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Foutgrense - graad 11

Ons het gesien dat getalle rasionaal of irrasionaal kan wees en ons het gesien hoe om getalle af te rond. In 'n berekening wat meer as een stap het, is dit beter om die afronding in die heel laaste stap te doen.

By voorbeeld, as jy gevra word om 3 3 + 12 te skryf as 'n desimale getal afgerond tot twee desimale plekke, is daar twee maniere om dit te doen.

Metode 1

3 3 + 12 = 3 3 + 4 · 3 = 3 3 + 2 3 = 5 3 = 5 × 1 , 732050808 ... = 8 , 660254038 ... = 8 , 66

Metode 2

3 3 + 12 = 3 × 1 , 73 + 3 , 46 = 5 , 19 + 3 , 46 = 8 , 65

In hierdie voorbeeld sien ons dat Metode 1 die antwoord gee as 8,66 terwyl Metode 2 die antwoord gee as 8,65. Die antwoord van Metode 1 is meer akkuraat want die uitdrukking is so ver as moontlik vereenvoudig voordat die antwoord afgerond is.

In die algemeen is dit beter om 'n uitdrukking sover as moontlik te vereenvoduig voordat jy jou sakrekenaar gebruik om die antwoord in desimale vorm te gee.

Vereenvoudiging en Akkuraatheid

Dit is nodig om alle uitdrukkings sover as moontlik te vereenvoudig voordat antwoorde afgerond word. Dit het 'n invloed op die akkuraatheid van jou antwoord.

Bereken 54 3 + 16 3 . Gee die antwoord korrek tot drie desimale plekke.

  1. 54 3 + 16 3 = 27 · 2 3 + 8 · 2 3 = 27 3 · 2 3 + 8 3 · 2 3 = 3 2 3 + 2 2 3 = 5 2 3
  2. 5 2 3 = 5 × 1 , 25992105 ... = 6 , 299605249 ... = 6 , 300
  3. 6 , 299605249 ... = 6 , 300 tot drie desimal plekke

    ∴ 54 3 + 16 3 = 6 , 300 tot drie desimale plekke.

Bereken x + 1 + 1 3 ( 2 x + 2 ) - ( x + 1 ) as x = 3 , 6 . Gee die antwoord korrek tot twee desimale plekke.

  1. x + 1 + 1 3 ( 2 x + 2 ) - ( x + 1 ) = x + 1 + 1 3 2 x + 2 - x - 1 = x + 1 + 1 3 x + 1 = 4 3 x + 1
  2. 4 3 x + 1 = 4 3 3 , 6 + 1 = 4 3 4 , 6 = 2 , 144761059 ... × 4 ÷ 3 = 2 , 859681412 ...
  3. 2 , 859681412 ... = 2 , 86 tot twee desimale plekke

    ∴ x + 1 + 1 3 ( 2 x + 2 ) - ( x + 1 ) = 2 , 86 (tot twee desimale plekke) as x = 3 , 6 .

Beduidende syfers

In enige getal is elke syfer wat nie nul is nie, 'n beduidende syfer. Nulle word slegs getel indien hulle tussen twee nie-nul syfers is of aan die einde van die desimale gedeelte van die getal.By voorbeeld, die getal 2000 het 1 beduidende syfer (die 2), maar 2000 , 0 het 5 beduidende syfers. Skatting van 'n getal word gedoen deur die beduidende syfers uit jou getal (begin by syfer aan regterkant) te verwyder totdat jy die verlangde aantal beduidende syfers het. Rond af soos jy voortgaan. By voorbeeld 6 , 827 het 4 beduidende syfers, maar as jy dit wil skryf as 'n getal met 3 beduidende syfers, beteken dit dat jy die 7 moet verwyder en oprond, so dit word 6 , 83 . Dit is belangrik om te weet wanneer om 'n getal te benader en wanneer nie. Dit is gewoonlik goeie praktyk om slegs getalle te benader wanneer dit absoluut noodsaaklik is, en liewer simbole te gebruik om sekere irrasionale getalle voor te stel (byvoorbeeld π ). Benadering gebeur eers aan die einde van die berekening. As dit nodig is om 'n getal in die middel van 'n berekening te benader, is dit dikwels goed genoeg om te benader tot 'n aantal desimale plekke.

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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 11). OpenStax CNX. Sep 20, 2011 Download for free at http://cnx.org/content/col11339/1.4
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