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Wiskunde

Gewone breuke

Opvoeders afdeling

Memorandum

24.1 ( )

or

÷

x

+

_

  • (i) = 7 2 size 12{ { { size 8{7} } over { size 8{2} } } } {} x 30 7 size 12{ { { size 8{"30"} } over { size 8{7} } } } {} - 10

= 210 14 size 12{ { { size 8{"210"} } over { size 8{"14"} } } } {} - 10

= 15 – 10

s = 5

  1. = 32 - 11 3 4 size 12{ { { size 8{3} } over { size 8{4} } } } {}

v = 20 1 4 size 12{ { { size 8{1} } over { size 8{4} } } } {}

  1. = 23 4 size 12{ { { size 8{"23"} } over { size 8{4} } } } {} x 4 7 size 12{ { { size 8{4} } over { size 8{7} } } } {}

= 23 7 size 12{ { { size 8{"23"} } over { size 8{7} } } } {}

p = 3 2 7 size 12{ { { size 8{2} } over { size 8{7} } } } {}

26. a) Metode: Kanselleer uit

Antwoord: 1 8 size 12{ { { size 8{1} } over { size 8{8} } } } {}

b) 1 25 size 12{ { { size 8{1} } over { size 8{"25"} } } } {}

Leerders afdeling

Inhoud

Aktiwiteit: gemengde bewerkings [lu 1.10]

24. GEMENGDE BEWERKINGS

24.1 Onthou jy nog?

  • Skryf gou die volgorde van bewerkings hier neer:

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24.2 Pas dit nou toe in die volgende:

(i) s = 3 1 2 × 4 2 7 2 3 of 15 size 12{s=3 { { size 8{1} } over { size 8{2} } } times 4 { { size 8{2} } over { size 8{7} } } - { { size 8{2} } over { size 8{3} } } `"of"`"15"} {}

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(ii) v = 1 2 of 64 7 1 2 + 4 1 4 size 12{v= { { size 8{1} } over { size 8{2} } } `"of"`"64" - left (7 { { size 8{1} } over { size 8{2} } } +4 { { size 8{1} } over { size 8{4} } } right )} {}

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(iii) p = 5 3 4 ÷ 3 1 8 1 3 4 size 12{p=5 { { size 8{3} } over { size 8{4} } } div left (3 { { size 8{1} } over { size 8{8} } } - 1 { { size 8{3} } over { size 8{4} } } right )} {}

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25. Tyd vir selfassessering

26. KOPKRAPPERS!

a) Wat sal die antwoord van die volgende wees? Probeer dit kry sonder ellelange bewerkings!

o = 1 1 2 × 1 1 3 × 1 1 4 × 1 1 5 × 1 1 6 × 1 1 7 × 1 1 8 size 12{o= left (1 - { { size 8{1} } over { size 8{2} } } right ) times left (1 - { { size 8{1} } over { size 8{3} } } right ) times left (1 - { { size 8{1} } over { size 8{4} } } right ) times left (1 - { { size 8{1} } over { size 8{5} } } right ) times left (1 - { { size 8{1} } over { size 8{6} } } right ) times left (1 - { { size 8{1} } over { size 8{7} } } right ) times left (1 - { { size 8{1} } over { size 8{8} } } right )} {}

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b) As jy so sou aanhou met die patroon tot by . . . 1 1 25 size 12{ times left (1 - { { size 8{1} } over { size 8{"25"} } } right )} {} , wat sal die antwoord dan wees?

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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 1.10: Dit is duidelik wanneer die leerder ‘n verskeidenheid strategieë gebruik om oplossings te kontroleer en die redelikheid daarvan te beoordeel.

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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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