Amistre64, it's me again. Need help with an exponent again!

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Amistre64, it's me again. Need help with an exponent again!

Mathematics
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didja try yelling at it?
jI even grounded it and took away it's cell phone and it won't cooperate! I can't understand...it works with my 15 year old daughter!
\[-3/2\times 2^{-2/9}\]

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dont punish them it makes it worse. They now think math is bad if you truly want to help arrange extra help and talk to the teacher
Did you see my expression I need to solve?
i see it
we wanna derive it?
No I need to figure out what it's = to.
2^-2/9 equals what?
\[\frac{-3}{2*2^{2/9}}\]
\[\frac{-3}{2^{(9/9 + 2/9)}}\]
we good so far?
yes, I got that far. I know that answer, but what i got wasn't right according to them Like they know any better! Who are they to tell me I am always wrong!?
; )
:) \[\frac{-3}{2^{11/9}}\] \[\frac{-3}{\sqrt[9]{2^{11}}}\]
\[2^{11} \iff 2^{9}*2^{2}\] \[\sqrt[9]{2^9.2^2}\iff \sqrt[9]{2^9}.\sqrt[9]{2^2}\] \[\frac{-3}{2\sqrt[9]{2^2}}\]
\[\sqrt[9]{2^4}*\sqrt[9]{2^5} = \sqrt[9]{2^9}\iff 2\] right?
yes, I remember that one from the s thing earlier! Yay me!
\[\frac{-3}{2\sqrt[9]{2^4}}*\frac{\sqrt[9]{2^5}}{\sqrt[9]{2^5}} = \frac{-3\sqrt[9]{64}}{4}\]
i typoed it dint i lol
9rt(2^2) not equal 9rt(2^4) lol
Actually we lucked out cuz they are leaving the radical in the denominator! The answer to the whole problem is cbrt 2 - 3/2ninth root of 4, just what you said! I am not surprised you got it right!
yay!! i was dreading having to retype thaqt lol
\[\sqrt[3]{2}- 3/2\sqrt[9]{4}\]
Again and again and again, thank you.
yw :)

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