Stuck on this problem

9514 1404 393

  -8,257,536·u^5·v^10

Step-by-step explanation:

The expansion of (a +b)^n is ...

  (c0)a^nb^0 +(c1)a^(n-1)b^1 +(c2)a^(n-2)b^2 +... +(ck)a^(n-k)b^k +... +(cn)a^0b^n

Then the k-th term is (ck)a^(n-k)b^k, where k is counted from 0 to n.

The value of ck can be found using Pascal's triangle, or by the formula ...

  ck = n!/(k!(n-k)!) . . . . where x! is the factorial of x, the product of all positive integers less than or equal to x.

This expansion has 11 terms, so the middle one is the one for k=5. That term will be ...

  5th term = (10!/(5!(10-5)!)(2u)^(10-5)(-4v^2)^5

  = (252)(32u^5)(-1024v^10) = -8,257,536·u^5·v^10

idk but i found this "the graph of y = f(x-2) is the graph of y = f(x) translated two units rightward.  

so, since the point (9,2) lies on the graph of y = f(x), the point (11,2) lies on the graph of y = f(x-2)."

step-by-step explanation:

link is

step-by-step explanation:

if x = 2 you have to find x to finish the problem