Answer:
The fourth term of the expansion is -220 * x^9 * y^3
Step-by-step explanation:
Question:
Find the fourth term in (x-y)^12
Solution:
Notation: "n choose k", or combination of k objects from n objects,
C(n,k) = n! / ( k! (n-k)! )
For example, C(12,4) = 12! / (4! 8!) = 495
Using the binomial expansion formula
(a+b)^n
= C(n,0)a^n + C(n,1)a^(n-1)b + C(n,2)a^(n-2)b^2 + C(n,3)a^(n-3)b^3 + C(n,4)a^(n-4)b^4 +....+C(n,n)b^n
For (x-y)^12, n=12, k=3, a=x, b=-y, and the fourth term is
C(n,3)a^(n-3)b^3
=C(12,3) * x^(12-3) * (-y)^(3)
= 220*x^9*(-y)^3
= -220 * x^9 * y^3
Answer:
Step-by-step explanation:
<u>Let the numbers be x, y and z</u>
- x + y + z = 24
- 2y = z + 2
- z = x + y
<u>Solving by substitution:</u>
- x + y + z = z + z = 24
- 2z = 24
- z = 12
- 2y = z + 2
- 2y = 12 + 2
- 2y = 14
- y = 7
- z = x + y
- x = z - y
- x = 12 - 7 = 5
<u>The answer:</u>
B the person above me is right
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Answer:
Step-by-step explanation:
A man steps out of a plane at 4,000m of height above the ground.The point at which he jumps out of the plane would make a good reference point. However, if his acceleration is going to change as a result of him opening his parachute 2000m above the ground, a good reference point would be there. Keep in mind though, that his velocity at that instant would need to be known for it to be useful- otherwise the airplane reference point would be just as good with appropriate modeling....