<u>Answer-</u>
At
the curve has maximum curvature.
<u>Solution-</u>
The formula for curvature =

Here,

Then,

Putting the values,

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

Now, equating this to 0






Solving this eq,
we get 
∴ At
the curvature is maximum.
So you need getting the coordinates of center of the circle x and y
- than you check this graph so you get x= -3 and y=-1
Answer: -16
Step-by-step explanation:
-12-4=-16
Making -16+4=-12
There’s an app called photo math where u can take a picture of the problem and it gives u the answer.
hope that helps lol
GCF 32 24 40
^ ^ ^
2 16 2 12 2 20
^ ^ ^
2 8 2 6 2 10
^ ^ ^
2 4 2 3 2 5
^
2 2
(2) (2 & 3) (2 & 5)