I might be wrong but it should be 1
Answer:
AB=3.16
BC=6.32
CD=3.16
AD6.32
Step-by-step explanation:
Answer:
See below.
Step-by-step explanation:
Here's an example to illustrate the method:
f(x) = 3x^2 - 6x + 10
First divide the first 2 terms by the coefficient of x^2 , which is 3:
= 3(x^2 - 2x) + 10
Now divide the -2 ( in -2x) by 2 and write the x^2 - 2x in the form
(x - b/2)^2 - b/2)^2 (where b = 2) , which will be equal to x^2 - 2x in a different form.
= 3[ (x - 1)^2 - 1^2 ] + 10 (Note: we have to subtract the 1^2 because (x - 1)^2 = x^2 - 2x + 1^2 and we have to make it equal to x^2 - 2x)
= 3 [(x - 1)^2 -1 ] + 10
= 3(x - 1)^2 - 3 + 10
= <u>3(x - 1)^2 + 7 </u><------- Vertex form.
In general form the vertex form of:
ax^2 + bx + c = a [(x - b/2a)^2 - (b/2a)^2] + c .
This is not easy to commit to memory so I suggest the best way to do these conversions is to remember the general method.
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:
12/5
Step-by-step explanation:
