In order for the ball to be used in the game, it must be able to meet the minimum and maximum weight requirements. These are the limits of the weight of the ball. If it exceeds the maximum limit of 16 ounces or below the minimum limit of 14 ounces, the ball will not be approved.
So, by adding 1.5 ounces, that would mean that the initial weight of the ball did not reach the minimum limit. The initial weight of the ball, denoted as x, have two possible values. The first value is the initial weight plus added with 1.5 ounces would reach 14 ounces.
x + 1.5 = 14
x = 12.5 ounces
The other scenario is when the initial weight is added to reach the maximum requirement of 16 ounces
x + 1.5 = 16
x = 14.5 ounces
From both answer, we could conclude that the initial weight has to be 12.5 ounces. If the initial weight were 14.5 ounces to begin with, there should be no need for air. It cans till be approved to be used.
Answer: 18 c
Step-by-step explanation: since they have the same variables you can easily just add them together
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:
3rd one, 4th one, 1st one, 2nd one
Step-by-step explanation:
1st one - 2.6
2nd one - (-3.125
)
3rd one - 10.35
4th one - -2.95
