The derivative of f(x) at x=3 is 2x=6 approaching from the left side (apply power rule to y=x^2). The derivative of f(x) at x=3 is m approaching from the right side. In order for the function to be differentiable, the limit of derivative at x=3 must be the same approaching from both sides, so m=6. Then, x^2=mx+b at x=3, plug in m=6, 9=18+b, so b=-9.
Can u send me a picture of the question instead please because i dont understand what ur trying to ask..
If y is directly proportional to x then y:x.
When y = 30, x = 6 then 30/6 = 5:1. Therefor for every 1x there is 5y vice versa. When x = 12, then y = 5 * 12 = 60.
Therefore y = 60 hence 60:12
The correct is going to be “Identify the Choices”.
Hope this helps! :))
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)
