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aroub

  • 2 years ago

14- Find a decimal approximation to the nearest tenth: i) √82 Is there ANY easy way other than: Find the perfect squares that is before and after the number 82 and then add the both numbers and blah..blah..

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  1. asnaseer
    • 2 years ago
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    since it is to the "nearest tenth" you won't have many numbers that you need to try. start with 9 since 9*9=81 and increase by one tenth until you get an answer that is closest to 82.

  2. aroub
    • 2 years ago
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    Can you show me?

  3. asnaseer
    • 2 years ago
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    another possible method might be to use the fact that:\[(a+b)^2=a^2+2ab+b^2\]so set a=9 and you need to find b such that:\[(9+b)^2=82\]

  4. asnaseer
    • 2 years ago
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    therefore:\[(9+b)^2=81+2*9*b+b^2=81+18b+b^2\]therefore:\[81+18b+b^2=82\]therefore:\[b^2+18b=1\]then try b=0.1, 0.2, ....

  5. aroub
    • 2 years ago
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    Oh cool =) Can you do this method if they asked you " Find a decimal approximation to the nearest hundredth" ? @asnaseer

  6. asnaseer
    • 2 years ago
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    yes, then you would need to try b=0.01, 0.02, etc. However, this might take too long, so a better strategy in this case might be to use a technique where you keep halving the interval, e.g.: try b=0.50 ---> if this is too low, then try b=0.75 ---> otherwise try b=0.25 that will converge quicker.

  7. asnaseer
    • 2 years ago
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    have you heard of the Newton-Raphson method?

  8. asnaseer
    • 2 years ago
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    if you have, then that will converge to the answer faster.

  9. aroub
    • 2 years ago
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    Actually no I havent

  10. asnaseer
    • 2 years ago
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    no worries - just use one (or more) of the techniques mentioned above to see which one you prefer.

  11. aroub
    • 2 years ago
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    Thank you so much =D

  12. asnaseer
    • 2 years ago
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    yw :)

  13. aroub
    • 2 years ago
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    Okay, I got a bit stuck using your method.. My method is SOOOOO complicated. Anyway, right you said you try b=0.1,0.2.. like this: b^2+18b=1 0.1^2=18(0.1)=1 ? and if it didnt work you try 0.2 and if didnt work you try 0.3 and so on? O.o

  14. aroub
    • 2 years ago
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    @asnaseer :)

  15. asnaseer
    • 2 years ago
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    you won't find an exact answer - you are supposed to find the nearest answer to one tenth

  16. aroub
    • 2 years ago
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    Yeah and I think you have to round it at the end right?

  17. asnaseer
    • 2 years ago
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    ok, let me try and explain. we said we can re-arrange the question to this form:\[b^2+18b=1\]

  18. aroub
    • 2 years ago
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    but my question is you substitute b with 0.1 like that? until you find an answer to the nearest tenth?

  19. asnaseer
    • 2 years ago
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    so what we need to do is to find what value of b will give an answer that is "closest" to 1. we can re-arrange further to get:\[b^2+18b-1=0\]so now we need to find a value for b that gives the closest answer to zero. so, lets try b=0.0, we get \(b^2+18b-1=-1\) so, lets try b=0.1, we get \(b^2+18b-1=0.81\) so, lets try b=0.2, we get \(b^2+18b-1=2.64\) so the "closest" answer is when b=0.1 therefore, the answer is 9.1 to nearest tenth.

  20. asnaseer
    • 2 years ago
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    does that make sense?

  21. aroub
    • 2 years ago
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    It makes sense till so the "closest" answer is when b=0.1 and then how did it become 9.1?

  22. asnaseer
    • 2 years ago
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    because we started this from:\[(9+b)^2=82\]now we have found 'b', answer is '9+b'

  23. asnaseer
    • 2 years ago
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    remember we said \(9^2=81\) and we are trying to find a number that, when squared equals 82. so that number must be 9 plus a little bit. we called that "little bit" 'b'

  24. aroub
    • 2 years ago
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    This is even more complicated lol but yeah i understood how =) Thank youu!!

  25. asnaseer
    • 2 years ago
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    Afwan :)

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