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Wiskunde

Gewone breuke

Opvoeders afdeling

Memorandum

  • b)
10 10
1 10 size 12{ { { size 8{1} } over { size 8{"10"} } } } {} 1 10 size 12{ { { size 8{1} } over { size 8{"10"} } } } {}

Answers is the same

(i) = 3 4 size 12{ { { size 8{3} } over { size 8{4} } } } {} x 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {}

x = 3 8 size 12{ { { size 8{3} } over { size 8{8} } } } {}

y = 18 2 3 size 12{ { { size 8{2} } over { size 8{3} } } } {}

(ii) = 7 x 8 3 size 12{ { { size 8{8} } over { size 8{3} } } } {}

= 56 3 size 12{ { { size 8{"56"} } over { size 8{3} } } } {}

  1. = 6 1 size 12{ { { size 8{6} } over { size 8{1} } } } {} x 5 4 size 12{ { { size 8{5} } over { size 8{4} } } } {}

= 30 4 size 12{ { { size 8{"30"} } over { size 8{4} } } } {}

m = 7 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {}

(iv) = 2 7 size 12{ { { size 8{2} } over { size 8{7} } } } {} x 1 9 size 12{ { { size 8{1} } over { size 8{9} } } } {}

n = 2 63 size 12{ { { size 8{2} } over { size 8{"63"} } } } {}

  • b)

(i) x = 3 8 size 12{ { { size 8{3} } over { size 8{8} } } } {} 9 24 size 12{ { { size 8{9} } over { size 8{"24"} } } } {}

= 3 8 size 12{ { { size 8{3} } over { size 8{8} } } } {} x 24 9 size 12{ { { size 8{"24"} } over { size 8{9} } } } {}

x = 1

(ii) k = 15 18 size 12{ { { size 8{"15"} } over { size 8{"18"} } } } {} 45 6 size 12{ { { size 8{"45"} } over { size 8{6} } } } {}

= 15 18 size 12{ { { size 8{"15"} } over { size 8{"18"} } } } {} x 6 45 size 12{ { { size 8{6} } over { size 8{"45"} } } } {}

k = 1 9 size 12{ { { size 8{1} } over { size 8{9} } } } {}

(iii) c = 7 9 size 12{ { { size 8{7} } over { size 8{9} } } } {} 5 6 size 12{ { { size 8{5} } over { size 8{6} } } } {}

= 7 9 size 12{ { { size 8{7} } over { size 8{9} } } } {} x 6 5 size 12{ { { size 8{6} } over { size 8{5} } } } {}

c = 14 15 size 12{ { { size 8{"14"} } over { size 8{"15"} } } } {}

(iv) f = 11 12 size 12{ { { size 8{"11"} } over { size 8{"12"} } } } {} 6 5 size 12{ { { size 8{6} } over { size 8{5} } } } {}

= 11 12 size 12{ { { size 8{"11"} } over { size 8{"12"} } } } {} x 5 6 size 12{ { { size 8{5} } over { size 8{6} } } } {}

= 55 72 size 12{ { { size 8{"55"} } over { size 8{"72"} } } } {}

23.3 c)

(i) b = 2 1 4 size 12{2 { { size 8{1} } over { size 8{4} } } } {} 3 2 size 12{ { { size 8{3} } over { size 8{2} } } } {}

= 9 4 size 12{ { { size 8{9} } over { size 8{4} } } } {} x 2 3 size 12{ { { size 8{2} } over { size 8{3} } } } {}

b = 1 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {}

(ii) e = 3 4 5 size 12{ { { size 8{4} } over { size 8{5} } } } {}  2 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {}

= 19 5 size 12{ { { size 8{"19"} } over { size 8{5} } } } {} x 2 5 size 12{ { { size 8{2} } over { size 8{5} } } } {}

e = 38 25 size 12{ { { size 8{"38"} } over { size 8{"25"} } } } {}

e = 1 13 25 size 12{ { { size 8{"13"} } over { size 8{"25"} } } } {}

  1. g = 3 4 7 size 12{ { { size 8{4} } over { size 8{7} } } } {}  1 2 7 size 12{ { { size 8{2} } over { size 8{7} } } } {}

= 25 7 size 12{ { { size 8{"25"} } over { size 8{7} } } } {} x 7 9 size 12{ { { size 8{7} } over { size 8{9} } } } {}

= 25 9 size 12{ { { size 8{"25"} } over { size 8{9} } } } {}

g = 2 7 9 size 12{ { { size 8{7} } over { size 8{9} } } } {}

(iv) r = 15 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {}  5 1 4 size 12{ { { size 8{1} } over { size 8{4} } } } {}

= 31 2 size 12{ { { size 8{"31"} } over { size 8{2} } } } {} x 4 21 size 12{ { { size 8{4} } over { size 8{"21"} } } } {}

= 62 21 size 12{ { { size 8{"62"} } over { size 8{"21"} } } } {}

r = 2 20 21 size 12{ { { size 8{"20"} } over { size 8{"21"} } } } {}

Leerders afdeling

Inhoud

Aktiwiteit: deling met breuke [lu 1.7.3, lu 2.1.5]

23. Kom ons kyk nou na DELING MET BREUKE!

23.1 Deling van heelgetalle deur breuke en andersom :

a) Werk saam met ’n maat en kyk goed na die volgende probleme.

Ma bak vyf koeke en wil graag vir jou en jou maats elkeen ’n halwe ( 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} ) stuk gee. Hoeveel maats kan van die koek eet?

  • Op ’n getallelyn lyk dit so:

Dus: 5 ÷ 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} = 1010 kinders kan elkeen 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} koek kry.

Ma bak weer, maar hierdie keer net een reghoekige koek. Sy besluit om die helfte daarvan tussen haar vyf kinders te verdeel. Watter breuk kry elkeen?

  • Kom ons maak ’n skets daarvan!

12345

Kan jy sien dat elke kind een tiende ( 1 10 size 12{ { { size 8{1} } over { size 8{"10"} } } } {} ) van die koek sal kry?Dus: 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} ÷ 5 = 1 10 size 12{ { { size 8{1} } over { size 8{"10"} } } } {}

b) Voltooi die tabel:

5 ÷ 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} = ............ 5 × 2 1 size 12{ { { size 8{2} } over { size 8{1} } } } {} = ............
1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} ÷ 5 1 size 12{ { { size 8{5} } over { size 8{1} } } } {} = ............ 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} × 1 5 size 12{ { { size 8{1} } over { size 8{5} } } } {} = ............

Wat merk jy op? ___________________________________________________

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c) Het jy geweet?

Enige deelsom met breuke kan in ’n vermenigvuldigingsom verander word! Ons doen dit deur die deler in sy resiprook te verander. Ons “keer dus die deler om”!

Dus:

÷ 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} =
´ 2 1 size 12{ { { size 8{2} } over { size 8{1} } } } {} = 10

d) Verbind kolom A met die korrekte antwoord in kolom B:

A B
÷ deur 5 × met 4 3 size 12{ { { size 8{4} } over { size 8{3} } } } {}
÷ deur 3 4 size 12{ { { size 8{3} } over { size 8{4} } } } {} × met 3
÷ deur 7 8 size 12{ { { size 8{7} } over { size 8{8} } } } {} × met 5
÷ deur 1 3 size 12{ { { size 8{1} } over { size 8{3} } } } {} × met 1 5 size 12{ { { size 8{1} } over { size 8{5} } } } {}
÷ deur 1 5 size 12{ { { size 8{1} } over { size 8{5} } } } {} × met 8 7 size 12{ { { size 8{8} } over { size 8{7} } } } {}

e) Bereken die volgende:

i) x = 3 4 ÷ 2 size 12{x= { { size 8{3} } over { size 8{4} } } div 2} {}

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ii) y = 7 ÷ 3 8 size 12{y=7 div { { size 8{3} } over { size 8{8} } } } {}

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iii) m = 6 ÷ 4 5 size 12{m=6 div { { size 8{4} } over { size 8{5} } } } {}

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iv) n = 2 7 ÷ 9 size 12{n= { { size 8{2} } over { size 8{7} } } div 9} {}

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23.2 Deling van breuke deur breuke:

a) Werk weer saam met ’n maat en bestudeer die volgende:

x = 6 25 ÷ 3 5 size 12{x= { { size 8{6} } over { size 8{"25"} } } div { { size 8{3} } over { size 8{5} } } } {}

Ek weet ek moet die volgende stappe volg:

1. Verander die ÷ in ×

2. Draai die breuk na die ÷ (deler) om – kry dus resiprook

3. Vermenigvuldig soos gewoonlik: teller × teller noemer × noemer size 12{ { { ital "teller" times ital "teller"} over { ital "noemer" times ital "noemer"} } } {}

Dus: 6 25 ÷ 3 5 = 6 25 × 5 3 size 12{ { { size 8{6} } over { size 8{"25"} } } div { { size 8{3} } over { size 8{5} } } = { { size 8{6} } over { size 8{"25"} } } times { { size 8{5} } over { size 8{3} } } } {}

Ek kanselleer waar ek kan:
2 6
5 25
×
5 1
3 1

Die antwoord is dus 2 × 1 5 × 1 = 2 5 size 12{ { { size 8{2 times 1} } over { size 8{5 times 1} } } = { { size 8{2} } over { size 8{5} } } } {}

b) Probeer die volgende op jou eie:

i) x = 3 8 ÷ 9 24 size 12{x= { { size 8{3} } over { size 8{8} } } div { { size 8{9} } over { size 8{"24"} } } } {}

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ii) k = 15 18 ÷ 45 6 size 12{k= { { size 8{"15"} } over { size 8{"18"} } } div { { size 8{"45"} } over { size 8{6} } } } {}

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iii) c = 7 9 ÷ 5 6 size 12{c= { { size 8{7} } over { size 8{9} } } div { { size 8{5} } over { size 8{6} } } } {}

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iv) f = 11 12 ÷ 6 5 size 12{f= { { size 8{"11"} } over { size 8{"12"} } } div { { size 8{6} } over { size 8{5} } } } {}

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23.3 Deling met gemengde getalle:

a) Kan jy die volgende probleem vir ’n maat verduidelik?

’n Gesin eet 1 en ’n halwe ( 1 1 2 size 12{1 { { size 8{1} } over { size 8{2} } } } {} ) pizza. As elkeen net een kwart ( 1 4 size 12{ { { size 8{1} } over { size 8{4} } } } {} ) van die pizza eet, uit hoeveel lede bestaan die gesin?

  • Ek moet 1 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} ÷ 1 4 size 12{ { { size 8{1} } over { size 8{4} } } } {} bereken.
  • Dis makliker as ek dit teken:

It’s easier if I draw it:

  • Die antwoord is dus 6.
  • Wiskundig skryf ek dit so:

y = 1 1 2 ÷ 1 4 3 2 ÷ 1 4 3 2 × 4 1 12 2 6 alignl { stack { size 12{y=1 { { size 8{1} } over { size 8{2} } } div { { size 8{1} } over { size 8{4} } } } {} #= { { size 8{3} } over { size 8{2} } } div { { size 8{1} } over { size 8{4} } } {} # = { { size 8{3} } over { size 8{2} } } times { { size 8{4} } over { size 8{1} } } {} #= { { size 8{"12"} } over { size 8{2} } } {} # =6 {}} } {}

  • Ek verkies om ’n getallelyn te gebruik:

b) Het jy geweet?

Ons verander gemengde getalle eers in onegte breuke voordat ons die antwoord bereken.

c) Probeer op jou eie:

i) b = 2 1 4 ÷ 3 2 size 12{b=2 { { size 8{1} } over { size 8{4} } } div { { size 8{3} } over { size 8{2} } } } {}

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ii) e = 3 4 5 ÷ 2 1 2 size 12{e=3 { { size 8{4} } over { size 8{5} } } div 2 { { size 8{1} } over { size 8{2} } } } {}

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iii) g = 3 4 7 ÷ 1 2 7 size 12{g=3 { { size 8{4} } over { size 8{7} } } div 1 { { size 8{2} } over { size 8{7} } } } {}

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iv) r = 15 1 2 ÷ 5 1 4 size 12{r="15" { { size 8{1} } over { size 8{2} } } div 5 { { size 8{1} } over { size 8{4} } } } {}

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Assessering

Leeruitkomste 1: 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.

Assesseringstandaard 1.7: Dit is duidelik wanneer die leerder skat en bereken deur geskikte bewerkings vir probleme wat die volgende behels, kies en gebruik:

1.7.3: optelling, aftrekking en vermenigvuldiging van gewone breuke.

Leeruitkomste 2: Die 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.

Assesseringstandaard 2.1: Dit is duidelik wanneer die leerder numeriese en meetkundige patrone ondersoek en uitbrei op soek na ‘n verwantskap of reëls, insluitend patrone;

2.1.5: voorgestel in tabelle.

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
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
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ramon Reply
Kristine 2*2*2=8
Bridget Reply
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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
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
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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 7. OpenStax CNX. Oct 21, 2009 Download for free at http://cnx.org/content/col11076/1.2
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