Answer:
C.4.18g
The amount will remain after 12 days is 4.18 g
Step-by-step explanation:
we can use formula
where
h is half life time
t is time in days
Po is initial quantity
P(t) is the quantity after t days
we are given
Curium-243 has a half-life of 28.5 days
In a sample of 5.6 grams of curium-243
so,
now, we can plug values
now, we can plug t=12
So,
The amount will remain after 12 days is 4.18 g
Rewrite the limand as
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = (1 - sin(<em>x</em>)) / (cos²(<em>x</em>) / sin²(<em>x</em>))
… = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / cos²(<em>x</em>)
Recall the Pythagorean identity,
sin²(<em>x</em>) + cos²(<em>x</em>) = 1
Then
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / (1 - sin²(<em>x</em>))
Factorize the denominator; it's a difference of squares, so
1 - sin²(<em>x</em>) = (1 - sin(<em>x</em>)) (1 + sin(<em>x</em>))
Cancel the common factor of 1 - sin(<em>x</em>) in the numerator and denominator:
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = sin²(<em>x</em>) / (1 + sin(<em>x</em>))
Now the limand is continuous at <em>x</em> = <em>π</em>/2, so