E^(xy) = 2
(xdy/dx + y)e^(xy) = 0
At point (1, ln2), dy/dx + ln2 = 0
dy/dx = -ln2
Answer:
167.27 mg.
Step-by-step explanation:
We have been given that the half-life of Radium-226 is 1590 years and a sample contains 400 mg.
We will use half life formula to solve our given problem.
, where N(t)= Final amount after t years,
= Original amount, t/2= half life in years.
Now let us substitute our given values in half-life formula.


Therefore, the remaining amount of Radium-226 after 2000 years will be 167.27 mg.
Answer:
(a).
=
Step-by-step explanation:
A =
=
=
Bottom Like They Go on the left horizontally
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Answer:192$
Step-by-step explanation: