1024
/ \
32 32
/ \ / \
8 4 8 4
/\ /\ /\ /\
(2) 4 (2) (2)(2) 4 (2) (2)
/\ /\
(2)(2) (2)(2)
2x2x2x2x2x2x2x2x2x2 or 2^10
*(There wasn’t a little 10 to use so I just put 2^10 which is two to the tenth power)
<span>Postulates are helpful because it helps you plan and solve geometry problems.
That's my guess. It doesn't tell me in the text either.</span>
lines with undefined slope are vertical
equation are in form of x constant
through the points ( 3 , -4)
the line is x=3
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)