Given expression is
![\sqrt[4]{\frac{16x^{11}y^8}{81x^7y^6}}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B11%7Dy%5E8%7D%7B81x%5E7y%5E6%7D%7D)
Radical is fourth root
first we simplify the terms inside the radical
![\frac{x^{11}}{x^7}=x^4](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B11%7D%7D%7Bx%5E7%7D%3Dx%5E4)
![\frac{y^(8)}{y^6}=y^2](https://tex.z-dn.net/?f=%5Cfrac%7By%5E%288%29%7D%7By%5E6%7D%3Dy%5E2)
So the expression becomes
![\sqrt[4]{\frac{16x^4y^2}{81}}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E4y%5E2%7D%7B81%7D%7D)
Now we take fourth root
![\sqrt[4]{16} = 2](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B16%7D%20%3D%202)
![\sqrt[4]{81} = 3](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B81%7D%20%3D%203)
![\sqrt[4]{x^4} = x](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E4%7D%20%3D%20x)
We cannot simplify fourth root (y^2)
After simplification , expression becomes
![\frac{2x\sqrt[4]{y^2}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2x%5Csqrt%5B4%5D%7By%5E2%7D%7D%7B3%7D)
Answer is option B
Does it tell you what (x) stands for
Answer:
c
Step-by-step explanation:
because I had a test on these
Answer: (4a - 3b)2
Step-by-step explanation:it stays the same because there is no like terms