Expand the binomial (2x^2+y^2)^4

the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a

Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$ (Eq. 1)

Where:

$\frac{dm}{dt}$ - First derivative of mass in time, measured in miligrams per year.

$\tau$ - Time constant, measured in years.

$m$ - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

$\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt$

$\ln m = -\frac{1}{\tau} + C$

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (Eq. 2)

Where:

$m_{o}$ - Initial mass of isotope, measured in miligrams.

$t$ - Time, measured in years.

And time is cleared within the equation:

$t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]$

Then, time constant can be found as a function of half-life:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

If we know that $t_{1/2} = 5730\,yr$ and $\frac{m(t)}{m_{o}} = 0.35$ , then:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.35$

$t \approx 8678.505\,yr$

The wood was cut approximately 8679 years ago.

5 0

3 years ago

What is the answer to this?

myrzilka [38]

Answer:r=3cm,diameter =6cm and C=18.84cm

Step-by-step explanation:

To find the radius,diameter and circumference according to the diagram drawn above

r=3cm

Diameter=2×3cm

d=6cm

To find the circumference

C=2πr

C=2×3.14×3cm

C=18.84cm

4 0

3 years ago

2. Solve each equation. Check your solutions.<br> b. 5 +<br> - 16

Ber [7]

Answer:

Step-by-step explanation:

-11

4 0

3 years ago

Your English teacher has decided to randomly assign poems for the class to read. The syllabus includes four poems by Shakespeare

Tresset [83]

The probability of event A and B to both occur is denoted as P(A ∩ B) = P(A) P(B|A). It is the probability that Event A occurs times the probability that Event B occurs, given that Event A has occurred.

So, to find the probability that you will be assigned a poem by Shakespeare and by Tennyson, let Event A = the event that a Shakespeare poem will be assigned to you; and let Event B = the event that the second poem that will be assigned to you will be by Tennyson.

At first, there are a total of 13 poems that would be randomly assigned in your class. There are 4 poems by Shakespeare, thus P(A) is 4/13.
After the first selection, there would be 13 poems left. Therefore, P(B|A) = 2/12
Based on the rule of multiplication,
P(A ∩ B) = P(A) P(B|A)P(A ∩ B) = 4/13 * 2/12
P(A ∩ B) = 8/156
P(A ∩ B) = 2/39

The probability that you will be assigned a poem by Shakespeare, then a poem by Tennyson is 2/39 or 5.13%.

6 0

3 years ago

The five-number summary for the number of stylists employed by each hair salon in Jamesville is shown in the following table:

asambeis [7]

Answer:

I think its 4

Step-by-step explanation:

don't really know how to but its simple really

6 0

3 years ago