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astraxan [27]
3 years ago
14

Part 3 please help ​

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0

I'm taking "part 3" to mean the 3rd of the posted questions.

If f(x)=(x^3+2x)\ln x, then

f'(x)=(x^3+2x)'\ln x+(x^3+2x)(\ln x)'=(3x^2+2)\ln x+\dfrac{x^3+2x}x

\implies f'(x)=(3x^2+2)\ln x+x^2+2

Then when x=1, we get f'(1)=3.

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Which of the following describes the sequence 1,1,2,3,5
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Answer:

Step-by-step explanation:

It starts at 1 adds 1 which got you 2 , then 1 plus 2, is 3, then 2 plus 3 is 5, you get it?

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An important tool in archeological research is radiocarbon dating, developed by the American chemist Willard F. Libby.3 This is
Radda [10]

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)                                    

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