Answer:
225.78 grams
Step-by-step explanation:
To solve this question, we would be using the formula
P(t) = Po × 2^t/n
Where P(t) = Remaining amount after r hours
Po = Initial amount
t = Time
In the question,
Where P(t) = Remaining amount after r hours = unknown
Po = Initial amount = 537
t = Time = 10 days
P(t) = 537 × 2^(10/)
P(t) = 225.78 grams
Therefore, the amount of iodine-131 left after 10 days = 225.78 grams
The answer would be 32,200 50*28*23
(x - 4)²
(x - 4)(x - 4)
x² - 4x - 4x + 16
x² - 8x +16
-3 and 3, both three away from 0
If
then the derivative is
Critical points occur where 
. This happens for
In the first case, we find
In the second,
So all the critical points occur at multiples of 
, or 
. (This includes all the even multiples of .)
