+101 \(\sqrt[3]{{(2{n}^{0})}^{-4}.2.{m}^{2}.{n}^{4}.{m}^{-1}.{n}^{3}\over {m}^{-4}.{n}^{4}}\)
\(=\sqrt[3]{{2}^{-4}.2.{m}^{2}.{m}^{4}.{n}^{4}.{m}^{-1}.{n}^{3}.{n}^{-4}}\)
\(=\sqrt[3]{{2}^{-3}.{m}^{2}.{m}^{4}.{m}^{-1}{n}^{4}.{n}^{3}.{n}^{-4}}\)
\(= \sqrt[3]{{2}^{-3}.{m}^{5}.{n}^{3}}\)
\(= \sqrt[3]{{n}^{3}\over{2}^{3}}*\sqrt[3]{{m}^{5}}\)
\(= \frac{n}{3}.{m}^{\frac{5}{3}}\)
.
+9673 \(\sqrt[3]{\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}}\\ =\left(\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}\right)^{1/3}\\ =\left(2^5\cdot m^5n^3\right)^{1/3}\\ =\sqrt[3]{32}\cdot m^{5/3}n\\ =2\sqrt[3]{4}\cdot m^{5/3}n\)
.
+118667 \sqrt[3]{\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}}\\
\(\sqrt[3]{\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}}\\ =\sqrt[3]{\dfrac{\left(2*1\right)^4\cdot 2m^{2+-1}n^{4+3}}{m^{-4}n^4}}\\ =\sqrt[3]{\dfrac{2^4\cdot 2^1m^{1}n^{7}}{m^{-4}n^4}}\\ =\sqrt[3]{\dfrac{2^5m^{1--4}n^{7-4}}{1}}\\ =\sqrt[3]{2^5m^{5}n^{3}}\\ =\sqrt[3]{2^3*2^2m^{3}m^2n^{3}}\\ =\sqrt[3]{2^3m^{3}n^{3}*2^2m^2}\\ =2mn\sqrt[3]{4m^2}\\\)
This is the same as Max's answer :)
+101 Emily, isn't it 2 left to the power -4? You and Max made 2 left to the power 4
+118667 Huh, where are you talking about Majid?
Oh yea - I copied the question wrong - you are of course correct :)
Good spotting :)
I tried to message you but this stupid site wont let me! It keeps giving me an error message.
I can message everyone else. I do not remember that ever happening before with anyone!
(yes I know you are a nice site, not stupid - please do not take offense and do any other horrible thing to me. )
+118667 Actually Max that was YOUR fault.
I copied and pasted the first line of your LaTex coding, that's how come I missed the -4 !
You might be the self proclaimed 'Smartest Cookie in the World" but you cannot claim to be the most careful one! :))
+9673 Oh. I see.
Oops.
Yes I am very careless :P
~The smartest cookie in the world
+9673 Btw no one is going to help me with my Calculus question :P
Nvm.
~The smartest cookie in the world
+118667 I did look at your calculus question Max but I'd have to look stuff up before I could have a hope of answering it.
It seemed like a lot of effort. Having said that, it doesn't look like it should be that hard. :/
It doesn't even look like calculus but that is probably just me displaying my total ignorance.....
Isn't there some simple formula you can use?
+118667 For those who may be interested, this is Max's question I am referring to
http://web2.0calc.com/questions/calculus-calculus
Actually I do not think anyone attempted Majid's last differential equations question either, though I could be wrong and I have no idea where it is anymore.
Maybe you could repost it Majid?
+101 \(\sqrt[3]{{(2{n}^{0})}^{-4}.2.{m}^{2}.{n}^{4}.{m}^{-1}.{n}^{3}\over {m}^{-4}.{n}^{4}}\)
\(=\sqrt[3]{{2}^{-4}.2.{m}^{2}.{m}^{4}.{n}^{4}.{m}^{-1}.{n}^{3}.{n}^{-4}}\)
\(=\sqrt[3]{{2}^{-3}.{m}^{2}.{m}^{4}.{m}^{-1}{n}^{4}.{n}^{3}.{n}^{-4}}\)
\(= \sqrt[3]{{2}^{-3}.{m}^{5}.{n}^{3}}\)
\(= \sqrt[3]{{n}^{3}\over{2}^{3}}*\sqrt[3]{{m}^{5}}\)
\(= \frac{n}{3}.{m}^{\frac{5}{3}}\)
