Answer:
![\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E6y%5E4%7D%7B81x%5E2y%5E8%7D%7D%5Crightarrow%5Cfrac%7B2x%7D%7B3y%7D%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B81x%5E2y%5E%7B10%7D%7D%7B81x%5E6y%5E6%7D%7D%20%5Crightarrow%5Cfrac%7B3y%7D%7B2x%7D%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7B64x%5E8y%5E7%7D%7B125x%5E2y%5E%7B10%7D%7D%7D%5Crightarrow%5Cfrac%7B4x%5E2%7D%7B5y%7D%5C%5C%5Csqrt%5B5%5D%7B%5Cfrac%7B243x%5E%7B17%7Dy%5E%7B16%7D%7D%7B32x%5E7y%5E%7B21%7D%7D%7D%5Crightarrow%5Cfrac%7B3x%5E2%7D%7B2y%7D%5C%5C%5Csqrt%5B5%5D%7B%5Cfrac%7B32x%5E%7B12%7Dy%5E%7B15%7D%7D%7B243x%5E7y%5E%7B10%7D%7D%7D%20%5Crightarrow%5Cfrac%7B2xy%7D%7B3%7D%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B10%7Dy%5E%7B9%7D%7D%7B256x%5E2y%5E%7B17%7D%7D%7D%5Crightarrow%5Cfrac%7Bx%7D%7B2y%7D)
Step-by-step explanation:
![\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E6y%5E4%7D%7B81x%5E2y%5E8%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B%282%5E4%29%28x%5E%7B6-2%7D%29%28y%5E%7B4-8%7D%29%7D%7B%283%5E4%29%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B2%5E4x%5E4y%5E%7B-4%7D%7D%7B3%5E4%7D%7D%20%3D%5Cfrac%7B2xy%5E%7B-1%7D%7D%7B3%7D%3D%5Cfrac%7B2x%7D%7B3y%7D)
![\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B81x%5E2y%5E%7B10%7D%7D%7B81x%5E6y%5E6%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B%283%5E4%29%28x%5E%7B2-6%7D%29%28y%5E%7B10-6%7D%29%7D%7B%282%5E4%29%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B3%5E4x%5E%7B-4%7Dy%5E%7B4%7D%7D%7B2%5E4%7D%7D%20%3D%5Cfrac%7B3x%5E%7B-1%7Dy%5E1%7D%7B3%7D%3D%5Cfrac%7B3y%7D%7B2x%7D)
![\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B64x%5E8y%5E7%7D%7B125x%5E2y%5E%7B10%7D%7D%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%284%5E3%29%28x%5E%7B8-2%7D%29%28y%5E%7B7-10%7D%29%7D%7B%285%5E3%29%7D%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4%5E3x%5E6y%5E%7B-3%7D%7D%7B5%5E3%7D%7D%20%3D%5Cfrac%7B4x%5E2y%5E%7B-1%7D%7D%7B5%7D%3D%5Cfrac%7B4x%5E2%7D%7B5y%7D)
![\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%5Cfrac%7B243x%5E%7B17%7Dy%5E%7B16%7D%7D%7B32x%5E7y%5E%7B21%7D%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B%283%5E5%29%28x%5E%7B17-7%7D%29%28y%5E%7B16-21%7D%29%7D%7B%282%5E5%29%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B3%5E5x%5E%7B10%7Dy%5E%7B-5%7D%7D%7B2%5E5%7D%7D%20%3D%5Cfrac%7B3x%5E2y%5E%7B-1%7D%7D%7B2%7D%3D%5Cfrac%7B3x%5E2%7D%7B2y%7D)
![\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%5Cfrac%7B32x%5E%7B12%7Dy%5E%7B15%7D%7D%7B243x%5E7y%5E%7B10%7D%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B%282%5E5%29%28x%5E%7B12-7%7D%29%28y%5E%7B15-10%7D%29%7D%7B%283%5E5%29%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B2%5E5x%5E%7B5%7Dy%5E%7B5%7D%7D%7B3%5E5%7D%7D%20%3D%5Cfrac%7B2x%5E1y%5E%7B1%7D%7D%7B3%7D%3D%5Cfrac%7B2xy%7D%7B3%7D)
![\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B10%7Dy%5E%7B9%7D%7D%7B256x%5E2y%5E%7B17%7D%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B%282%5E4%29%28x%5E%7B10-2%7D%29%28y%5E%7B9-17%7D%29%7D%7B%284%5E4%29%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B2%5E4x%5E%7B8%7Dy%5E%7B-8%7D%7D%7B4%5E4%7D%7D%20%3D%5Cfrac%7B2x%5E%7B1%7Dy%5E%7B-1%7D%7D%7B4%7D%3D%5Cfrac%7Bx%7D%7B2y%7D)
Thus,
![\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E6y%5E4%7D%7B81x%5E2y%5E8%7D%7D%5Crightarrow%5Cfrac%7B2x%7D%7B3y%7D%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B81x%5E2y%5E%7B10%7D%7D%7B81x%5E6y%5E6%7D%7D%20%5Crightarrow%5Cfrac%7B3y%7D%7B2x%7D%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7B64x%5E8y%5E7%7D%7B125x%5E2y%5E%7B10%7D%7D%7D%5Crightarrow%5Cfrac%7B4x%5E2%7D%7B5y%7D%5C%5C%5Csqrt%5B5%5D%7B%5Cfrac%7B243x%5E%7B17%7Dy%5E%7B16%7D%7D%7B32x%5E7y%5E%7B21%7D%7D%7D%5Crightarrow%5Cfrac%7B3x%5E2%7D%7B2y%7D%5C%5C%5Csqrt%5B5%5D%7B%5Cfrac%7B32x%5E%7B12%7Dy%5E%7B15%7D%7D%7B243x%5E7y%5E%7B10%7D%7D%7D%20%5Crightarrow%5Cfrac%7B2xy%7D%7B3%7D%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B10%7Dy%5E%7B9%7D%7D%7B256x%5E2y%5E%7B17%7D%7D%7D%5Crightarrow%5Cfrac%7Bx%7D%7B2y%7D)
Answer: B. 2 2/5
Step-by-step explanation: Hope this help :D
Answer:
triangle AEH, triangle CFG, triangle BFG, and triangle DEG.
so b,c,e,f
this is right for e2020.
The answer would be c- 2/6
there are six cups total and three (blake plus two others) friends
there are two ways you can solve this…
1. by making groups
if you look at the picture, you can split it into three groups because that is the number of people, then count how many cups there are in one group which will be your numerator for the denominator six which is the total
2. simple division
divide the total number of cups buy the number of people which is 3
you get two and follow the same process above
Answer:
Domain = (-∞, -2) ∪ (-2, ∞).
Range = (-∞, 0) ∪ (0, ∞).
Step-by-step explanation:
t can't have the value -2 because that would make 1/(t + 2) = 1/0 which is undefined.
So the domain is All real values of t. except t = -2.
As t approaches infinity or negative infinity f(t) approaches zero, and as t approaches -2 from below or above f(t) approaches negative infinity or positive infinity.