xerxes

I know, I realised that. Another thing I leant today - however, I am pretty sure they taught me that you can't get a root if the radicant is negative.

it would just work with odd denominators, wouldn't it? 

d = 12 472km

r = 6 236km

p = d * pi = $${\mathtt{12\,472}}{\mathtt{\,\times\,}}{\mathtt{\pi}} = {\mathtt{39\,181.943\: \!575\: \!571\: \!901\: \!270\: \!1}}$$

Let's add a meter:

(1m = 1/1000km)

$${\mathtt{39\,181.943\: \!58}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{1\,000}}}} = {\mathtt{39\,181.944\: \!58}}$$

And back to the diameter:

$${\frac{{\mathtt{39\,181.944\: \!58}}}{{\mathtt{\pi}}}} = {\mathtt{12\,472.000\: \!319\: \!719\: \!393\: \!786\: \!5}}$$

We have ~3.197 Meter more.

We can lift it now by 1.5985 Meter on each side. 

It is indeed, and you learn a lot by doing this. Mathematicians in the making is a nice word, btw.

I think Badinage's answer is great, I am now able to accept the .2 as a proper number. Without the 0. This nice construction is quite fascinating 😁

I know 

But the number itself still contains the digit 0, doesn't it? 

xy contains a times sign as well, doesn't it? Even tough we don't have to write it?

Thanks Melody, I appreciate your support :) 

However, I do not like to do the homework for other people. Helping is fine tough. If someone asks a short question, he or she gets probably a short answer back 

I really like the way how Melody is able to get a solution out of this question. 

But in my opinion, it is a irrational number as a result. Or, if you "don't know" these irrational numbers, it is simply not defined. Why? Because of this:

$${\left(-{\mathtt{1.42}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{7}}}}\right)} = {\sqrt[{{\mathtt{{\mathtt{7}}}}}]{-{\mathtt{1.42}}}}$$

And the result of any root with a negative radicant is either not defined or a irrational number. 

(I am pretty sure that you can not pull the negative sign out of the brackets. Even if there is a nice result ;) )

Always ways up for comments!

I am ashamed  It is probably getting too late for me. 

I thought about 1dm^3 being 1litre. But that's alright. 

$$1dm^3 = 1litre$$

If the pool would be 10cm high, we would have 30dm^3 of content. We are now looking for a height. 10cm of height means we have 30 litre of content.  

Hence that:

$${\frac{{\mathtt{126}}}{{\mathtt{30}}}} = {\frac{{\mathtt{21}}}{{\mathtt{5}}}} = {\mathtt{4.2}}$$

As we calculate in dm:

42cm

and if we just fill the pool by 2/3s:

the height should be 63cm then. 

However, same result, as before. Thanks for pointing this out anyways, Melody. 

I think you can not take the number 0.22, as the 0 itself is a digit, isn't it? And the 0 is not a 2, Even Rough You don't need write it  

But correct me if I am wrong 

$${\mathtt{AG}} = {\sqrt{{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{6}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{8}}}^{{\mathtt{2}}}}}$$

Please mark this post as "answer" if it helped you! 

Just enter this in the calculator.

$${\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{7}}}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{7}}}}{\mathtt{\,\times\,}}{\mathtt{2}} = {\mathtt{15.874\: \!507\: \!866\: \!387\: \!543\: \!5}}$$

or

$${\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{7}}}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{7}}}} = {\mathtt{6}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{7}}}}$$

Please mark this post as "answer" if it helped you and solved your problem