The answer should be 31, if I’m doing this correctly.
6(3)2-5 (multiply 6 and 3, you get 18)
18(2)-5 (multiply 18 and 2, you get 36)
36-5 (subtract 36 and 5)
31
Answer: A'(-3, 7), B'(3, 6), C'(-7, -2)
Step-by-step explanation:
Reflecting over y = x maps (x,y) onto (y,x).
You could make 15 lines and cross out 7 and see what you end up with or make a rectangle and make 14 lines so there are 15 lines/boxes then shade 7 and see how many are not shaded...... the answer is 8
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)
