<span>The best way to solve each equation is:
</span> 1) 5x2 + 12x - 3 = 0 -----> solve by quadratic formula
2) 4x2 - 25 = 0 -----------> solve by square root method
3) x2 - 5x + 6 = 0 --------> solve by factoring
4) x2 - 4x = 8 -------------> solve by completing the square
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:
-39
Step-by-step explanation:
![\left(-3\right)^3-\sqrt[3]{27}-4^3\left(\sqrt[3]{64}-\left(2\right)\left(2\right)\right)-\frac{\left(3\right)^3\left(2\right)}{6}](https://tex.z-dn.net/?f=%5Cleft%28-3%5Cright%29%5E3-%5Csqrt%5B3%5D%7B27%7D-4%5E3%5Cleft%28%5Csqrt%5B3%5D%7B64%7D-%5Cleft%282%5Cright%29%5Cleft%282%5Cright%29%5Cright%29-%5Cfrac%7B%5Cleft%283%5Cright%29%5E3%5Cleft%282%5Cright%29%7D%7B6%7D)
Which gives us -39.
