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
That means there would be 28 in California and still 12 in Nevada or if California takes from Nevada it would be California 28 and Nevada 2
Answer:
The answers are, "John saves $35 dollars the first week" and "John is saving at a faster rate than Barb"
Answer:
Step-by-step explanation:
volume of box=6×6×7 in³
he fits gold 2×2×2
so he fits it in 6×6×6
so number of gold boxes=(6×6×6)/(2×2×2)=27
remaining space=1×6×6
number of silver boxes=(1×6×6)/(1×1×1)=36
total boxes he took=27+36=63
Answer:
34
Step-by-step explanation:
First we need to do 2/3*21, which equals 14 as 3 and 21 can be simplified to 1 and 7. 7 * 2/1= 14. Then we add the 20 to 14, 20+14= 34