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Kaylis [27]
2 years ago
13

The triangles are similar.

Mathematics
2 answers:
erik [133]2 years ago
6 0
5×4=20 , 12×4=48
53÷4 = 13 then
x=13
pishuonlain [190]2 years ago
5 0

Answer : The value of x is, 13

Step-by-step explanation :

As we know that when two object have same shape then they are similar. That means, when two triangles are similar then their ratios of lengths of their corresponding sides are equal.

Let the name of the big triangle be, PQR and small triangle be, XYZ.

Given:

PQR is similar to triangle XYZ. That means,

\frac{PR}{PQ}=\frac{XZ}{XY}

\frac{52}{48}=\frac{x}{12}

x = 13

Therefore, the value of x is, 13

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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
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Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

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