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  • GETSenPhaseMaths display tagshide tags

    This collection is included inLens: Siyavula: Mathematics (Gr. 7-9)
    By: Siyavula

    Collection Review Status: In Review

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Breuke - Gemengde Bewerkings

Module by: Siyavula Uploaders. E-mail the author

WISKUNDE

Gewone Breuke

OPVOEDERS AFDELING

Memorandum

24.1 ( )

or

÷

x

+

_

  • (i) = 7272 size 12{ { { size 8{7} } over { size 8{2} } } } {} x 307307 size 12{ { { size 8{"30"} } over { size 8{7} } } } {} - 10

= 2101421014 size 12{ { { size 8{"210"} } over { size 8{"14"} } } } {} - 10

= 15 – 10

s = 5

  1. (i) = 32 - 11 3434 size 12{ { { size 8{3} } over { size 8{4} } } } {}

v = 20 1414 size 12{ { { size 8{1} } over { size 8{4} } } } {}

  1. (i) = 234234 size 12{ { { size 8{"23"} } over { size 8{4} } } } {} x 4747 size 12{ { { size 8{4} } over { size 8{7} } } } {}

= 237237 size 12{ { { size 8{"23"} } over { size 8{7} } } } {}

p = 3 2727 size 12{ { { size 8{2} } over { size 8{7} } } } {}

26. a) Metode: Kanselleer uit

Antwoord: 1818 size 12{ { { size 8{1} } over { size 8{8} } } } {}

b) 125125 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=312×427−23of15s=312×427−23of15 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=12of64−712+414v=12of64−712+414 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=534÷318−134p=534÷318−134 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

Figure 1
Figure 1 (graphics1.png)

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 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 . . . 1−1251−125 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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