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2 years ago

Monu

Mathematics

2 answers:

aalyn [17]2 years ago

8 0

Length=L=8m
Base=B=4m
Height=h=5m

Now

$\\ \sf\longmapsto Volume=V=LBH$

$\\ \sf\longmapsto V=8(4)(5)$

$\\ \sf\longmapsto V=32(5)$

$\\ \sf\longmapsto V=160m^3$

Mama L [17]2 years ago

3 0

Step-by-step explanation:

The volume of rectangular prism:

V = lwh

Given:

l = 8 m, w = 4 m, h = 5 m

The volume is:

V = 8*4*5 = 160 m³

Correct choices are:

V = 32 × 5
V= 8 × 4 × 5

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Evgen [1.6K]

Answer:

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6 0

3 years ago

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:

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

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$m_{o}$ - 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:

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

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The wood was cut approximately 8679 years ago.

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3 years ago

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xxMikexx [17]

Answer:

Step-by-step explanation:

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Brrunno [24]

Answer:

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3 years ago

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MissTica

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3 years ago