Please help me with this!

  -540

Step-by-step explanation:

The coefficient of the k-th term of (a+b)^n is nCk. (k = 0, 1, 2, ..., n)

Here, we want the coefficient of a^3b^3 for n=6, so k=3 and the coefficient is ...

  6C3 = 6!/(3!(6-3)!) = 5·4 = 20

So for a=3x and b=-y, we want ...

  20a^3b^3 = 20(3x)^3(-y)^3 = -540x^3y^2

The coefficient you're looking for is -540.

the formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. if a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point.