Answer:
Given that,
The number of grams A of a certain radioactive substance present at time, in years
from the present, t is given by the formula

a) To find the initial amount of this substance
At t=0, we get


We know that e^0=1 ( anything to the power zero is 1)
we get,

The initial amount of the substance is 45 grams
b)To find thehalf-life of this substance
To find t when the substance becames half the amount.
A=45/2
Substitute this we get,


Taking natural logarithm on both sides we get,







Half-life of this substance is 154.02
c) To find the amount of substance will be present around in 2500 years
Put t=2500
we get,




The amount of substance will be present around in 2500 years is 0.000585 grams
I need the answer because i need to ask a question sry!
Answer:
-9A · √(5yA)
Step-by-step explanation:
The coefficient -3 stays the same.
45 factors into 5·9, which is helpful because 9 is a perfect square.
Thus, √45 = 3√5.
y cannot be factored. It stays under the radical.
A³ can be factored into A² (a perfect square) and A.
Thus,
-3√(45yA³) = -3 · 3√5 · √y · A · √A, or
= (-3)(3)(A) · √(5yA), or
= -9A · √(5yA)
An equavalent ratio of 12\24 is 3\4
1.<u>3 </u>x 1= 1.<u>3 </u>hope this helped