Answer:
The independent quantity in the situation is the length of the diameter.
Step-by-step explanation:
Consider a relationship between two variables.
Of the two variables one variable is dependent upon the other.
Dependent variables are those variables that are under study, i.e. they are being observed for any changes when the other variable values are changed.
Independent variables are the variables that are being altered to see a proportionate change in the dependent variable.
In this case, it is provided that Grissom knows there is a relationship between the volume of the sphere and the length of its diameter.
With every sphere that Grissom draws, the volume of the sphere changes according to its diameter length.
That is the volume of the sphere depends upon the length of its diameter.
Thus, the independent quantity in the situation is the length of the diameter.
Answer:
b) 16%
Step-by-step explanation:
add 36 to 48 (84) then subtract that from 100. 16% played both games
28.27 is the area bc i had this as an example before lol
Given:
Replace f(x) by f(x - h).
To find:
The effect on the graph of replacing f(x) by f(x - h).
Solution:
Horizontal shift is defined as:
If the graph f(x) shifts h units left, then f(x+h).
If the graph f(x) shifts h units right, then f(x-h).
Where, h is a constant that represents the horizontal shift.
In the given problem f(x) is replaced by f(x - h) and we need to find the effect on the graph.
Here, we have x-h in place of x.
Therefore, the graph of f(x) shifts h units right to get the graph of f(x-h).
We have the equation:
![x^2+y^2-8x-6y+24=0](https://tex.z-dn.net/?f=x%5E2%2By%5E2-8x-6y%2B24%3D0)
By arranging this equation in terms of x and y, we have:
![x^2-8x+y^2-6y=-24 \\ \\](https://tex.z-dn.net/?f=x%5E2-8x%2By%5E2-6y%3D-24%20%5C%5C%20%5C%5C)
By using the method of completing the square, we have:
![x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}](https://tex.z-dn.net/?f=x%5E2-8x%2B%5Cmathbf%7B%5Cleft%28%5Cfrac%7B8%7D%7B2%7D%5Cright%29%5E2%7D%2By%5E2-6y%2B%5Cmathbf%7B%5Cleft%28%5Cfrac%7B6%7D%7B2%7D%5Cright%29%5E2%7D%3D-24%2B%5Cmathbf%7B%5Cleft%28%5Cfrac%7B8%7D%7B2%7D%5Cright%29%5E2%7D%2B%5Cmathbf%7B%5Cleft%28%5Cfrac%7B6%7D%7B2%7D%5Cright%29%5E2%7D%20%5C%5C%20%5C%5C%20x%5E2-8x%2B%5Cmathbf%7B16%7D%2By%5E2-6y%2B%5Cmathbf%7B9%7D%3D-24%2B%5Cmathbf%7B16%7D%2B%5Cmathbf%7B9%7D%20%5C%5C%20%5C%5C%20%5Cboxed%7B%28x-4%29%5E2%2B%28x-3%29%5E2%3D1%7D)
The center of this circle is:
![(h,k)=(4,3)](https://tex.z-dn.net/?f=%28h%2Ck%29%3D%284%2C3%29)
So the equation that fulfills the statement is:
![(x-4)^2+(y-3)^2=2^2](https://tex.z-dn.net/?f=%28x-4%29%5E2%2B%28y-3%29%5E2%3D2%5E2)
Finally, the right answer is c)