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Readme [11.4K]
3 years ago
12

Help asap!!! compute $\left(-3\sqrt{128}\right)\left(-4\sqrt{50}\right)$

Mathematics
1 answer:
labwork [276]3 years ago
5 0

(-3\sqrt{128})(-4\sqrt{50})\qquad \begin{cases} 128=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\\ \qquad 2^6\cdot 2\\ \qquad (2^3)^2\cdot 2\\ 50=5\cdot 5\cdot 2\\ \qquad 5^2\cdot 2 \end{cases} \\\\\\ (-3\sqrt{(2^3)^2\cdot 2})(-4\sqrt{5^2\cdot 2})\implies 12\sqrt{(2^3)^2\cdot 2}\cdot \sqrt{5^2\cdot 2} \\\\\\ 12\cdot 2^3\sqrt{2}\cdot 5\sqrt{2}\implies 12\cdot 2^3\cdot 5\cdot \sqrt{2^2}\implies 12\cdot 2^3\cdot 5\cdot 2\implies 960

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An important tool in archeological research is radiocarbon dating, developed by the American chemist Willard F. Libby.3 This is
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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

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The expression for Q(t).

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- 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)                                    

7 0
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