The answer is 76/15. The answer can also be written as 5 1/15 or 5.06
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
Answer:
x=2 and y=7
Step-by-step explanation:
Step: Solvey=2x+3for y:
y=2x+3
Step: Substitute2x+3foryiny=3x+1:
y=3x+1
2x+3=3x+1
2x+3+−3x=3x+1+−3x(Add -3x to both sides)
−x+3=1
−x+3+−3=1+−3(Add -3 to both sides)
−x=−2
−x
−1
=
−2
−1
(Divide both sides by -1)
x=2
Step: Substitute2forxiny=2x+3:
y=2x+3
y=(2)(2)+3
y=7(Simplify both sides of the equation)
Answer:
1 pound?
Step-by-step explanation:
4/4 = 1 * 1 = 1