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

Mathematics
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\[\frac{3^3-2^2}{\sqrt{4}+\sqrt{49}}\]
It is not written as a fraction
You have to use BIDMAS in the process

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Other answers:

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