Answer:
Q(t) = Q_o*e^(-0.000120968*t)
Step-by-step explanation:
Given:
- The ODE of the life of Carbon-14:
Q' = -r*Q
- The initial conditions Q(0) = Q_o
- Carbon isotope reaches its half life in t = 5730 yrs
Find:
The expression for Q(t).
Solution:
- Assuming Q(t) satisfies:
Q' = -r*Q
- Separate variables:
dQ / Q = -r .dt
- Integrate both sides:
Ln(Q) = -r*t + C
- Make the relation for Q:
Q = C*e^(-r*t)
- Using initial conditions given:
Q(0) = Q_o
Q_o = C*e^(-r*0)
C = Q_o
- The relation is:
Q(t) = Q_o*e^(-r*t)
- We are also given that the half life of carbon is t = 5730 years:
Q_o / 2 = Q_o*e^(-5730*r)
-Ln(0.5) = 5730*r
r = -Ln(0.5)/5730
r = 0.000120968
- Hence, our expression for Q(t) would be:
Q(t) = Q_o*e^(-0.000120968*t)
Answer:
Using the derived property of 2 similar triangle.
x/28 = 30/25
=> x = 30x28/25 = 33.6
D is correct
D is your answer.
Hope this helps & good luck. :)
Answer:
plug in 2 , answer is -1
Step-by-step explanation:
4-5= -1
Answer:
The approximate area of the circle is <u>2462 centimeters²</u>.
Step-by-step explanation:
Given:
John and Becky measured the circle-shaped part of a sun they drew on the sidewalk. The diameter of the circle is 56 centimeters.
Now, to get the area of the circle approximately.
Diameter of the circle = 56 centimeters.
<em>So, the radius </em><em>(r)</em><em> of the circle</em> = 
Using 3.14 for π.
Now, to get the area of circle by putting formula:
<u><em>Approximately the area of the circle = 2462 centimeters².</em></u>
Therefore, the approximate area of the circle is 2462 centimeters².
