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3. Kan jy die volgende doen?

3.1 8 - 4 3 7 size 12{ { { size 8{3} } over { size 8{7} } } } {}

3.2 3 1 9 size 12{ { { size 8{1} } over { size 8{9} } } } {} - 1 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {}

  • Belangrik : Voordat jy breuke kan optel of van mekaar aftrek, moet die noemers dieselfde wees.
  • bv. 2 4 7 size 12{ { { size 8{4} } over { size 8{7} } } } {} - 1 6 7 size 12{ { { size 8{6} } over { size 8{7} } } } {}
  • 2 – 1 = 1 en
  • 4 7 size 12{ { { size 8{4} } over { size 8{7} } } } {} - 6 7 size 12{ { { size 8{6} } over { size 8{7} } } } {} ( 4 – 6 --- dit kan nie. Gaan vat een hele : 1 = 7 7 size 12{ { { size 8{7} } over { size 8{7} } } } {} )
  • ( 4 + 7 = 11 --- ja, 11 – 6 = 5)Antwoord: 5 7 size 12{ { { size 8{5} } over { size 8{7} } } } {}
  • Jy kan ook alle gemengde getalle herlei na onegte breuke en dan die noemers dieselfde maak.
  • bv. 18 7 13 7 = 5 7 size 12{ { { size 8{"18"} } over { size 8{7} } } - { { size 8{"13"} } over { size 8{7} } } = { { size 8{5} } over { size 8{7} } } } {} (18 – 13 = 5: Die noemers is dieselfde. Trek die tellers van mekaar af.)

4. Doen nou die volgende:

4.1 4 1 7 size 12{ { { size 8{1} } over { size 8{7} } } } {} + 4 16 42 size 12{ { { size 8{"16"} } over { size 8{"42"} } } } {}

4.2 36 - 15 6 11 size 12{ { { size 8{6} } over { size 8{"11"} } } } {}

4.3 1 8 + 0, 625 3 8 size 12{ { { size 8{1} } over { size 8{8} } } +0,"625" - { { size 8{3} } over { size 8{8} } } } {}

4.4 4 5 10 + 7 1 2 + 6 3 4 size 12{4 { { size 8{5} } over { size 8{"10"} } } +7 { { size 8{1} } over { size 8{2} } } +6 { { size 8{3} } over { size 8{4} } } } {}

4.5 7 1 3 size 12{ { { size 8{1} } over { size 8{3} } } } {} - 4 7 8 size 12{ { { size 8{7} } over { size 8{8} } } } {}

4.6 7 a - a 4 size 12{ { { size 8{a} } over { size 8{4} } } } {} a / 4

4.7 9 a + 6 ab 3 b size 12{ { { size 8{9} } over { size 8{a} } } + left ( { { size 8{6} } over { size 8{ ital "ab"} } } - { { size 8{3} } over { size 8{b} } } right )} {}

4.8 - 6 + 2 6 7 size 12{ { { size 8{6} } over { size 8{7} } } } {}

4.9 5 - (4 4 9 size 12{ { { size 8{4} } over { size 8{9} } } } {} + 2 2 3 size 12{ { { size 8{2} } over { size 8{3} } } } {} )

4.10 3 1 3 size 12{ { { size 8{1} } over { size 8{3} } } } {} a - 2 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} a

AKTIWITEIT 5

Vermenigvuldiging en deling van rasionale getalle (breuke)

LU 1.2.6 LU 1.6.2
  • Jy het dit in graad 7 gedoen – kom ons verfris net die geheue.

1. Vermenigvuldiging:

  • Belangrik : Alle gemengde getalle moet as breuke geskryf word.Dan kan jy oorkruis kanselleer.
  • Probeer nou die volgende:
  • 1 1 4 size 12{ { { size 8{1} } over { size 8{4} } } } {} × 2 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} × 4

2. Deling:

  • Die resiprook speel ‘n groot rol by deling van breuke.
  • Maak gebruik van ‘n voorbeeld om die term te verduidelik.

bv. 1 3 ÷ 2 3 size 12{ { { size 8{1} } over { size 8{3} } } div { { size 8{2} } over { size 8{3} } } } {}

  • Albei getalle is breuke
  • Maak ÷ ‘n × - teken en kry resiprook van die deler (breuk ná die ÷-teken).
  • Kanselleer net soos by vermenigvuldiging.

3. Doen nou die volgende:

3.1 8 ÷ 8 11 size 12{ { { size 8{8} } over { size 8{"11"} } } } {}

3.2 18 ÷ 7 8 size 12{ { { size 8{7} } over { size 8{8} } } } {}

3.3 5 6 ÷ 5 2 size 12{ { { size 8{5} } over { size 8{6} } } div { { size 8{5} } over { size 8{2} } } } {}

3.4 -2 2 3 size 12{ { { size 8{2} } over { size 8{3} } } } {} ÷ -1 7 9 size 12{ { { size 8{7} } over { size 8{9} } } } {}

3.5 6 3 4 size 12{ { { size 8{3} } over { size 8{4} } } } {} mn ÷ -6 m 3

3.6 4 xy 3 ab ÷ 2x 3a size 12{ { { size 8{ - 4 ital "xy"} } over { size 8{3 ital "ab"} } } div { { size 8{ - 2x} } over { size 8{3a} } } } {} -

Assessering

Leeruitkomstes(LUs)
LU 1
Getalle, Verwerkings en Verwantskappe Die 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:
1.2.2 desimale, breuke en persentasies;
1.2.5 optel- en vermenigvuldiginginverses;
1.2.6 veelvoude en faktore;
1.2.7 irrasionele getalle in die konteks van meting (bv. vierkants- en derdemagwortels van nie perfekte vierkante en derdemagte?);
1.3 ekwivalente vorms van die bogenoemde rasionale getalle herken en kan gebruik;
1.6 skat en bereken deur stappe te kies wat geskik is om probleme op te los wat die volgende behels:
1.6.1 afronding;
1.6.2 veelvoudige stappe met rasionele getalle (insluitend deling met breuke en desimale);
1.7 ‘n reeks tegnieke gebruik om berekeninge te doen, insluitend:
1.7.1 die gebruik van kommutatiewe, assosiatiewe en distributiewe eienskappe met rasionale nommers;
1.7.2 die gebruik van ‘n sakrekenaar;
1.9 die volgende herken en gebruik:
1.9.1 algoritmes vir die vind van ekwivalente breuke;
1.9.2 die kommutatiewe, assosiatiewe en distributiewe eienskappe met rasionale getalle (die ver­wagting is dat leerders hierdie eienskappe sal kan gebruik sonder om noodwendig die name van die eienskappe te ken).

Memorandum

AKTIWITEIT 1

1. Natuurlike getalle

Telgetalle

Heelgetalle

Rieële getalle

2. a b size 12{ { {a} over {b} } } {} ; b ≠ 0

2 size 12{ sqrt {2} } {}
  • Q
  • Q 1

4.

  • 1 + 4 size 12{ sqrt {4} } {} ; -4
  • 2 3 size 12{ { { - 2} over {3} } } {} ; 12 1 5 size 12{ { {1} over {5} } } {}
  • 9 + 4 size 12{ sqrt {9+4} } {} ; 1 + 2 2 size 12{ { {1+ sqrt {2} } over { sqrt {2} } } } {}

6. Gelyk in waarde

7. 4 14 size 12{ { {4} over {"14"} } } {} = 6 24 size 12{ { {6} over {"24"} } } {} ens.

  • Egte breuk
  • Onegte breuk
  • Gemengde getal
  • Desimale getal
  • Repeterende desimale getal
  • Persentasie

AKTIWITEIT 2

1. 2,15

  • 0,625
  • 3,25
  • 5,75
  • 2,875
  • 6, 000 7 size 12{ { {6,"000"} over {7} } } {} = 0,8571 . . . ≈ 0,86
  • 7, 000 9 size 12{ { {7,"000"} over {9} } } {} = 0,777 . . . =
    of 0,8
  • 6 8 1000 size 12{ { {8} over {"1000"} } } {} = 6 1 125 size 12{ { {1} over {"125"} } } {}
  • 4 65 100 size 12{ { {"65"} over {"100"} } } {} = 4 13 20 size 12{ { {"13"} over {"20"} } } {}
  • 375 1000 size 12{ { {"375"} over {"1000"} } } {} = 3 8 size 12{ { {3} over {8} } } {}
  • 7 75 1000 size 12{ { {"75"} over {"1000"} } } {} = 7 3 40 size 12{ { {3} over {"40"} } } {}
  • 13 65 100 size 12{ { {"65"} over {"100"} } } {} = 13 13 20 size 12{ { {"13"} over {"20"} } } {}
  • 125 1000 size 12{ { {"125"} over {"1000"} } } {} = 1 8 size 12{ { {1} over {8} } } {}

7.1 3 9 size 12{ { {3} over {9} } } {} = 1 3 size 12{ { {1} over {3} } } {}

7.2 45 99 size 12{ { {"45"} over {"99"} } } {} = 5 11 size 12{ { {5} over {"11"} } } {}

7.3 23 990 size 12{ { {"23"} over {"990"} } } {}

7.4 3 900 size 12{ { {3} over {"900"} } } {} = 1 300 size 12{ { {1} over {"300"} } } {}

9. 0, 4 size 12{ {4} cSup { size 8{ cdot } } } {} 5 size 12{ {5} cSup { size 8{ cdot } } } {} = x

x = 0,4545 . . . 

100 x = 45,4545 . . .

  • –  99 x = 45

x = 45 99 size 12{ { {"45"} over {"99"} } } {} = 5 11 size 12{ { {5} over {"11"} } } {}

AKTIWITEIT 3

2.1 17 x5 20 x5 size 12{ { {"17"x5} over {"20"x5} } } {} = 85%

2.2 19 40 size 12{ { {"19"} over {"40"} } } {} × 100 1 size 12{ { {"100"%} over {1} } } {} = 47,5%

2.3 38 x2 50 x2 size 12{ { {"38"x2} over {"50"x2} } } {} = 76%

2.4 45 60 size 12{ { {"45"} over {"60"} } } {} × 100 1 size 12{ { {"100"%} over {1} } } {} = 75%

3.1 55 100 size 12{ { {"55"} over {"100"} } } {} = 11 20 size 12{ { {"11"} over {"20"} } } {}

3.2 15 , 5 100 size 12{ { {"15",5} over {"100"} } } {} = 0,155 = 155 1000 size 12{ { {"155"} over {"1000"} } } {} = 31 200 size 12{ { {"31"} over {"200"} } } {}

3.3 33 200 size 12{ { {"33"} over {"200"} } } {}

3.4 2 0 30 { 0 size 12{ { {2 { {0}}} over {"30 {"{0}}} } } {} = 2 30 size 12{ { {2} over {"30"} } } {}

4.a) 33 800 size 12{ { {"33"} over {"800"} } } {} × 25500 1 size 12{ { {"25500"} over {1} } } {} size 12{ approx } {} 1 052

b) 3 5 size 12{ { {3} over {5} } } {} × 25500 1 size 12{ { {"25500"} over {1} } } {} = 15 300

c) 85 1000 size 12{ { {"85"} over {"1000"} } } {} × 25500 1 size 12{ { {"25500"} over {1} } } {} = 2 167,5 size 12{ approx } {} 2 168

  • (14,5) 15300 1052 size 12{ { {"15300"} over {"1052"} } } {} = 7650 526 size 12{ { {"7650"} over {"526"} } } {} = 3825 263 size 12{ { {"3825"} over {"263"} } } {}
  • 25 500 – 18 520 = 6 980

4.4

4.5 3 5 size 12{ { {3} over {5} } } {} x 2 1 size 12{ { {2} over {1} } } {} = 6 5 size 12{ { {6} over {5} } } {} = 1 1 5 size 12{1 { {1} over {5} } } {}

AKTIWITEIT 5

1. 5 1 4 size 12{ { {5} over { {} rSub { size 8{1} } { {4}}} } } {} × 5 2 size 12{ { {5} over {2} } } {} x 4 1 1 size 12{ { { { {4}} rSup { size 8{1} } } over {1} } } {} = 25 2 size 12{ { {"25"} over {2} } } {} = 12 1 2 size 12{"12" { {1} over {2} } } {}

3.1 8 1 size 12{ { {8} over {1} } } {} ÷ 8 11 size 12{ { {8} over {"11"} } } {} = 8 1 1 size 12{ { { { {8}} rSup { size 8{1} } } over {1} } } {} × 11 8 1 size 12{ { {"11"} over { { {8}} rSub { size 8{1} } } } } {} = 11

3.2 18 1 size 12{ { {"18"} over {1} } } {} × 8 7 size 12{ { {8} over {7} } } {} = 144 7 size 12{ { {"144"} over {7} } } {} = 20 4 7 size 12{"20" { {4} over {7} } } {}

3.3 5 1 6 3 size 12{ { { { {5}} rSup { size 8{1} } } over { { {6}} rSub { size 8{3} } } } } {} × 2 1 5 1 size 12{ { { { {2}} rSup { size 8{1} } } over { { {5}} rSub { size 8{1} } } } } {} = 1 3 size 12{ { {1} over {3} } } {}

3.4 8 1 3 1 size 12{ { { - { {8}} rSup { size 8{1} } } over { { {3}}"" lSub { size 8{1} } } } } {} × 9 3 1 6 2 size 12{ { { - { {9}} rSup { size 8{3} } } over { { {1}} { {6}} rSub { size 8{2} } } } } {} = 3 2 size 12{ { {3} over {2} } } {} = 1 1 2 size 12{1 { {1} over {2} } } {}

3.5 2 7 9 mn 4 size 12{ { { { {2}} { {7}} rSup { size 8{9} } ital "mn"} over {4} } } {} × 1 6 2 m 3 size 12{ { {1} over { - { {6}}"" lSub { size 8{2} } m rSup { size 8{3} } } } } {} = 9n 8m 2 size 12{ { { - 9n} over {8m rSup { size 8{2} } } } } {}

3.6 4 2 xy 3 1 a b size 12{ { { - { {4}} rSup { size 8{2} } ital "xy"} over { { {3}}"" lSub { size 8{1} } { {a}}b} } } {} × 3 a 2 x size 12{ { { { {3}} { {a}}} over { - { {2}} { {x}}} } } {} = 2y b size 12{ { {2y} over {b} } } {}

Questions & Answers

how do you translate this in Algebraic Expressions
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why surface tension is zero at critical temperature
Shanjida
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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what is biological synthesis of nanoparticles
Sanket 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
I'm interested in nanotube
Uday
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
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
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
What is lattice structure?
s. 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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