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3 years ago

Find the area of a triangle with sides lengths 17, 18, and 21.4 units. Round the answer to the nearest hundredth.

1 answer:

8 0

To calculate for the area of the triangle given all its side, use the Heron's formula which is,

A = sqrt ((s)(s - a)(s - b)(s - c))

where A is the area. a, b, and c are the measure of the sides and s the semi - perimeter or half of the perimeter.

With the given above, s = (17 + 18 + 21.4) / 2 = 28.2

Substituting the known dimensions to the Heron's formula,

A = sqrt ((28.2)(28.2 - 17)(28.2 - 18)(28.2 - 21.4)) = 148 units^3

Thus, the area of the triangle is 148 units^3.

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During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time

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The population is of 500 after 10.22 hours.

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The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

$\frac{dP}{dt} = Pr$

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

$\frac{dP}{P} = r dt$

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Applying the exponential to both sides:

$P(t) = Ke^{rt}$

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At time t = 0 hours, the population is 300.

This means that K = 300. So

$P(t) = 300e^{rt}$

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

$P(t) = 300e^{rt}$

$1000 = 300e^{24r}$

$e^{24r} = \frac{1000}{300}$

$e^{24r} = \frac{10}{3}$

$\ln{e^{24r}} = \ln{\frac{10}{3}}$

$24r = \ln{\frac{10}{3}}$

$r = \frac{\ln{\frac{10}{3}}}{24}$

$r = 0.05$

$P(t) = 300e^{0.05t}$

At what time t is the population 500?

This is t for which P(t) = 500. So

$P(t) = 300e^{0.05t}$

$500 = 300e^{0.05t}$

$e^{0.05t} = \frac{500}{300}$

$e^{0.05t} = \frac{5}{3}$

$\ln{e^{0.05t}} = \ln{\frac{5}{3}}$

$0.05t = \ln{\frac{5}{3}}$

$t = \frac{\ln{\frac{5}{3}}}{0.05}$

$t = 10.22$

The population is of 500 after 10.22 hours.

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