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moongazer
question below :)
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@myininaya could you help me ?
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\[\LARGE{\frac{\sqrt{(95t)^{3}}}{\sqrt[3]{(27s^{3}t^{-4}}){^2}}*(\frac{3s^{-2}}{4\sqrt[3]{t}})^{-1}}\] Hard to read, but is that the equation?
No, The 5 there should be s
Is this correct? \[\LARGE{\frac{\sqrt{(9st)^{3}}}{\sqrt[3]{(27s^{3}t^{-4}}){^2}}*(\frac{3s^{-2}}{4\sqrt[3]{t}})^{-1}} \]
could you help me? because I think I am getting the wrong answer
Blah, gotta use pen and paper for this one lol, getting some crazy answer.
My teacher always gives us crazy equations :)
\[\LARGE{(\frac{3s^{-2}}{4\sqrt[3]{t}})^{-1} = \frac{(3s^{-2})^{-1}}{(4\sqrt[3]{t})^{-1}}} = \frac{4\sqrt[3]{t}}{3s^{-2}} = \frac{4s^{2}\sqrt[3]{t}}{3} \]
\[\Large{\sqrt{(9st)^{3}} = \sqrt{(9st)^{2}*(9st)} = 9st*\sqrt{9st}}\]
I understand it, please continue :)
\[\Large{\sqrt[3]{(27s^{3}t^{-4})^{-2}} = \sqrt[3]{\frac{1}{(27s^{3}t^{-4})^{2}}}}\] Ack, gotta go!
just continue it tomorrow if you want to :) what answer did you get ?
Didn't, that's all I did so far. Does it match up with what you have?
I did another method tonight. But I already did that kind of method although but I didn't finish answering it. I'll just try answering this with that kind of way and the other way so I can check my answers :)
final answer i am getting as : \(4st^4\sqrt{st}\)
\[\Large{\sqrt[3]{\frac{1}{27^{2}(s^{3})^{2}(t^{-4})^{2}}} = \sqrt[3]{\frac{t^{8}}{9^{3}(s^{2})^{3}}} = \frac{t^{2}}{9s^{2}}\sqrt[3]{t^{2}}}\]
\[\LARGE{\frac{9st*\sqrt{9st}}{\frac{t^{2}}{9s^{2}}*\sqrt[3]{t^{2}}}*\frac{4s^{2}\sqrt[3]{t}}{3}} = \LARGE{\frac{108s^{5}\sqrt{9st}}{t\sqrt[3]{t}}}\] No idea if that's right or not. Didn't check every step.