To simplify
![\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B24x%5E6y%7D%7B128x%5E4y%5E5%7D%7D)
we need to use the fact that
![\sqrt[4]{x^4}=|x|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E4%7D%3D%7Cx%7C)
Why the absolute value? It's because 
.
We start by rewriting as
![\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E6y%7D%7B2%5E6x%5E4y%5E5%7D%7D)
![\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E4x%5E2y%7D%7B2%5E42%5E2x%5E4y%5E4y%7D%7D)
Since 
, we have 
, and the above reduces to
![\sqrt[4]{\dfrac{3x^2y}{2^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7B2%5E4y%5E4y%7D%7D)
Then we pull out any 4th powers under the radical, and simplify everything we can:
![\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Csqrt%5B4%5D%7B2%5E4y%5E4%7D%7D%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7By%7D%7D)
![\dfrac1{|2y|}\sqrt[4]{3x^2}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%7C2y%7C%7D%5Csqrt%5B4%5D%7B3x%5E2%7D)
where 
allows us to write 
, and this also means that 
. So we end up with
![\dfrac{\sqrt[4]{3x^2}}{2y}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B4%5D%7B3x%5E2%7D%7D%7B2y%7D)
making the last option the correct answer.
See it is < and not ≤ so we don't include -4
put a circle around -4 but don't shade it in
to get there, move 4 units left from 0
then we see x is less than -4
so shade to the left of the circle
Answer:
i think this is the answer not 100 percentage sure V≈84.82cm³
p=3h + 3u subtract 3u from both sides
p-3u=3h now divide both sides by 3
h = p/3 - u
