Answer:
Go back to connections accademy
Step-by-step explanation:
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
Answer:
there answer is either a or c
Answer:
300
(second answer choice)
Step-by-step explanation:
Recall that the Z-distribution is defined as :
and therefore, to obtain a Z = 2.5 when the mean value 
=200, and the standard deviation 
= 40 , we solve for "X":
Because its not quite clear if the 5 is negative you might want to try once with a positive and once with it negative. anyway, when you only have multiplication like this, so no addition or subtraction, you can "distribute"the exponent on the outside of the parenthesis. so the first two terms become 8^(-2/3)(w^7)^(-2/3) hopefully you can figure out the rest. from here, with w if you have something to an exponent which is also to an exponent, you can multiply the exponents. so that's 7*(-2/3) which is-14/3. let me know if you have trouble with fraction multiplication. now negative exponents have something special about them. if you have a negative exponent you can write it as one over that term with a positive exponent. so the first term becomes 1/(8^(2/3)) which is 1/4.
