If you know this pls help ):
1 answer:
From the right hand side, we will need to find a way to rewriting 3x²y in terms of cube roots.
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
![\sqrt[3]{162x^{c}y^{5}} = \sqrt[3]{(3x^{2}y)^{3}} * \sqrt[3]{6y^{d}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B162x%5E%7Bc%7Dy%5E%7B5%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%283x%5E%7B2%7Dy%29%5E%7B3%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B6y%5E%7Bd%7D%7D)
![= \sqrt[3]{162x^{6}y^{3+d}}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B162x%5E%7B6%7Dy%5E%7B3%2Bd%7D%7D)
Hence, c = 6 and d = 2
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
Hence, c = 6 and d = 2
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Answer:
Step-by-step explanation:
Water is being pumped into an inflated rubber sphere at a constant rate of 0.03 cubic meters per second.
So, dV/dt = 0.03.
We want to show that dS/dt is directly proportional to 1/r.
In other words, we want to establish the relationship that dS/dt = k(1/r), where k is some constant.
First, the volume of a sphere V is given by:
Therefore:
Next, the surface area of a sphere S is given by:
Therefore:
We can divide both sides by 2:
We can substitute this into dV/dt. Rewriting:
So:
Since dV/dt = 0.03 or 3/100:
Therefore:
Where k = 3/50.
And we have shown that dS/dt is directly proportional to 1/r.