Answer:
not a function - it does not pass the vertical line test.
Step-by-step explanation:
Answer:
The answer is 1/2 inches
Step-by-step explanation:
Using the given center and point at the circumference, the equation of the circles are:
- a)

- b)

- c)

<h3>
What is the equation of a circle?</h3>
- The equation of a circle of radius r and center
is given by:

Item a:
- Center C(5,2), hence
.
The point at the circumference is A(11,10), hence:



Hence:

Item b:
- Center C(-2,-5), hence
.
The point at the circumference is A(3,-17), hence:



Hence:

Item c:
- Center C(5,-1), hence
.
The point at the circumference is A(-2,-5), hence:



Hence:

You can learn more about equation of a circle at brainly.com/question/24307696
Find the LCD (least common denominator)

it would be 12 (6x2=12 and 4x3=12)
so you multiply

by 2 and get

then multiply

by 3 and get

you get 3 WHOLE servings from

cups of icecream
x-coordinates for the maximum points in any function f(x) by f'(x) =0 would be x = π/2 and x= 3π/2.
<h3>How to obtain the maximum value of a function?</h3>
To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
we want to find x-coordinates for the maximum points in any function f(x) by f'(x) =0
Given f(x)= 4cos(2x -π)

In general 
from x = 0 to x = 2π :
when k =0 then x = π/2
when k =1 then x= π
when k =2 then x= 3π/2
when k =3 then x=2π
Thus, X-coordinates of maximum points are x = π/2 and x= 3π/2
Learn more about maximum of a function here:
brainly.com/question/13333267
#SPJ4