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
2,2.06,2.7
Look at this
This is the reason
2, 2.06, 2.70
What figure?? Please elaborate
Answer:
y=-2x+4
Explanation:
Subtract 6x from both sides of the equation.
3y=12−6x
Divide each term by 3 and simplify.
Divide each term in 3y=12−6x by 3.
Cancel the common factor of 3.
Simplify