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2.11 Breuke - vermenigvuldiging van breuke  (Page 2/2)

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Dit behels die deling van die teller EN die noemer deur ’n gemeenskaplike faktor

Nog korter lyk dit so:

Ons kan soos volg kanselleer: i) oorkruis ii) vertikaal

f) Bereken die volgende deur van kansellering gebruik te maak:

(i) k = 15 8 × 4 5 size 12{k= { { size 8{"15"} } over { size 8{8} } } times { { size 8{4} } over { size 8{5} } } } {}

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(ii) c = 18 25 × 40 27 size 12{c= { { size 8{"18"} } over { size 8{"25"} } } times { { size 8{"40"} } over { size 8{"27"} } } } {}

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18.3 Vermenigvuldiging van gemengde getalle

a) LET OP!

Ons moet eers die gemengde getal na ’n onegte breuk herlei en dan vermenigvuldig.

Bv. 6 × 1 3 4 = 6 1 × 7 4 42 4 10 2 4 10 1 2 alignl { stack { size 12{6 times 1 { { size 8{3} } over { size 8{4} } } = { { size 8{6} } over { size 8{1} } } times { { size 8{7} } over { size 8{4} } } } {} #= { { size 8{"42"} } over { size 8{4} } } {} # ="10" { { size 8{2} } over { size 8{4} } } {} #="10" { { size 8{1} } over { size 8{2} } } {} } } {}

Onthou om altyd die antwoord te vereenvoudig!

b) Bereken die volgende en vereenvoudig waar moontlik:

(i) m = 8 3 4 × 16 5 size 12{m=8 { { size 8{3} } over { size 8{4} } } times { { size 8{"16"} } over { size 8{5} } } } {}

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(ii) n = 5 2 3 × 9 10 size 12{n=5 { { size 8{2} } over { size 8{3} } } times { { size 8{9} } over { size 8{"10"} } } } {}

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(iii) p = 3 3 5 × 2 2 3 size 12{p=3 { { size 8{3} } over { size 8{5} } } times 2 { { size 8{2} } over { size 8{3} } } } {}

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19.1 Kan jy die antwoorde van die volgende ook bereken?

a) h = 3 4 × 4 3 size 12{h= { { size 8{3} } over { size 8{4} } } times { { size 8{4} } over { size 8{3} } } } {}

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b) f = 7 9 × 9 7 size 12{f= { { size 8{7} } over { size 8{9} } } times { { size 8{9} } over { size 8{7} } } } {}

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c) e = 8 9 × 9 8 size 12{e= { { size 8{8} } over { size 8{9} } } times { { size 8{9} } over { size 8{8} } } } {}

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d) b = 12 15 × 15 12 size 12{b= { { size 8{"12"} } over { size 8{"15"} } } times { { size 8{"15"} } over { size 8{"12"} } } } {}

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19.2 Wat merk jy op as jy na die antwoorde van bogenoemde kyk?

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19.3 Het jy geweet?

As die produk van twee breuke 1 is, is die een breuk die resiprook van die ander. Om dus die resiprook van ’n breuk te kry, ruil ons net die teller en die noemer om!

19.4 Bepaal nou die resiprook van die volgende:

a) 17 20 size 12{ { { size 8{"17"} } over { size 8{"20"} } } } {} = ___________________________________________________

b) 40 = ___________________________________________________

c) 6 1 5 size 12{6 { { size 8{1} } over { size 8{5} } } } {} = ___________________________________________________

d) 9 1 8 size 12{9 { { size 8{1} } over { size 8{8} } } } {} = ___________________________________________________

19.5 Verduidelik aan ’n maat hoe jy die antwoorde van c en d hierbo gekry het. Skryf eers jou antwoord hier neer.

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20. KOPKRAPPER!

Jodi is 1 2 3 size 12{1 { { size 8{2} } over { size 8{3} } } } {} meter lank. Sandy is die helfte van haar lengte. Grace is een derde van Sandy se lengte.

a) Hoe lank is Sandy?

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b) Hoe lank is Grace?

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21. Tyd vir selfassessering!

  • Merk die toepaslike blokkie:
 JA NEE
Ek kan breuke korrek optel
Ek kan die kleinste gemene veelvoud van twee noemers bepaal.
Ek kan breuke korrek aftrek
Ek kan breuke met natuurlike getalle vermenigvuldig
Ek kan breuke met breuke vermenigvuldig
Ek kan breuke met gemengde getalle vermenigvuldig
Ek weet hoe om van kansellering gebruik te maak as ek vermenigvuldig
Ek weet hoe om te vereenvoudig
Ek kan die resiprook van ’n breuk bepaal

22. Kom ons kyk nou eers weer hoe flink jy nog kan dink! Kyk of jy die volgende hoofrekentoets binne 2 minute kan voltooi:

a) 17 + 15 = ............ i) 1 10 size 12{ { { size 8{1} } over { size 8{"10"} } } } {} van 70 = ............
b) 27 + ............ = 42 j) 1 10 size 12{ { { size 8{1} } over { size 8{"10"} } } } {} van 1 minuut is ............ sekondes
c) 52 – 27 = ............ k) 1 4 size 12{ { { size 8{1} } over { size 8{4} } } } {} van ’n dag = ............ uur
d) 72 – 47 = ............ l) 56 ÷ 7 = ............
e) 82 – 37 = ............ m) 72 ÷ 9 = ............
f) 10 × 1 5 size 12{ { { size 8{1} } over { size 8{5} } } } {} = ............ n) 48 ÷ 6 = ............
g) 1 4 size 12{ { { size 8{1} } over { size 8{4} } } } {} van 32 = ............ o) 10 4 ÷ 10 2 = ............
h) 1 8 size 12{ { { size 8{1} } over { size 8{8} } } } {} van 56 = ............

(15)

  • Kleur in:

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.

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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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