anonymous
  • anonymous
Solve: 3^2 - 2^2 / square root of 4 + square root of 49 using BIDMAS. Cheers
Mathematics
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SOLVED
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chestercat
  • chestercat
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Mertsj
  • Mertsj
\[\frac{3^3-2^2}{\sqrt{4}+\sqrt{49}}\]
anonymous
  • anonymous
It is not written as a fraction
anonymous
  • anonymous
You have to use BIDMAS in the process

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More answers

Mertsj
  • Mertsj
9-4/2+7= 9-2+7=7+7=14
anonymous
  • anonymous
Why can't it be 9-9=0 if addition and subtraction according to BIDMAS is of equal priority?
Mertsj
  • Mertsj
Because 4/2 means 4 divided by 2 and you must do the division before the addition and subtraction.
anonymous
  • anonymous
what the monkey is a bidmas?
Mertsj
  • Mertsj
brackets, indices, division multiplication, addition, subtraction
anonymous
  • anonymous
oooh sort of like "please excuse my senile aunt sally"
Mertsj
  • Mertsj
Precisely.
anonymous
  • anonymous
That wasn't my question. I said why did you do subtraction before addition?
anonymous
  • anonymous
If you did it the other way you would have got 9-9=0
anonymous
  • anonymous
\[3^2 - 2^2 \div (\sqrt{4}) + \sqrt{49} \implies 9 - 4 \div 2 + 7\] According to BIDMAS rule, here you will divide first...
anonymous
  • anonymous
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?
AravindG
  • AravindG
what is I in BIDMAS ?
AravindG
  • AravindG
Oh its above :P ....i got the same doubt as satellite ..we usually use the term BODMAS
hartnn
  • hartnn
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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