# 2.1 Vergelykings  (Page 5/5)

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6. Laat n enige natuurlike getal wees. As 3 die tienesyfer in n² is, bepaal die enesyfer in n².

7. Die gemiddelde van drie heelgetalle is 86. As een van hulle 70 is, wat is die gemiddelde van die ander twee?

8. Saretha het ’n R10- en R20-noot by ABSA omgeruil vir ’n gelyke aantal 50c-, 20c-, en 5c-stukke. Hoeveel muntstukke het Saretha gekry?

## 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. - 5. Algemeen

6.1 1 – a = 7 a = –6

• a = 10
• 42 = a
• a = 11
• –30 = a
• a = 6
• 5 – 12 + 3 a = 5 a + 5 + 2

3 a – 5 a = 5 + 2 + 12 – 5

–2 a = 14

a = –7

• 2 a – 3 A = 24

a = 24

$\therefore$ a = –24

8.3 $\frac{a}{4}$ = $\frac{5}{1}$ a = 20

8.4 2 a = 16

a = 8

• a = –9
• 9 + 3 a + 3 = 8 – 8 a

11 a = –4

$\therefore$ a = $\frac{-4}{\text{11}}$ (–2 $\frac{3}{8}$ )

8.7 $\frac{-\text{12}}{a}$ = $\frac{-\text{24}}{1}$ –24 a = –12

a = $\frac{1}{2}$

• 10 a + 5 = 8 a + 12

2 a = 7

a = 3 $\frac{1}{2}$

8.9 –6 a + 6 = 6 a + 24

–12 a = 18

a = $\frac{-\text{18}}{\text{12}}$ $\frac{-3}{2}$ (–1 $\frac{1}{2}$ )

• Eie keuse

KLASWERKOPDRAG 2

1. 20: 1. Nommer 1: x

Nommer 2: 15 – x

2. Kinders: x 20 x

Grootmense: (140 – x ) 45(140 – x )

20 x + 45(140 – x ) = 5 580

20 x + 6 300 – 45 x = 5 580

–25 x = –720

x = 28,8 28 / 29

Grootmense: 140 – 28 = 112

of 140 – 29 = 111

HUISWERKOPDRAG 1 EN 2

• –3 a = 3

a = –1

• a = –4
• 8 x – 21 x + 10 = 28

8(3) – 2(3) + 10 = 28

– 6 + 10 = 28 √

2.2 5 x – 10 = 10 x – 10

5 x – 10 x = –10 + 10

–5 x = 0

x = 0

x  1R √

3.1 $\frac{1}{z}$ = $\frac{1}{\text{18}}$ $z$ = 18

3.2 1 – 5 z = 11

–5 z = 10

z = –2

3.2.1 Geen oplossing

3.2.2 z = –2

3.3 z + 3[ z + 2 z – 12] = 45

z + 3[3 z – 12] = 45

z + 9 z – 36] = 45

10 z = 81

z = 8,1

3.4 24 z – 32 – 2 z – 14 = 37

22 z = 83

z  3,77

3.5 z – 5 z + 40 = 48

–7 z = –88

z = 22

• 6 x – 8 = 55

6 x = 63

x = 10 $\frac{1}{2}$ (10,5)

4.2 x + 9 x = –64

10 x = –64

x = –6,4

4.3 x + x 1 x + 2 = –90

3 x = –93

x = –31 –31; –30; 29

4.4 Lemoene: 3 x x 45 36

Piesangs: x x 18 12

18 x + 135 x = 1 836

153 x = 1 836

x = 12

• Nou –6

Cameron: x + 8 [ 18 ]; x + 8 –6

Liam: x [ 10 ]; x – 6

3( x – 6) = x + 2

3 x – 18 = x + 2

2 x = 20

x = 10

4.6 Seëls: R1,20 : x 50 – x

R2,40 : 50 – x of x

120 x + 240(50 – x ) = 5 880 1,20(50 – x ) + 2,40 x = 58,8

120 x + 12 000 – 240 x ) = 5 880 60 – 1,20 x ) + 2,40 x = 58,8

–120 x = –6 120 of 1,20 x = 1,20

x = 51 x = 1

R1,20 4

4.7 Een gedeelte: x

Ander gedeelte: x + 550

x + 2( x + 550) = 18 000

x 2 x + 1 100 = 18 000

3 x = 16 900

4.8 Massa

3 792 kgVrouens: 18 3 x

Dogters: 25 x

25 x + 3 x (18) = 3 792

25 x + 54 x = 3 792

79 x = 3 792

x = 48

Dogters: 48 kg elk

Vrouens: 144 kg elk

Nommer12 x 4.9 Ene: 2 x 8

Tiene x x 10 4

omgeruil: 21 x

2 x – 36 = 12 x

9 x = 36

x = 4

4.10 Gr. 8: x + 3 2 257

Gr. 9: x 225

Gr. 10: x + 25 250

x + 32 + x + x + 25 = 732

3 x = 675

x = 225

TUTORIAAL 2

• Ya √ = Kan x se waarde uitwerk

[5 x = 5 0 . x = 0]

• x = 0 √
• a = 12 √√
• a = 7 √√
• a = $\frac{7}{3}$ (2 $\frac{1}{3}$ ) √√
• a = 3 √√
• a = 4 √√
• 32 = 4(2(9) – p ) √

32 = 4(18 – p )

8 = 18 – p

p = 10 √

• 45,67 = $\frac{p}{\text{21},3}$

p = 972,771 √

4.1 $\frac{7a}{1}$ + $\frac{a}{3}$ = 30

$\frac{\text{22}a}{3}$ = $\frac{\text{30}}{1}$

22 a = 90 √

a = $\frac{\text{90}}{\text{22}}$

= 4 $\frac{2}{\text{22}}$ = 4 $\frac{1}{\text{11}}$

4 a 2 – 3 a = 4 a 2 √

0 = 3 a

1. = a
2. –4 a + 8 = 3 a – 12 √

–7 a = –20 √

a = $\frac{\text{20}}{7}$ = 2 $\frac{6}{7}$ (of  2,86) √

• 5 a + 15 + 4 a + 5 = 2 a – 14 √

7 a = –34 √

a = 4,9 √

• 5 a = –2 a + 6 √

7 a = 6 √

a = $\frac{6}{7}$ (0,86) √

• 9 x – 28 = 5 x

4 x = 28

x = 7 √

5.2.1 x + 4 2 x + 2( x + 4) = 108 √

2 x + 2 x + 8 = 108

4 x = 100

x = 25√

√ √

$\therefore$ l = 29 cm b = 25 cm

5.2.2 A = 29 x 25 √

= 725 cm 2

6. x + x + 2 + x + 4 + x + 6 = 112 √

4 x = 112 – 12

4 x = 100

x = 25 √

$\therefore$ Nommers: 25; 27; 29; 31 √√

7. $\sqrt{\text{6162}}$  78 √

$\therefore$ 6 162 ÷ 78 = 79

$\therefore$ 78; 79 √√

TOETS 1

1.11 x = 34

1.12 –3 x = 1

–3 x = –6

$\therefore$ x = 2

1.13 2 x + 10 = 18

2 x = 8

$\therefore$ x = 4

1.14 2 x = 40 – 8

2 x = 32

$\therefore$ x = 16

• x – 6 – x – 1 = 5 x – 4

2 x x – 5 x = –4 + 1 + 6

–4 x = 3

x = $\frac{-3}{4}$

• 3 x = 15 – 6

3 x = 9

$\therefore$ x = 3

1.7 x = 4 x 2 = 8

1.8 6 x + 48 = 114

6 x = 66

$\therefore$ x = 11

1.9 x – 7 x = 30 – 9

3 x = 21

$\therefore$ x = 7

1.10 x – 2 – 2 x 3 = 7

x = 7 + 3 + 2

x = 12

$\therefore$ x = –12

• $\left(\frac{x-5}{3}\right)$ = 4

x = (4 × 3) + 5

= 17

• x + x 1 + x + 2 = 66

3 x = 63

x = 21

• 2 x + 2( x + 5,5) = 27

2 x + 2 x + 11 = 27

4 x = 16 x 4 cm

x = 4 x + 5,5 9,5 cm

2.4 Jonte: 5 x 50 5 x + x = 60

Dogters: x 10 6 x = 60

x = 10

2.5 Nou +5

Gareth 5 x 5 x + 5

Seun x x + 5

$\therefore$ Gareth: 25

$\therefore$ Seun: 5 x + 5 = 3( x + 5)

5 x + 5 = 3 x + 15

2 x = 10

x = 5

3. $\left(\frac{1}{2a}$ + 2 $a\right)$ 2 = 7 2

$\frac{1}{{4a}^{2}}$ + 2 + 4 $a$ 2 = 49

$\frac{1}{{4a}^{2}}$ + 4 $a$ 2 = 47

where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
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learn
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Damian
yes that's correct
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I think
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why?
what school?
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biomolecules are e building blocks of every organics and inorganic materials.
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how did you get the value of 2000N.What calculations are needed to arrive at it
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