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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

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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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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