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Stells [14]
3 years ago
13

17-3(3-4x)-8=5(x+9)-6

Mathematics
1 answer:
VashaNatasha [74]3 years ago
6 0
X=59/12 or x= 4 11/12
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Someone help me figure this out
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-6, because you can see a pattern in which the x coordinate of each pair is reversed to the opposite value, from positive to negative.

5 0
3 years ago
Read 2 more answers
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.

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3 years ago
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6 is the critical point
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Anthony bought an 8-foot board. He cut off 3/4 of the board to build a shelf and gave 1/3 of the rest to his brother for an art
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I’m pretty sure the answer is 8 inches, mark as brainleist pls :)
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2 years ago
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