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Heelgetalle en die ordening daarvan

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5. Die laaste syfer van 3 1993 is ...

6. Jy ry teen ’n konstante spoed van 105 km per uur verby telefoonpale wat ewe ver van mekaar af staan. As dit 72 sekondes neem om van die eerste tot die vyftiende paal te reis, bereken die afstand, (in meter) tussen twee opeenvolgende pale.

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:
1.2 die volgende getalle kan herken, klassifiseer en voorstel om hulle te beskryf en te vergelyk:
  • heelgetalle;
  • desimale breuke en persentasies;
1.2.5 optelling- en vermenigvuldiginginverses;
1.7 ’n reeks tegnieke gebruik om berekeninge te doen, wat die volgende insluit:1.7.1 die gebruik van kommutatiewe, assosiatiewe en distributiewe eienskappe met rasionale getalle;1.7.2 die gebruik van ’n sakrekenaar;
1.8 ’n reeks strategieë gebruik om oplossings te kontroleer, en die korrektheid van oplossings beoordeel.
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:
2.5 vergelykings oplos deur inspeksie, toets-en-verbeter- of algebraïese prosesse (optelling- en vermenigvuldiginginverses) en die oplossings deur vervanging toets;
2.8 konvensies van algebraïese noterings en die wisselbare, verenigbare en verspreibare wette gebruik om:2.8.4 algebraïese uitdrukkings wat in hakienotasie met een of twee stelle hakies en twee tipe bewerkings gegee word, te vereenvoudig;2.8.6 algebraïese uitdrukkings, formules of vergelykings binne konteks in eenvoudiger of meer bruikbare vorms te skryf.

Memorandum

1. “Minder as “0” nie positief nie

2.

-2 -1 0

  1. Temperature; bankbalanse; ens.

4. Getalle met geen breuke of desimale daarby nie, bv. 2 nie 2½ of 2,5 nie

5. Z

KLASOPDRAG 2

2. 4 0 – 8 0 = –4 0 C

  • –13
  • –5
  • –27
  • –8 – 5 + 7 = –6

4. Tel alle (+) getalle en (–) getalle op. Trek hulle van mekaar af.

  • –9 + 6 = –3
  • –18 – 13 + 7 = –14
  • 20 – 75 = –55
  • 10 – (–2) = 10 + 2 = 12

6. –31 – (–17) = –31 + 17 = –14

7.1 –6 + (–3) = –9

7.2 5 – (–5) = 10

HUISWERKOPDRAG 2

  • 13 – 18 + 4 – 17 = –18
  • –9 – (–8) + (–16)

–9 + 8 – 16 = –17

  • – (–16) 2 + (–3) 2

= –256 + 9

= –247

  • (–13) 2 – (–13)

= 169 + 13

= 179

  • a – b + b = a
  • a – b – a = – b
  • – b – b + a = –2 b
  • y 2 − x 2 − x 2 = y 2 size 12{y rSup { size 8{2} } - x rSup { size 8{2} } - x rSup { size 8{2} } =y rSup { size 8{2} } } {}
  • waar
  • − x − y ≠ − x + y size 12{ - x - y<>- x+y} {} fals
  • y + z = z + y size 12{y+z=z+y} {} waar
  • − x + y = − x + y size 12{ - x+y= - x+y} {} waar

3.1 a = –2

3.2 a = 12

3.3 a = –3

3.4 –8 = a

4. R615 – R(46 + 480 + 199)

= R615 – R725

= R110 (–) Verlies

KLASOPDRAG 3

1. ( )

2. of

3. × of ÷ : van links na regs

4. + of – : van links na regs

  • –42
  • 36 + 34 = 70
  • 35
  • 3 x (–1) + 6 = –3 + 6 = 3
  • –24 × 25 = –600
  • (–2) 3 = –8
  • (–64) – (+2) = –64 – 2

= –66

  • (15 – 9) 2 = (6) 2 = 36
  • (–6) 2 = 36
  • –2(9) = –18
  • − 11 3 size 12{ { { - "11"} over {3} } } {} = –3 1 3 size 12{ { {1} over {3} } } {}
  • 24 − 12 + 2 size 12{ { {"24"} over { - "12"+2} } } {} = 24 − 10 size 12{ { {"24"} over { - "10"} } } {} = –2,4
  • –6 x 5 7 size 12{ { {5} over {7} } } {} = − 30 7 size 12{ { { - "30"} over {7} } } {} = –4 2 7 size 12{ { {2} over {7} } } {}
  • 53 − 25 size 12{ { {"53"} over { - "25"} } } {} = –2 3 25 size 12{ { {3} over {"25"} } } {} of –2,12

1.5 –50 ÷ 5 = –10

2. p = (–2) × (3) ÷ (–2) 2

= –6 ÷ 4

= − 6 4 size 12{ { { - 6} over {4} } } {} = –1 1 2 size 12{ { {1} over {2} } } {} / –1,5

  • p = 4(–2)(3) ÷ (–2)(3)

= –24 ÷ (–6)

= 4

HUISWERKOPDRAG 2

  • (13) 2 – (–13) 2 – 13 2

= 169 – 169 – 169 = –169

  • (7 – 8) 2 – (8 – 7) 2 – 8 2 – 7 2

= (–1) 2 – (1) 2 – 64 – 49

= +1 – 1 – 64 – 49

= –113

  • (5)3 – 33 – 22

= 15 – 55

= –40

2. − 147 − 21 size 12{ { { - "147"} over { - "21"} } } {} – (–55)

= 7 + 55

= 62

3. 17 x (–15) ÷ (–7)

= –255 ÷ (–7)

= 36,4

4. (–88 + 7) – (–58)

= –81 + 58

= –23

5. –7 – (–5 × 17)

= –7 + 85

= 78

  • p = –60
  • p = –8
  • p = 7
  • 2 p + 6 = –4

p = –5

7. a = 4

8. –(–3) = 3

  • {–1; 0; 1; 2; 3}
  • {2; 3; 4; 5}
  • {–1; –2; –3}

Tutoriaal 1

1. 30; − 53 7 size 12{ { { - "53"} over {7} } } {} = –7,6 √; 10; –145 √

2.1 η size 12{η} {} = 27 4 size 12{ { {"27"} over {4} } } {} = 6 3 4 size 12{ { {3} over {4} } } {}

η size 12{η} {} >6 3 4 size 12{ { {3} over {4} } } {}

η size 12{η} {} size 12{ in } {} {7; 8; 9; . . . } √√

  • {5; 4; 3; 2; 1} √
  • √√
  • √√
  • [–(4)] 3 √ = –64 √
  • –8 – 9 + 8 + 9 √ = 0 √
  • 15 + (–40) √ + (–12) √ = –37 √
  • 6 1 11 size 12{ { { { {6}} rSup { size 8{1} } } over {"11"} } } {} x 1 − 2 4 4 size 12{ { {1} over { - { {2}} { {4}} rSub { size 8{4} } } } } {} √

= – 1 44 size 12{ { {1} over {"44"} } } {} √

  • (0,09) √ x (–0,04) = –0,0036 √
  • –1 √√
  • 87 √√

√ √ √

4.8 –0,225 a 5

4.9 − 1 2 3 a a b b 4 b 3 4 a 5 b size 12{ left ( { { - { {1}} { {2}} rSup { size 8{3} } { {a}} rSup { size 8{a { {b}}} } { {b}} rSup { size 8{ { {4}}b rSup { size 6{3} } } } } over { { {4}} { {a}} rSup { { {5}}} { { size 12{b}}}} } right )} {}

√ √ √

= 9 a 2 b

  • [3(–1) – 3{–2)] 2 √

= [–3 + 6] 2

= 9 √

  • –3(–2) 3 + 3(–1) 2 √

= –3(–8) + 3(1) √

= 24 + 3

= 27√

5.3 3(–2) 2 √

= 3(4)

= 12 √

TOETS

  • 0 √
  • 1 √
  • − 8 4 M 6 3 2 M 3 size 12{ { { - { {8}} rSup { size 8{4} } { {M}} rSup { size 8{ { {6}} rSup { size 6{3} } } } } over { { {2}} { {M}} rSup { { {3}}} } } } {} - –4 M 3 √√

√ √

  • –8 c 12 d 9

√ √ √

  • 30 p 5 q 6
  • − 6 3 a 8 6 − 2 a 2 size 12{ { { - { {6}} rSup { size 8{3} } a rSup { size 8{ { {8}} rSup { size 6{6} } } } } over { - { {2}}a rSup { { {2}}} } } } {} + 12 a 6

√ √ √

= 3 a 6 + 12 a 6 = 15 a 6

  • –2 + 3 + 4 + 1 √ = 6 √
  • –6 a 3 – 2 a 2 b – 4 a 3 – 5 ab 2 √

= –10 a 3 – 2 a 2 b – 5 ab 2 √√

√√√1.9 k 4 3M 9 size 12{ { {k rSup { size 8{4} } } over {3M rSup { size 8{9} } } } } {}

√ √ √

  • –3 a 2 b 2 + 6 ab 2 + 4 ab

√ √ √

  • –3 a 4 + 1 [ 12 a 6 − 4a 2 size 12{\[ { {"12"a rSup { size 8{6} } } over { - 4a rSup { size 8{2} } } } } {} – 4a 2 − 4a 2 ] size 12{ { {4a rSup { size 8{2} } } over { - 4a rSup { size 8{2} } } } \]} {}
  1. –2(2 p – 3 q – 4 r ) – 3(–2 p + 3 r – 4 q ) √

= –4 p + 6 q + 8 r + 6 p – 9 r + 12 q √√

= 2 p + 18 q – r √

3. 5 a 3 b – 10 a 3 b 3 – [–6 a 2 b 2 (–3 a + 12 ab )] √

5 a 3 b – 10 a 3 b 3 – [18 a 3 b 2 – 72 a 3 b 3 ] √

5 a 3 b – 10 a 3 b 3 – 18 a 3 b 2 – 72 a 3 b 3 √

5 a 3 b – 18 a 3 b 2 + 62 a 3 b 3 √

4. − 2 ( a + b ) − 3a size 12{ { { - 2 \( a+b \) } over { - 3a} } } {} = − 2a − 2b − 3a size 12{ { { - 2a - 2b} over { - 3a} } } {} √

= − 2 a − 3 a size 12{ { { - 2 { {a}}} over { - 3 { {a}}} } } {} – 2b − 3a size 12{ { {2b} over { - 3a} } } {}

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

√ √

√ √

5. x ( x + 1)( x + 2) + 1

= ( x 2 + x )( x + 2) + 1

= x 3 + 2 x 2 + x 2 + 2 x + 1

= x 3 + 3 x + 2 x + 1

2(3)(4) + 1 = 25

4(5)(6) + 1 = 35 Onwaar

5(6)(7) + 1 = 211 Onwaar

Verrykingsoefening

1. 1 3x − 1 3 size 12{ { {1} over { { {3x - 1} over {3} } } } } {}

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

2 1 size 12{ { {2} over {1} } } {} = 3x − 1 3 size 12{ { {3x - 1} over {3} } } {}

6 = 3 x + 1

7 = 3 x

(2 1 3 size 12{ { {1} over {3} } } {} ) 7 3 size 12{ { {7} over {3} } } {} = x

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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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