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aliina [53]
3 years ago
15

Cosmic ray bombardment of the atmosphere produces neutrons, which in turn react with nitrogen to produce radioactive carbon-14 p

oint Radioactive carbon-14 enters all living tissue through carbon dioxide (via plants). As long as a plant or animal is alive, carbon-14 is maintained in the organism at a constant level. Once the organism dies, however, carbon-14 decays exponentially into carbon-12 point By comparing the amount of carbon-14 to the amount of carbon-12 comma one can determine approximately how long ago the organism died. (Willard Libby won a Nobel Prize for developing this technique for use in dating archaeological specimens). The half-life of carbon-14 is about 5730 years. In answering the following questions, assume that the initial quantity of carbon-14 is 500 milligrams. a. Construct an exponential function that describles the relationship between Upper A comma the amount of carbon-14 in milligrams, and t comma the number of 5730-year time periods.
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
8 0
I think I have the answer let me just double check my work okay!
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Find the value of a $3,000 certificate in 2 years, if the interest rate is 10% compounded annually.
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Answer:

600$

Step-by-step explanation:

3000×2×.10=600

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Estimate the quotient what is 467.6 divided by 8
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Answer:

<h2>        64 units³</h2>

Step-by-step explanation:

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the value of a collectible coin can be represented by the equation y=2x+9.74 where x represents the number of years that consuel
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Answer:

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Step-by-step explanation:

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4 0
3 years ago
A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria
Kaylis [27]

Answer:

A) The expression for the number of bacteria is P(t) = 400e^{0.7783t}.

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

P' = kP

where P stands for the population of bacteria.

Writing P' as \frac{dP}{dt}, we get

\frac{dP}{dt} = kP.

Notice that this is a separable equation, so

\frac{dP}{P} = kdt.

Then, integrating in both sides of the equality:

\int\frac{dP}{P} = \int kdt.

We have,

\ln P = kt+C.

Now, taking exponential

P(t) = Ce^{kt}.

The next step is to find the value for the constant C. We do this using the initial condition P(0)=400. Recall that this is the initial population of bacteria. So,

400 = P(0) = Ce^{k0}=C.

Hence, the expression becomes

P(t) = 400e^{kt}.

Now, we find the value for k. We are going to use that P(4)=9000. Notice that

9000 = 400e^{k4}.

Then,

\frac{90}{4} = e^{4k}.

Taking logarithm

\ln\frac{90}{4} = 4k, so \frac{1}{4}\ln\frac{90}{4} = k.

So, k=0.7783788273, and approximating to the fourth decimal place we can take k=0.7783. Hence,

P(t) = 400e^{0.7783t}.

B) To find the number of bacteria after 5 hours, we only need to evaluate the expression we have obtained in the previous exercise:

P(5) =400e^{0.7783*5} = 19593.723 \approx 19593.  

C) In this case we want to do the reverse operation: we want to find the value of t such that

30000 = 400e^{0.7783t}.

This expression is equivalent to

75 = e^{0.7783t}.

Now, taking logarithm we have

\ln 75 = 0.7783t.

Finally,

t = \frac{\ln 75}{0.7783} \approx 5.55.

So, after 5.55 hours the population of bacteria will reach 30000.

6 0
3 years ago
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