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
Answer:
y=3x-1
Step-by-step explanation:
28: 1, 2, 4, 7, 14, 28
40: 1, 2, 4, 5, 8, 10, 20, 40
The common factors are 1, 2, and 4.
You can use a factor rainbow to help you find the factors (:
Remember, order of operations
exponent before multiply
so 4x^5/6=4 times x^5/6
simplify the x^5/6 first
remember
![x^\frac{m}{n}=\sqrt[n]{s^m}](https://tex.z-dn.net/?f=x%5E%5Cfrac%7Bm%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Bs%5Em%7D)
so
![x^\frac{5}{6}=\sqrt[6]{x^5}](https://tex.z-dn.net/?f=x%5E%5Cfrac%7B5%7D%7B6%7D%3D%5Csqrt%5B6%5D%7Bx%5E5%7D)
so
![4x^\frac{5}{6}=4\sqrt[6]{x^5}](https://tex.z-dn.net/?f=4x%5E%5Cfrac%7B5%7D%7B6%7D%3D4%5Csqrt%5B6%5D%7Bx%5E5%7D)
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