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1.5 Deling in algebra  (Page 3/3)

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3.1 Hoeveel terme het die bogenoemde uitdrukking? ......................... [1]

3.2 Skryf die koëffisiënt van p neer. ......................... [1]

3.3 Skryf die eksponent van p neer. ......................... [1]

3.4 Wat is -5 se resiprook? ......................... [1]

[4]

4.

4.1 Bereken hoeveel 7 a + 5 b + 9 c groter is as 2 c + 3 a - 7 c . [3]

4.2 Bereken hoeveel n + 3 m + 3 k kleiner is as -3 k - 7 m + 2 n [3]

4.3 Trek x size 12{x} {} ² - 2 x size 12{x} {} + 4 af van 5 c ² + 6 x size 12{x} {} - 9 [3]

[9]

5. Skryf ‘n algebraïese uitdrukking vir elk van die volgende neer.

5.1 23 m verminder met die vierkant van n ........................... [2]

5.2 Trek die produk van m en n af van die verskil tussen m en n . [2]

5.3 Jy het twintig muntstukke. y van hulle is vyftigsentstukke en die res is tiensentstukke.

5.3.1 hoeveel tiensentstukke het jy? (In terme van y ) [1]

5.3.2 Wat is die totale waarde van jou geld? [2]

[7]

6. Voltooi die volgende tabelle en gee by elk die formule y = ......

6.1 y = .................................................. [3]

x size 12{x} {} 3 5 7 9 10 15
y 9 25

6.2 y = .................................................. [4]

x size 12{x} {} 3 5 7 9
y 11 23 38

[7]

7. Bonuspuntvraag

Bestudeer die patroon:

1 = 1 x 1

1 + 3 = 2 x 2

1 + 3 + 5 = 3 x 3

1 + 3 + 5 + 7 = 4 x 4

Bereken nou:

1 + 3 + 5 + 7 + 9 + ..... + 21 + 23 + 25 + 27 + 29

[2]

Verrykingsoefening vir die vinnige werker

1. As p q beteken 3 p + q 2 , dan is (3  4)  5 gelyk aan ...

a) 60 b) 100 c) 87 d) 72 e) 91

2. In die vermenigvuldiging getoon is S en T verskillende syfers tussen 1 en 9. Die waarde van S + T is dus ....

S 6

x 2 T

2 1 5 0

a) 13 b) 14 c) 15 d) 16 e) 17

3. Hoeveel syfers is daar in 5 8 ?

a) 2 b) 5 c) 6 d) 8 e) 40

4. Die gemiddelde van die drie getalle 0,1 ; 0,11 en 0,111 is ...

a) 0,041 b) 0,107 c) 0,11 d) 0,1111 e) 0,17

5. In ‘n sekere klas is 1 / 3 van die leerlinge meisies. Ses seuns is ‘n kwart van die seuns in die klas. Hoeveel leerlinge is daar in die klas?

a) 18 b) 24 c) 27 d) 32 e) 36

6. As ½ - x = x - 1 / 3 , dan is x gelyk aan ...

a) 1 / 12 b) 1 / 6 c) 5 / 12 d) 5 / 6 e) 3 / 12

7. Bepaal die waarde van x indien .....

4 + x 3 = 2 size 12{ nroot { size 8{3} } {4+ sqrt { size 11{x}} } size 12{ {}=2}} {}

a) 4 b) 2 c) 64 d) 16 e) 0

Assessering

Leeruitkomstes(LUs)
LU 1
Getalle, Verwerkings en VerwantskappeDie leerder is in staat om getalle en die verwantskappe daarvan te herken, te beskryf en voor te stel, en om tydens probleemoplossing bevoeg en met selfvertroue te tel, te skat, te bereken en te kontroleer.
Assesseringstandaarde(ASe)
Dit word bewys as die leerder:
1.1 die historiese en kulturele ontwikkeling van getalle kan beskryf en illustreer;
1.2 die volgende getalle kan herken, klassifiseer en voorstel om hulle te beskryf en te vergelyk:1.2.3 getalle wat in eksponensiële vorm geskryf is, insluitend vierkante en derdemagte van natuurlike getalle en hul vierkants- en derdemagwortels;1.2.6 veelvoude en faktore;1.2.7 irrasionele getalle in die konteks van meting (bv. π size 12{π} {} en vierkants- en derdemagwortels van nie-perfekte vierkante en derdemagte);
1.6 skat en bereken deur stappe te kies wat geskik is om probleme op te los wat die volgende behels:1.6.2 veelvoudige stappe met rasionale getalle (insluitend deling met breuke en desimale);1.6.3 eksponente.
LU 2
Patrone, Funksies en AlgebraDie leerder is in staat om patrone en verwantskappe te herken, te beskryf en voor te stel, en probleme op te los deur algebraïese taal en vaardighede te gebruik.
Dit word bewys as die leerder:
2.1 numeriese en geometriese patrone ondersoek en uitbrei om te soek vir verwantskappe of reëls, insluitend patrone wat:2.1.1 in fisiese of diagrammatiese vorm voorgestel is;2.1.2 nie beperk is tot reeks met konstante verskil of verhouding;2.1.3 in natuurlike en kulturele kontekste gevind word; 2.1.4 die leerder self geskep het;2.1.5 in tabelle weergegee word;2.1.6 algebraïes weergegee word;
2.2 waargenome verwantskappe of reëls in eie woorde of in algebra kan beskryf, verduidelik en verantwoord;
2.3 verwantskappe tussen veranderlikes voorstel en gebruik om op verskeie wyses inset- en/of uitset- waardes te bepaal deur gebruik te maak van:2.3.1 verbale beskrywings;2.3.2 vloeidiagramme;2.3.3 tabelle;2.3.4 formules en vergelykings;
2.4 wiskundige modelle bou wat oplossings vir probleemsituasies weergee, beskryf en verskaf, terwyl verantwoordelikheid teenoor die omgewing en gesondheid van ander getoon word (insluitende probleme die konteks van menseregte, sosiale ekonomiese, kulturele en omgewingskontekste);
2.7 die gelykwaardigheid van verskillende beskrywings van dieselfde verwantskap of reël wat soos volg voorgestel word kan bepaal, analiseer en interpreteer:2.7.1 verbaal;2.7.2 in vloeidiagramme;2.7.3 in tabelle;2.7.4 deur vergelykings of uitdrukkings om sodoende die mees bruikbare voorstelling vir ‘n gegewe situasie te kies;
2.8 konvensies van algebraïese noterings en die wisselbare, verenigbare en verspreibare wette kan gebruik om:2.8.1 terme soos gelyk en ongelyk te klassifiseer en om die klassifikasie te verantwoord;2.8.2 gelyke terme te versamel;2.8.3 ‘n algebraïese uitdrukking met een, twee of drie terme met ‘n eenterm te vermenigvuldig of deel;
2.8.4 algebraïese uitdrukkings wat in hakienotasie met een of twee stelle hakies en twee tipe bewerkings gegee word, te vereenvoudig;2.8.5 verskillende weergawes van algebraïese uitdrukkings met een of meer bewerkings te vergelyk en om dié wat ekwivalent is te selekteer en die keuse te motiveer;2.8.6 algebraïese uitdrukkings, formules of vergelykings binne konteks in eenvoudiger of meer bruikbare vorms te skryf;
2.9 die volgende algebra-woordeskat binne konteks kan interpreteer en gebruik: term, uitdrukking, koëffisiënt, eksponent (of indeks), basis, konstante, veranderlike, vergelyking, formule (of reël).

Memorandum

KLASOPDRAG 1

  • 2 m 3
  • 5y 4 p 3 size 12{ { {5y rSup { size 8{4} } } over {p rSup { size 8{3} } } } } {}
  • 8 m 2 size 12{ { {8} over {m rSup { size 8{2} } } } } {}
  • 4 32 m 10 2m 4 size 12{ { {4 left ("32"m rSup { size 8{"10"} } right )} over {2m rSup { size 8{4} } } } } {} = 64 m 6

1.5 a 4

  • 20 b 2 c 3 size 12{ { {"20"b rSup { size 8{2} } c} over {3} } } {}

2. 4a 3 size 12{ { {4a} over {3} } } {} – 2 b

2.1.1 1a 5 size 12{ { {1a} over {5} } } {}

2.1.2 2a 2 5 size 12{ { {2a rSup { size 8{2} } } over {5} } } {} – 3

  • a – 4 b + 6 c
  • 2 b 2 size 12{ { {2} over {b rSup { size 8{2} } } } } {} – 5 a 2 b 2 c 4 + 9b 5 c 7 a 2 size 12{ { {9b rSup { size 8{5} } c rSup { size 8{7} } } over {a rSup { size 8{2} } } } } {}
  • − mpq 5 n 3 size 12{ { { - ital "mpq" rSup { size 8{5} } } over {n rSup { size 8{3} } } } } {} + 7m 5 n 4 q 4 size 12{ { {7m rSup { size 8{5} } n rSup { size 8{4} } } over {q rSup { size 8{4} } } } } {} + 5p 6 q 8 mn 3 size 12{ { {5p rSup { size 8{6} } q rSup { size 8{8} } } over { ital "mn" rSup { size 8{3} } } } } {}

HUISWERKOPDRAG 1

1.1 7p 7 q 3 mn 3 size 12{ { {7p rSup { size 8{7} } q rSup { size 8{3} } } over { ital "mn" rSup { size 8{3} } } } } {}

1.2 − ac 2 b size 12{ { { - ital "ac" rSup { size 8{2} } } over {b} } } {} + 12 a 3 b 5 c 2 size 12{ { {"12"a rSup { size 8{3} } b rSup { size 8{5} } } over {c rSup { size 8{2} } } } } {} – 8a 2 c 2 size 12{ { {8a rSup { size 8{2} } } over {c rSup { size 8{2} } } } } {}

1.3 5a 2 2 size 12{ { {5a rSup { size 8{2} } } over {2} } } {} – 25 b 2 size 12{ { {"25"b} over {2} } } {}

= 5a 2 − 25 b 2 size 12{ left ( { {5a rSup { size 8{2} } - "25"b} over {2} } right )} {}

1.4 a 6 b 6 . a 4 b 8 a 2 b 3 size 12{ { {a rSup { size 8{6} } b rSup { size 8{6} } "." a rSup { size 8{4} } b rSup { size 8{8} } } over {a rSup { size 8{2} } b rSup { size 8{3} } } } } {}

= a 8 b 11 size 12{a rSup { size 8{8} } b rSup { size 8{"11"} } } {}

1.5 3 16 k 2 p 8 2k 3 p 2 size 12{ { {3 left ("16"k rSup { size 8{2} } p rSup { size 8{8} } right )} over {2k rSup { size 8{3} } p rSup { size 8{2} } } } } {}

= 24 p 6 k size 12{ { {"24"p rSup { size 8{6} } } over {k} } } {}

  • 6 ab 2 + 12 a 2 – (6 ab )
  • 3b 2 size 12{ { {3b} over {2} } } {} + 3a b size 12{ { {3a} over {b} } } {}
  • 3 ab 2 + 6a 2 + 2 ab 4 ab size 12{ { {3 ital "ab" rSup { size 8{2} } +6a rSup { size 8{2} } +2 ital "ab"} over {4 ital "ab"} } } {}

= 3b 4 size 12{ { {3b} over {4} } } {} + 3a 2b size 12{ { {3a} over {2b} } } {} + 1 2 size 12{ { {1} over {2} } } {}

3. − 5 ab + 15 a 4 b 7 5a 3 b 2 size 12{ { { - 5 ital "ab"+"15"a rSup { size 8{4} } b rSup { size 8{7} } } over {5a rSup { size 8{3} } b rSup { size 8{2} } } } } {} = – 1 a 2 b size 12{ { {1} over {a rSup { size 8{2} } b} } } {} + 3 ab 5

TUTORIAAL

VRAAG 1

A.1. Onwaar: x

  1. Waar
  2. Waar
  3. Waar
  4. Onwaar: 2
  5. Waar
  6. Onwaar: 5 grondtal
  7. Waar
  8. Waar
  9. Onwaar: 4 y 2
  10. Onwaar: 7 x – 14 y
  11. Onwaar: 5 x 2
  12. Waar
  13. Waar

B .1. 3 b

  1. a + b

VRAAG 2

1. 7; 10; 16; 28

  1. (150 x 3) + 1 = 451
  2. (1 000 – 1) ÷ 3 = 333

VRAAG 3

A.1. 2(3)[3(–2) + 5(6)]

= 6[–6 + 30]

= 6[+24] = 144

2. 3 − ( − 2 ) − ( 6 ) ( − 2 ) − ( 6 ) size 12{ { {3 - \( - 2 \) - \( 6 \) } over { \( - 2 \) - \( 6 \) } } } {} of 3 + 2 − 6 − 2 − 6 size 12{ { {3+2 - 6} over { - 2 - 6} } } {}

= − 1 − 8 size 12{ { { - 1} over { - 8} } } {} = 1 8 size 12{ { {1} over {8} } } {}

B.1. 6 a – 5 b

2. 2 x – 4 y – 1

VRAAG 4

  1. 4 a 2 bc – ab 2 c
  2. 15 a 5 – 6 a 5 = 9 a 5
  3. 12 a 4 b 6 (– a 3 b 3 )

= –12 a 7 b 9

  1. –15 a 3 b 2 + 20 ab 3
  2. 14 a 2 + 8 a – 3 a – 9 a 2 + 4 a 2 + a

= 9 a 2 + 6 a

6. 9a 2 b 2 . 8a 3 4a 2 b 2 size 12{ { {9a rSup { size 8{2} } b rSup { size 8{2} } "." 8a rSup { size 8{3} } } over {4a rSup { size 8{2} } b rSup { size 8{2} } } } } {} = 18 a 3

7. 1 4 2 a a 4 b b 3 b c 1 size 12{ { { { {1}} { {4}} rSup { size 8{2} } { {a}} rSup { size 8{a { {4}}} } { {b}} rSup { size 8{b { {3}}} } } over { { {b}} { {c}} rSub { size 8{1} } } } } {} x 1 2 2 c 4 c2 − 7 a 3 b 2 size 12{ { { { {1}} { {2}} rSup { size 8{2} } { {c}} rSup { size 8{ { {4}}c2} } } over { - { {7}} { {a}} rSup { size 8{ { {3}}} } { {b}} rSup { size 8{ { {2}}} } } } } {}

= -4 abc 2

VRAAG 5

5.1 a 2 b + 5 ab 2 − 6 ab ab size 12{ { {a rSup { size 8{2} } b+5 ital "ab" rSup { size 8{2} } - 6 ital "ab"} over { ital "ab"} } } {}

= a + 5 b – 6

5.2 3a 2 b + 15 ab 2 − 18 ab ab size 12{ { {3a rSup { size 8{2} } b+"15" ital "ab" rSup { size 8{2} } - "18" ital "ab"} over { ital "ab"} } } {}

= 3 a + 15 b – 18

5.3 a 2 b + 5 ab 2 − 5 ab ab size 12{ { {a rSup { size 8{2} } b+5 ital "ab" rSup { size 8{2} } - 5 ital "ab"} over { ital "ab"} } } {}

= a + 5 b – 5

TOETS

  • 4 a 2 bc – 17 ab 2 c
  • 15 a 5 – 15 a 5 = 0
  • 7 ab + 4 ab = 11 ab
  • 3(8 a 6 b 9 ) (– a 6 b 6 )

= –24 a 12 b 15

  • –7 a 3 b 2 + 35 ab 3
  • 15 a 2 + 25 a – 4 a –8 a 2 + 6 a 2

=13 a 2 + 21 a

  • 3(3)[2(–2) + 4(6)]

= 9 [–4 + 24]

= 180

2.2 3 − ( − 2 ) − ( 6 ) ( − 2 ) + ( 6 ) size 12{ { {3 - \( - 2 \) - \( 6 \) } over { \( - 2 \) + \( 6 \) } } } {}

= 3 + 2 − 6 − 2 + 6 size 12{ { {3+2 - 6} over { - 2+6} } } {}

= 0 4 size 12{ { {0} over {4} } } {} = 0

2.3 3 ( 6 ) ( 3 ) ( − 2 ) ( 6 ) size 12{ { {3 \( 6 \) } over { \( 3 \) \( - 2 \) \( 6 \) } } } {}

= 18 − 36 size 12{ { {"18"} over { - "36"} } } {}

= – 1 2 size 12{ { {1} over {2} } } {}

2.4 9 ( 3 ) 5 3 2 − 3 ( 3 ) 3 size 12{ { {9 \( { {3}} \) rSup { size 8{ { {5}}} rSup { size 8{ {} rSup { size 6{3 rSup {2} } } } } } } over { - 3 \( { {3}} \) rSup { { {3}}} } } } {}

= 9 ( 9 ) 3 − 3 size 12{ { {9 \( { { {9}} \) } cSup { size 8{3} } } over { - { {3}}} } } {}

= –27

  • 3
  • –5
  • 1
  • − 1 5 size 12{ - { {1} over {5} } } {}
  • 7 a + 5 b + 9 c – (2 c + 3 a – 7 c )

= 7 a + 5 b + 9 c – 2 c + 3 a – 7 c

= 4 a + 5 b + 14 c

  • –3 k – 7 m + 2 n – ( n + 3 m 3 k )

= –3 k – 7 m + 2 n – n + 3 m 3 k

= –6 k – 10 m + n

  • 5 x 2 + 6 x – 9 – ( x 2 – 2 x + 4)

= 5 x 2 + 6 x – 9 – x 2 – 2 x + 4

= 4 c 2 + 8 x – 13

  • 23 m – n 2
  • ( m – n ) – mn
  • 20 – y
  • 50 y – 10(20 – y )

= 50 y – 200 + 10 y

= 60 y – 200

  • y = x 2

49; 81; 100; 225

  • y = 3 x + 2

17; 29; 18

7. 15 2 = 225

Verrykingsoefening

  1. b 2. a 3. c

4. b 5. e 6. c 7. d

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Source:  OpenStax, Wiskunde graad 8. OpenStax CNX. Sep 11, 2009 Download for free at http://cnx.org/content/col11033/1.1
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