Answer:
![\dfrac{2x}{5}\sqrt[3]{50x^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B2x%7D%7B5%7D%5Csqrt%5B3%5D%7B50x%5E2%7D)
Step-by-step explanation:
Since the root indices are the same, the fraction can be combined under one radical. Then you want to do two things:
- factor out perfect cubes
- make the denominator a perfect cube (so it can be factored out)
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![\displaystyle\dfrac{\sqrt[3]{16x^8y^2}}{\sqrt[3]{5x^3y^2}}=\sqrt[3]{\dfrac{16x^8y^2}{5x^3y^2}}=\sqrt[3]{\dfrac{16x^5}{5}}=\sqrt[3]{\dfrac{(2x)^3\cdot2\cdot5^2x^2}{5^3}}\\\\=\boxed{\dfrac{2x}{5}\sqrt[3]{50x^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cdfrac%7B%5Csqrt%5B3%5D%7B16x%5E8y%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B5x%5E3y%5E2%7D%7D%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B16x%5E8y%5E2%7D%7B5x%5E3y%5E2%7D%7D%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B16x%5E5%7D%7B5%7D%7D%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B%282x%29%5E3%5Ccdot2%5Ccdot5%5E2x%5E2%7D%7B5%5E3%7D%7D%5C%5C%5C%5C%3D%5Cboxed%7B%5Cdfrac%7B2x%7D%7B5%7D%5Csqrt%5B3%5D%7B50x%5E2%7D%7D)
Answer:
4) x=2.52 (3 s.f.)
5) x= -43
6) x= -184
Step-by-step explanation:
Please see attached picture for full solution.
Answer:
7/45
Step-by-step explanation:
If you can use your calculator*
Just press 0.1555555555555555555555555555 repeatedly
If you cant:
I dont know sorry
Answer:
Part 1
s'(t)=1

Part 2
![s'(t)=\frac{1}{3\sqrt[3]{t^2}}](https://tex.z-dn.net/?f=s%27%28t%29%3D%5Cfrac%7B1%7D%7B3%5Csqrt%5B3%5D%7Bt%5E2%7D%7D)
V'(t)=1
Step-by-step explanation:
As stated in the question, if s is the length of a cube, its volume is

When s changes in time, the volume will change too. The challenge here is to find the growth of the volume in terms of those changes in length
Part 1

The change of s with respect to time is found by differentiating the relation to get

The volume will also change, and its derivative is

Since s'=1

How s=t

Part 2

Replacing this into the formula for V(s)

So we have
![s=\sqrt[3]{t}=t^{\frac{1}{3}}](https://tex.z-dn.net/?f=s%3D%5Csqrt%5B3%5D%7Bt%7D%3Dt%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
Computing the derivatives:
V(t)=t =>
V'(t)=1
![s'(t)=\frac{1}{3}t^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{t^2}}](https://tex.z-dn.net/?f=s%27%28t%29%3D%5Cfrac%7B1%7D%7B3%7Dt%5E%7B-%5Cfrac%7B2%7D%7B3%7D%7D%3D%5Cfrac%7B1%7D%7B3%5Csqrt%5B3%5D%7Bt%5E2%7D%7D)