −4 −3 0

• '(&", -%"8 &"'" → 2 =m +"d" 1 (,) "d" "& "

→ 2 =m +"d" 2 (,) "d" "& y

C!*!1 2

" −3 −2 −1 0 1 2 3

y −1 −3 1 −1 −3 1 1

2


30/41


• '(&", -%"8 &"'" → 2 =m +"d" 1 (,) "d" "& "

→ 2 =m +"d" 5 (,) "d" "& y

30


31/41


(7) IN$EKS

→ < indeks = dan < >ukum 1ndeks =

an 3 2 1 0 1 2 3 4 5

261

41

21

1 2 4 6 1 32 4 126

3 21 1

31 1 3 2 61 243 2

5125

1251

51

1 5 25 125 25 3125

3431

41

1

1 4 343 2401

21

4 * 4 *

210

10001

1001

101

1 10 100 1000

am × an * am % n am ÷ an * am − n n

m

a

a * am − n Bamn * a m× n

an × bn * Ba × bn n

n

b

a *

n

b

a

n

b

a −

#

n

a

b

a−n *

n

a

1 / an *

na−1

na

1 * n a n

m

a *

n

ma

1 * n ma < n

m

a * ( )mna

1*

( )mn a

a0 * 1

→ pelbagai contoh

• P+%m(d"$", p5 × p−3

* p5 B−3

* p5 − 3

* p2

• C"% ,'" ; 23 ÷ 2−1

* 23 − B−1

* 23 1

* 24

* 1

*2

1

−

×

k

k k

* k 1 2

* k

* e3 × d 6e4

* e3 4 d 6

* ed 6

* 5−1 × 22

*51 ×

4

*5

4

• C"% ,'" ; 52

2

33

÷−

*52

C2B2

3

3

−

* 34 − 2 5

* 31

* 3

• N'"", ; 232

C6B × 4−5 × 1

* 24 × 2−10 × 24

* 24 B−10 4

* 24 −10 4

• C"% ,'" ; 32 × 21

16 ×

2

3

2 • N'"", ;

3

2

2

1

2

1

6

322 ×

*

×

322

2

1

2

C322B

31


32/41


* 2−2

*4

1

*2

2

1

2

4

*4

6

* 2

32


33/41


• P+%m(d"$", ;

B3 # 5 g 2 × B # 4 −3 ÷ # −2 g

* 32 # 10 g 2 × # −12 ÷ # −2 g

* # 10 B−12 − B−2 g 2 −

* # 10 −12 2 g −5

* # 0 g −5

* g −5

• S+'+&"", ; 3 " − 1 * 61

3 " − 1 * 34

" − 1 * 4

" * 4 1 " * 5

• S+'+&"", ; 2n − 3 × 6n * 32

2n − 3 × 23Bn * 25

n − 3 3n * 5

n 3n * 5 3

4n * 6

n *4

6

n * 2

() TRIGONOMETRI

→ menukar unit sudut daripada


34/41



35/41


C!*!1 3 ;

/awapan

C!*!1 9 ;

/awapan

B"

∴ &, θ *210

*13

5

B!

• )",

∠ 9:; * 4

3

9; * 16

9H * 16 − 10 * 6

B"

)", "° *125

B!

• 9& y° * 53

PQ *

PQR * 12 * 21

→ penelesaian masalah

cari sudut cari panjang

(AA) NISBAH' KA$AR $AN KA$ARAN

→ mempermudahkan nisbah dua kuantiti kepada sebutan terendah + kalkula!"

B1 50- ; 1 - 50-

• 50- ; 1050-

1 ; 21

B2 2.4 ; 0.0

• 40 ; 1

B3 2 ;4

3

• 6 ; 3

35


36/41


→ mempermudahkan nisbah tiga kuantiti kepada sebutan terendah + kalkulator

B1 40ml 100

1l 0.1l

• 40ml ; 10ml ; 100ml

4 ; 1 ; 10

B2 1 ;4

3 ;

3

11

→ menentukan sama ada suatu kuantiti berkadar dengan kuantiti ang lain apabila diberi dua nilai bagi

setiap kuantiti tersebut

B1 R""$ d !"#"$ m+,(,(", ,'" " d",

,'" y.

Ad""$ ,'" y !+%"d"%", d+,-", ,'" " F

• tidak

B2 R""$ d !"#"$ m+,(,(", !'",-", d(%",

d", $"%-",:".

Ad""$ $"%-" d(%", !+%"d"%", d+,-", !'",-",,:" F

• a

→ masalah melibatkan nisbah bagi dua kuantiti

B1 " ; y * 2 ; y * 26 7 " * F

B2 " ; y * 5 ; " y * 46 " * F

B3 " ; y * 3 ; 5 y − " * 12 7 " % y * F

B4 RM 40 d9,-& ",)"%" S"8" d", R+",m+,-() ,&!"$ n ; 3. J" !"$"-", S"8""'"$ RM400 ="% ,'" n.

→ masalah melibatkan nisbah bagi tiga kuantiti

B1 " ; y * 3 ; 2 d", y ; z * 6 ; 5 7 " ; y ; z * F B2 " ; y * 2 ; 5 d", z ; y * 1 ; 3 7 " ; y ; z * F

3


37/41


B3 C'+m+,) E,+' d", N(% Im", m+,("' (9,m+,-() ,&!"$ 4 ; 5 ; 3. J" E,+' m+,d"")

RM12 '+!$ !",:" d"%"d" N(% Im", !+%"" ("'", :",- C'+m+,) d"")

B4 RM 50 d!"$"- ",)"%" M+%%: J+88%9, d",+" m+,-() ,&!"$ m ; 5 ; 6. !+%

M+%%: m+,d") RM 100 ="% ,'" m .

→ kadar

B1 R""$ d !"#"$ m+,(,(", )-" + "=",- P Q d", R. :",- !+%'",", &m d", $"%-".

P+ "=",- m",""$ :",- "',- m"$"' F

• P * g

R7

100

0.0 * RM 0.00 +% -%"m

• Q * g

R7

150

0.0 * RM 0.00 +% -%"m

• R * g

R7

200

20.1 * RM 0.00 +% -%"m

∴ P

B2 J"d("' d !"#"$ m+,(,(", "d"% !":"%",:",- d+,"", (,)( +%$dm")", m+'":"%,)+%,+) d &+!("$ (&") 9m()+%.

T.-,!1 -a+a Kaa" ,." =a-

S")( "m +%)"m" RM 5.00

(" "m !+%(),:" RM 4.00

S+)" "m !+%(),:" RM 3.00

C")$+%,+ m+,--(,"", +%$dm")", )( d"%10.00 ".m. $,--" 4.00 .m. "d" &(")( $"%

)+%)+,)(. H)(,- (m'"$ "m"(, :",- +%'(A&:"$ !":"% (,)( +%$dm")", )(.

• "m 100 − "m 1000 * "m

• * 1 2 3

• 1B5 2B4 3B3 * RM 22

B3 J"d("' d !"#"$ m+,(,(", "d"% !":"%", "%+% !"- &+!("$ +%+)" d &+!("$ 9m'+&.

9asa Kadar

S")( "m +%)"m" 2.00

&+)"21

"m

!+%(),:"1.50

C$%&)+ m+'+)" +%+)",:" &+'"m" 21

"m

$)(,- (m'"$ #",- :",- +%'( d!":"%,:".

• 21

* 1 11 B21

• 2.00 11B1.50 * RM 4F50

B4 J"d("' d !"#"$ m+,(,(", "d"% $"%-"(,)( m+m!+' &(")( %9d(.

100 (,) +%)"m" 20 &+, &+(,)

200 (,) !+%() 15 &+, &+(,)

(,) &+)+%(&,:" 12 &+, &+(,)

S$+%':,+D" m+m!+' 560 (,) %9d( )(.H)(,- (m'"$ :",- +%'( d!":"% 9'+$,:".

• 560 * 100 200 260

• 100B0.20 200B0.15 260 B0.12

* RM43F0

3


38/41


→ menukar daripada satu unit laju kepada unit laju ang lain:

B1 0 m−1 * 777. m&−1

• j

km

1

0 *

s

m

C3001B

C10000B

×

×

* 25 m&−1

B2 00 m m,−1 * 77 m−1

• m,1

00 m *

j

km

01

1000

00

* 54 m−1

B3 50 m&−1 * 777 m−1

• s

m

1

50

* j

km

00

1

1000

50

×

* 160 m−1

→ laju, masa, jarak

laju 5masa

jarak masa 5

laju

jarak

jarak 5 laju × masa Purata laju 5 masa jumlah

jarak jumlah

B1 J"%" ",)%" K9)" K,"!"'( d", R","( "'"$260 m. S+!("$ !"& !+%'+"& d"% K9)"K,"!"'( + R","( "d" ((' 10.45 ".m.d", )!" d "d" ((' 2.15 .m. "d" $"%

:",- &"m". H)(,-", '"( d"'"m m−1

&+'(%($ +%"'",", )(.

•

10451415

260

jam jam

km

−

# 60

m/

B2 J"%" d"% K9)" >+'() + K(d") "'"$ 140 m.S+!("$ !"& !+%)9'" d"% K9)" >+'() "d" "m

0630. P(%")" '"( !"& )( "'"$ 60 m$−1.

P((' !+%"""$ d"'"m &:&)+m 24 "m !"&

)( )!" d K(d") F

• m"&"


39/41


pecutan 5masa

a>al lajua>al laju −

B1 J9+:=+ m+m",d( d+,-", '"( &+%"-"m 0

m−1. S+'+"& 40 m,) d" m+m",d( d+,-",

'"( 110 m−1. C"% +=(")", d"'"m m−2

+%+)" )(.

• it

jkm

m,40

/C0110B − * 5 m/2

B2 S+!("$ )" !+%'+"& d+,-", '"( 20 m&−1 d",

'"(,:" !+%(%",- d+,-", &+%"-"m &+$,--" " !+%$+,) d"'"m m"&" 40 &""). H)(,-",

,:"$+=()", d"'"m m&−2 )" )(.

• +=()", *.40

C200B 1

s

ms −− * −0.5

m&−1

• ,:"$+=()", * 0.5 m&−1

(BB) KOOR$INAT

→ koordinat

P B1 3 Q B−4 −2

"&"'", O B0 0

∗ )+%'+)" "d" "& y

→ 99%d,") " * 0

∗ )+%'+)" "d" "& "

→ 99%d,") y * 0

→ titik tengah Q * jarak Q pada garis ang selari dengan paksi)

&+'"% d+,-", "& " ⇒ k!!"*a Q &"m"

titik tengah *

+ y

" "

2

21

jarak 5 "2 −

"1

B1 K99%d,") !"- )) P d", Q m"&,-m"&,-

"'"$ B−1 2 d", B 2. C"% 99%d,")

)) )+,-"$ !"- PQ.

•

−

+−2

2

1 * B3 −2

B2 H)(,-", "%" ",)"%" )) A B−1 : d",

5 B− :.

• −1 − B− * 4

B3 !+% -"%& '(%(& PQ "d"'"$ &+'"% d+,-", "& ". T) P B−4 2 "d"'"$ 5 (,) d"%"d" )) Q.

C"%", )))) :",- m(,-, !"- Q.

3


40/41


∴ B− 2 B1 2

→ titik tengah Q * jarak Q pada garis ang selari dengan paksi)

&+'"% d+,-", "& y ⇒ k!!"*a Q &"m"

titik tengah 5

+

2

21 y y " jarak 5 y2 − y1

B1 K99%d,") )) P d", Q m"&,-m"&,-"'"$ B5 14 d", B5 2. C"%", ))

)+,-"$ !"- PQ.

•

+

2

2145 * B5 6

B2 !+% P B0 − d", Q B0 6 ="% ",",-

PQ F

• 6 − B− * 5

B3 J"%" ",)"%" S B 9 n d", T B 9 −2 "'"$ (,). C"%", ,'",'" !"- n.

→ titik tengah Q * jarak Q bagi suatu titik dengan asalan

titik tengah 5

2

2

11 y " jarak 5 22 y " +

B1 C"% 99%d,") )) )+,-"$ !"- )) :",-m+,:"m!(,-", )) B12 1 d+,-", "&"'",.

•

2

1

2

12 * B 6

B2 A,)"%" :",- !+%() )) m",""$ :",- "',- d+") d+,-", "&"'", F

→ titik tengah Q * jarak Q bagi dua titik (tidak mempunai koordinat ang sama!

titik tengah *

++2

2

2121 y y " " jarak * 2122

12 CBCB y y " " −+−

40


41/41


B1 K99%d,") P "'"$ B−4 −2 d", 99%d,") Q

"'"$ B2 −6. K99%d,") )) )+,-"$ !"- -"%&

'(%(& PQ.

•

−+−+−

2

C6B2

2

24 * B−1 −5

B2 C"% "%" ",)"%" K B−4 d", L B20 −1.

* 22 CC1BBC204B −−+−−

* 25

* 25

B3 "'"m %""$ d !"#"$ Q "'"$ )) )+,-"$

-"%& '(%(& PR.

H)(,-", ,'" " d", y.

B4 J"%" ",)"%" ) B 1 d",

express ulangkaji mathe pt3

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