Melody> And yes xy is x*y  so if you are not allowed to use a multiply sign

But you are allowed to use brackets, so it's permissible to write x*y as x(y)

divide both sides by -2

Thanks you xerxes for answering questions and for getting so involved in the forum :)

I take your point but you did not give a hint on how to do the question.  

You gave a final answer that I do not think any student could have learned from.  

I really like it when people give hints without giving an answer but giving an answer without any explanation is usually pretty pointless.

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

I am glad that you are happy xerxes, Badinage's answer is great and playing with numbers is always fund for mathematicians and mathematicans in the making :)

Odds of picking the first one are 0.7. Now there are 69 left out of 70 for the second choice,so odds of picking second good one are 0.69.Similarly for the third choice,the odds are 0.68. 

With each choice,multiply with odds of picking the next,and the result is 0.7x0.69x0.68=32.844% likelihood that they are all good.

I think your answer wins Badinage         Your answer puts mine to shame.    LOL 

Come on Badinage.  You have to tell us what tht equals!

Use the forum calc like I did   

Xerxes,  I suppose it depends on your take.   

Tell me, does 2 have a zero in it?   It is exactly the same as 2.0  so by your logic 2 must also have a zero.

Zeros are often necessary in numbers because they are place holders.  BUT 2 and .2 do not need any zeros as place holders because they are right up against the decimal point so their place value is indisputable.

And yes xy is x*y  so if you are not allowed to use a multiply sign, does that mean that you cannot multiply OR does it mean that you cannot put a * in the display.  I guess that is up to the individuals interpretation.  

There is a problem here xerxes.   

1cm^3 hold 1ml

so there are 1000cm^3 in a litre

$$\textcolor[rgb]{\mathbf{0,0,1}{.2^{2^{22}}}}$$$$\textcolor[rgb]{0,0,1}{\mathbf{.2^{2^{22}}}}$$

  evaluates to ∼ 2x10^-2931693

⛵⛵⛵⛵⛵⛵

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 did not write it the calc put it in there 0.2 = .2     the 0 is not needed.

 It is usually added just so that the . is not mistaken for a speck of dirt.

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!

Triangle EGH is a right angled triangle.  EG is the hypotenuse.   SO

$$EG=\sqrt{6^2+8^2}=\sqrt{100}=10cm$$

Now triangle AEG is a right angled triangle and  AG is the hypotenuse

so

$$\\AG=\sqrt{10^2+2^2}=\sqrt{104}=\sqrt{4*26}=2\sqrt{26}cm\\
AG\approx 10.2cm\qquad $correct to 2 decimal places$$$

Lets compare this to xerxes's answer.

$$\\AG^2=AE^2+EG^2\\
\\AG^2=2^2+\left(\sqrt{(EH^2+HG^2}\right)^2\\
\\AG^2=2^2+(EH^2+HG^2)\\
\\AG^2=2^2+(8^2+6^2)\\
\\AG^2=2^2+8^2+6^2\\
\\AG=\sqrt{2^2+8^2+6^2}\\$$

So you see, Xerxes wass correct in the fist place.  Maybe he should have explained him/herself better but I think he should get positive points not negative ones.

I only have 3 points that i can give you, I am sorry that i cannot give you any more.  :)

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. 

Mathcad, I have transfered your answer over onto the original thread.

Badinage has asked the question exactly as I have requested but it is better if the answers are all together on the original thread. 

Thank you   

Also answered here by Mathcad

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 

Thanks Mathcad, that looks interesting :)

Thanks, how do I post this as answer?

One last attempt at posting this.The software on this system is hopeless.

We can write 36x27!+25 as (36x31!+(31x30x29x28)x25)/31x30x29x28 and can now factor out 31 from the numerator.So the expression is divisible by 31 and we can write

36x27!+25=31n for some n.

then,to use Wilson,we can write 36x27!+25=31!n/30!   and since 31!=30mod 31 ,rearrange to get

n=30!(36x27!+25)/30modulo31  which is divisible by 31.

hope this helps.

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

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

$${\mathtt{\,-\,}}\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right) = {\mathtt{\,-\,}}{\frac{{\mathtt{5}}}{{\mathtt{3}}}} = -{\mathtt{1.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$