Answer:
the figure could be a rhombus because the 2 lines of symmetry bisect the angles
Step-by-step explanation:
Your answer is in the attached picture.
Hope this helps,
♥Nikki♥
Answer:

Step-by-step explanation:
Domain:

![\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{3y-x}=\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{-(x-3y)}\\\\=\dfrac{x^2+9y^2}{x-3y}-\dfrac{6xy}{x-3y}=\dfrac{x^2+9y^2-6xy}{x-3y}\\\\=\dfrac{x^2-2(x)(3y)+(3y)^2}{3y-x}=\dfrac{(x-3y)^2}{3y-x}\\\\=\dfrac{\bigg[-1(3y-x)\bigg]^2}{3y-x}=\dfrac{(-1)^2(3y-x)^2}{3y-x}\\\\=\dfrac{1(x-3y)(x-3y)}{x-3y}=x-3y](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2%2B9y%5E2%7D%7Bx-3y%7D%2B%5Cdfrac%7B6xy%7D%7B3y-x%7D%3D%5Cdfrac%7Bx%5E2%2B9y%5E2%7D%7Bx-3y%7D%2B%5Cdfrac%7B6xy%7D%7B-%28x-3y%29%7D%5C%5C%5C%5C%3D%5Cdfrac%7Bx%5E2%2B9y%5E2%7D%7Bx-3y%7D-%5Cdfrac%7B6xy%7D%7Bx-3y%7D%3D%5Cdfrac%7Bx%5E2%2B9y%5E2-6xy%7D%7Bx-3y%7D%5C%5C%5C%5C%3D%5Cdfrac%7Bx%5E2-2%28x%29%283y%29%2B%283y%29%5E2%7D%7B3y-x%7D%3D%5Cdfrac%7B%28x-3y%29%5E2%7D%7B3y-x%7D%5C%5C%5C%5C%3D%5Cdfrac%7B%5Cbigg%5B-1%283y-x%29%5Cbigg%5D%5E2%7D%7B3y-x%7D%3D%5Cdfrac%7B%28-1%29%5E2%283y-x%29%5E2%7D%7B3y-x%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%28x-3y%29%28x-3y%29%7D%7Bx-3y%7D%3Dx-3y)
Used:
The distributive property: a(b + c) = ab + ac
(a - b)² = a² - 2ab + b²
It's 10 because I'm in 5th I know