Melody

Yes I get the same as Mellie - I put my answer on the original question :)

The number of possibilities if all the dice look different is 6^4

Only 6 of those will have every cie the sames

So   P(all the same number )=     $${\frac{{\mathtt{6}}}{{{\mathtt{6}}}^{{\mathtt{4}}}}} = {\frac{{\mathtt{1}}}{{\mathtt{216}}}} = {\mathtt{0.004\: \!629\: \!629\: \!629\: \!629\: \!6}}$$

P (they are not all the same)  =   $${\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{6}}}{{{\mathtt{6}}}^{{\mathtt{4}}}}}\right) = {\frac{{\mathtt{215}}}{{\mathtt{216}}}} = {\mathtt{0.995\: \!370\: \!370\: \!370\: \!370\: \!4}}$$

Hey Mellie, what was the address of the post that we all got different answers on?

It takes a genius.   LOL

Thanks Bertie.   

Thanks Alan and Bertie :)

You cannot solve it because there is not = sign.

You can only simplify it.

$$\\\dfrac{(ab^2)^3*b^{-7}}{(a^{-1}*b)^{-2}}\\\\
=\dfrac{a^3b^6b^{-7}}{a^{-1*-2}*b^{-2}}\\\\
=\dfrac{a^3b^6b^{-7}}{a^{2}b^{-2}}\\\\
=\dfrac{a^3b^{6-7}}{a^{2}b^{-2}}\\\\
=\dfrac{a^3b^{-1}}{a^{2}b^{-2}}\\\\
=\dfrac{a^{3-2}b^{-1- -2}}{1}\\\\
=\dfrac{a^{1}b^{1}}{1}\\\\
=ab$$

$$\\V=\frac{(2.21x)}{cos(a)}*\sqrt{-y+tan(a)*x}\\
V^2=\frac{(2.21x)^2}{cos^2(a)}*(-y+tan(a)*x)\\
V^2cos^2a=(2.21x)^2*(-y+tan(a)*x)\\
V^2cos^2a=-(2.21x)^2y+(2.21x)^2tan(a)*x\\
V^2cos^2a+(2.21x)^2y=(2.21x)^2tan(a)*x\\
V^2cos^2a+(2.21x)^2y=(2.21x)^2\sqrt{\frac{1}{cos^2a}-1}*x\\
V^4cos^4a+2*V^2cos^2a*(2.21x)^2y+(2.21x)^4y^2=(2.21x)^4*(\frac{1}{cos^2a}-1)}*x^2\\
V^4cos^4a+2*V^2cos^2a*(2.21x)^2y+(2.21x)^4y^2=(2.21x)^4*(\frac{1-cos^2a}{cos^2a})*x^2\\
V^4cos^6a+2*V^2cos^4a*(2.21x)^2y+(2.21x)^4y^2cos^2a=(2.21x)^4*({1-cos^2a})*x^2\\$$

I am not getting anywhere with this.  Sorry.

Who are we going to believe.  Alan or our friendly neighbourhood troll ?     

Alan has back up evidence... The troll has googly eyes....  Mmm it IS a tough one!!

Anyway I think i will give you thumbs up anyway Mr Troll.  You are entertaining.  :))

Hi   Zacismyname

When you make a post there is a btton in the ribbon called  [Math Formula] 

if you open that and type in   3*sqrt(5) this will be your output.     $${\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}$$

$$\\3^{x+1}=6^{4-x}\\\\
ln[3^{x+1}]=ln[6^{4-x}]\\\\
(x+1)ln3=(4-x)ln6\\\\
xln3+ln3=4ln6-xln6\\\\
xln3+xln6=ln6^4-ln3\\\\
x(ln3+ln6)=ln1296-ln3\\\\
xln18=ln432\\\\
x=\frac{ln432}{ln18}\\\\$$

This answer is identical to Geno's