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Aktiwiteit 4

Om uitdrukkings en vergelykings te onderskei

[lu 2.1, 2.6]

  • Uitdrukkings is kombinasies van letters ( a , b , х , y , ens.), bewerkings (+, –, ×, ) en getalle (1, –5, π, ½ , ens.), asook hakies en ander tekens. Dit sluit nie gelykaantekens in nie.
  • ‘n Uitdrukking is nogal soos ‘n woord of ‘n frase – dit het nie ‘n werkwoord nie.
  • ‘n Paar voorbeelde: х , х 3 , 5½ , 2πr, 5( ab bc ), 5 a 3 – 3 a 2 + a – 3, 2 a + b size 12{ sqrt {2 left (a+b right )} } {} , 5a 4 2a 2 size 12{ { {5a - 4} over {2a rSup { size 8{2} } } } } {} , ens.
  • ‘n Uitdrukking kan slegs gemanipuleer word, gewoonlik om dit te vereenvoudig. Dit kan nie opgelos word nie; dit het nie ‘n oplossing nie. Jy kan jou werk slegs kontroleer deur agteruit te werk om te sien of jy by die begin uitkom.
  • ‘n Vergelyking is doodgewoon twee uitdrukkings met ‘n gelykaanteken tussenin!
  • Dit is soos ‘n sin met ‘n werkwoord; dit maak ‘n stelling. Byvoorbeeld 2 х – 3 = 45 sê dubbel ‘n sekere getal, met drie verminder, is gelyk aan 45. Dis ons werk om daardie getal te bepaal.
  • Vergelykings word opgelos; hulle het oplossings wat bevestig kan word.
  • Ons vereenvoudig heelwat tydens die oplos van vergelykings, maar ons doen meer – ons word toegelaat om meer te doen. Onthou dat ons terme kan bytel of aftrek, as ons dit net aan beide kante doen! Ons kan met faktore deel of vermenigvuldig, as ons dit net aan beide kante doen. Omdat ‘n uitdrukking nie twee kante het nie, kan ons hierdie bewerkings nie op uitdrukkings toepas nie. Moenie uitdrukkings en vergelykings verwar nie, en oefen totdat jy instinktief weet wat om te doen.

Aktiwiteit 5

Om twee vergelykings gelyktydig op te los

[lu 2.4, 2.9]

1. Die lyn in diagram 1 het definisie-vergelyking y = 2.

Vraag: Lê die punt (1 ; 1) op die lyn?

Antwoord: Ons kan die antwoord grafies (deur die grafiek te bekyk) oplos. Dis duidelik dat die punt nie op die lyn lê nie, en dus is die antwoord nee .

Ons kan die antwoord algebraïes oplos, soos volg: Substitueer die punt (1 ; 1) vir ( х ; y ) in die vergelyking. Doen LK en RK apart soos voorheen.

LK: y = ( 1 ) = 2 RK: 2 LK ≠ RK – die punt (1 ; 1) lê nie op y = 2 nie.

Vraag: Lê die punt (–2 ; 2) op die lyn?

Grafies : Ja.

Algebraïes : LK: y = ( 2 ) = 2 RK: 2 LK = RK; Ja.

Vraag: Lê die punt (1½ ; 2) op die lyn? Bepaal die antwoord beide grafies en algebraïes .

2. Die lyn in diagram 2 word gedefinieer deur die vergelyking y = 2 х – 1.

Vrae: Lê die punt (0 ; 0) op die lyn?

Lê die punt (1 ; 1) op y = 2 х – 1?

Lê die punt (1½ ; 2) op die lyn?

3. In diagram 3 is dieselfde twee lyne saam op een stel asse getrek.

Bepaal grafies: Watter punt lê op beide lyne? Die antwoorde op vrae 1 en 2 hierbo sal help.

Dit is ooglopend uit diagram 3 dat die enigste punt op beide lyne (1½ ; 2) is.

  • So bepaal ons dit algebraïes:

Die vergelyking y = 2 gee y die waarde 2. Substitueer nou hierdie waarde in y = 2 х – 1.

As ons dan die vergelyking oplos, kry ons die waarde van х . So:

Substitueer: ( 2 ) = 2 х – 1 en los op vir х :

2 = 2 х – 1 х - terme na links

–2 х + 2 = –1 konstante terme na regs

–2 х = –2 – 1 vereenvoudig

–2 х = –3 deel beide kante deur –2

х = –3  –2 vereenvoudig

х = 1½

Dit toon die punt waar die lyne mekaar sny: ( х ; y ) = (1½ ; 2).

  • In hierdie metode het ons die twee vergelykings gelyktydig opgelos om die waardes van beide veranderlikes te vind wat beide vergelykings waar maak. As ‘n vergelyking slegs een veranderlike het, benodig ons slegs een vergelyking om daardie waarde van die veranderlike te vind wat die vergelyking waar maak. As ons twee veranderlikes het, benodig ons twee vergelykings om op te los vir die twee veranderlikes.


1 Los algebraïes op vir a en b : 2 a – 3 b = 0 en a = 6

2 Waar sny die lyne y = – х + 5 en y = –1? Bepaal die antwoord algebraïes.

3 Lê die punt (3 ; 4) op beide lyn y = 4 en lyn y = – х + 1? Doen algebraïes.

4 Sny die lyne y = –2 en y = 2? Bepaal die antwoord algebraïes.

Aktiwiteit 6

Om eenvoudige eksponiensiële vergelykings op te los

[lu 2.4, 2.8]

Probleme en sommige antwoorde.

1 Ek dink aan ‘n getal waarvan die kwadraat 100 is. Wat is die getal?

Die getal kan 10 wees, want 10 2 = 100. Maar is –10 nie ook ‘n korrekte antwoord nie?

Ja, hierdie probleem het twee geldige antwoorde!

Maak ‘n vergelyking uit hierdie stelling: Gestel die getal is х .

х 2 = 100

х 2 = 10 2 of х 2 = (–10) 2 Die hakies is noodsaaklik – sien jy dit?

х = 10 of х = –10 Beide antwoorde is geldig.

2 Ek dink aan ‘n negatiewe getal waarvan die kwadraat 25 is. Wat is dit?

Laat die getal y wees

y 2 = 25

y 2 = (5) 2 of y 2 = (–5) 2

y = 5 of y = –5 is die twee oplossings verskaf deur die vergelyking.

Die probleemstelling bevestig egter dat y = –5 die enigste geldige antwoord is.

3 Vind daardie getal wat ‘n derdemag van 27 het.

Laat die getal х wees

х 3 = 27  х 3 = 3 3 х = 3.

Hoekom kan х nie –3 wees nie?

4 Die derdemag van ‘n sekere getal is –8. Wat is die getal?

5 Los op vir х , en bevestig jou antwoord met die LK/RK metode:

a) х 2 = 64

b) х 2 = 36

c) х 2 = –100

d) х 2 – 49 = 0

e) х 2 = 12,25

f) 3 х 2 = 12

g) 2 х 2 – 10,58 = 0

6 Los op vir a en kontroleer jou antwoord:

a) a 3 = 64

b) a 3 + 1 = 0

c) 2 a 2 = 16

d) a 4 = 81

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
Idrissa Reply
im all ears I need to learn
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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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
Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
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J, combine like terms 7x-4y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
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types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
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
anybody can imagine what will be happen after 100 years from now in nano tech world
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
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
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
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Source:  OpenStax, Wiskunde graad 9. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col11055/1.1
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