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.
Y - y1 = m (x + x1)
Solve for m by subtracting the y's and dividing them by the difference of the 2 x's.
-32 - 1 = -33
-8 - 3 = -11
Divide the two to get 3.
Use the first point (as instructed) and plug it into the equation.
y - (-32) = 3 (x - (-8))
y + [32] = [3] (x + [5])
The brackets are the fill in the blanks.
the blank is 162
since 81 and 81 are both perfect squares the middle term (blank) will be the product of their roots times 2
√81=9
√81 =9
9 x 9 x 2
= 162
Answer:
2/22， 3/33， 4/44
Step-by-step explanation:
Answer:
10.12 
Step-by-step explanation:
Given:
- 0.07 cm thick = 0.007m and it is dr (the change in radius)
As we know that, the volume of a hemispherical dome with diameter 48 m is:
- V =
π
Take the derivative of V we have:
= 2π
<=> dV = 2π*dr
<=> dV = 2π*
*0.0007 = 10.12
Hope it will find you well.
