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Chad123

  • 3 years ago

Solve: 3^2 - 2^2 / square root of 4 + square root of 49 using BIDMAS. Cheers

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  1. Mertsj
    • 3 years ago
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    \[\frac{3^3-2^2}{\sqrt{4}+\sqrt{49}}\]

  2. Chad123
    • 3 years ago
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    It is not written as a fraction

  3. Chad123
    • 3 years ago
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    You have to use BIDMAS in the process

  4. Mertsj
    • 3 years ago
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    9-4/2+7= 9-2+7=7+7=14

  5. Chad123
    • 3 years ago
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    Why can't it be 9-9=0 if addition and subtraction according to BIDMAS is of equal priority?

  6. Mertsj
    • 3 years ago
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    Because 4/2 means 4 divided by 2 and you must do the division before the addition and subtraction.

  7. anonymous
    • 3 years ago
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    what the monkey is a bidmas?

  8. Mertsj
    • 3 years ago
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    brackets, indices, division multiplication, addition, subtraction

  9. anonymous
    • 3 years ago
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    oooh sort of like "please excuse my senile aunt sally"

  10. Mertsj
    • 3 years ago
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    Precisely.

  11. Chad123
    • 3 years ago
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    That wasn't my question. I said why did you do subtraction before addition?

  12. Chad123
    • 3 years ago
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    If you did it the other way you would have got 9-9=0

  13. waterineyes
    • 3 years ago
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    \[3^2 - 2^2 \div (\sqrt{4}) + \sqrt{49} \implies 9 - 4 \div 2 + 7\] According to BIDMAS rule, here you will divide first...

  14. Chad123
    • 3 years ago
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    I know that part but why subtract and then do the division when it's supposed to be the opposite. Can't it be 9-9=0?

  15. AravindG
    • 3 years ago
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    what is I in BIDMAS ?

  16. AravindG
    • 3 years ago
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    Oh its above :P ....i got the same doubt as satellite ..we usually use the term BODMAS

  17. hartnn
    • 3 years ago
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    it can't be 9-9 because the order of operation (evaluation) is from left to right. so, 9-2 will be evaluated first, not 2+7

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