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pogonyaev
3 years ago
10

The floor of the school art closet is being tiled. The closet is shaped like a parallelogram. The parallelogram has a base of 17

.5 feet and the height of 6 feet. Each package of tile will cover 5 ft.² how many packages of tile will be needed to tile the music room floor?
Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
5 0
First we need to find the total area of the floor. Dividing the total area of floor by area of one package of tile will give us the number package of tiles that are needed to cover the floor.

The shape of floor is parallelogram with base 17.5 feet and height of 6 feet.
Area of parallelogram is the product of its base and height.

So,

Floor Area = 17.5 x 6 = 105 ft²

Area covered by 1 package of tile = 5 ft²

So, number of package of tiles that are needed = 105/5 = 21 package of tiles.

21 packages of tiles will be needed to fully tile the room.
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Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot
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Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

\frac{dm}{dt} = -\frac{t}{\tau}

Where:

m - Current mass of the isotope, measured in kilograms.

t - Time, measured in days.

\tau - Time constant, measured in days.

The solution of the differential equation is:

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

Where m_{o} is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}

\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }

The half-life of a isotope (t_{1/2}) as a function of time constant is:

t_{1/2} = \tau \cdot \ln2

t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2

The half-life difference between isotope B and isotope A is:

\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|

If \frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9, t_{A} = 33\,days and t_{B} = 43\,days, the difference in the half-lives of the isotopes is:

\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|

\Delta t_{1/2} \approx 65.788\,days

The approximate difference in the half-lives of the isotopes is 66 days.

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