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muminat
2 years ago
6

I need it rn plzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

Mathematics
1 answer:
Triss [41]2 years ago
4 0

9514 1404 393

Answer:

  ∠B = ∠C = 138°

  ∠D = 42°

Step-by-step explanation:

Angles between the parallel bases are supplementary:

  ∠B = 180° -∠A = 180° -42° = 138°

Right-side angles are the same measure as their mirrors across the line of symmetry.

  ∠C = ∠B = 138°

  ∠D = ∠A = 42°

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There you go

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The population of a certain town was 10,000 in 1990. The rate of change of the population, measured in people per year, is model
Katena32 [7]

The question is incomplete. The complete question is :

The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?

Solution :

According to the question,

The rate of change of population is given as :

$\frac{dP(t)}{dt}=200e^{0.02t}$  in 1990.

Now integrating,

$\int_0^{20}\frac{dP(t)}{dt}dt=\int_0^{20}200e^{0.02t} \ dt$

                    $=\frac{200}{0.02}\left[e^{0.02(20)}-1\right]$

                   $=10,000[e^{0.4}-1]$

                    $=10,000[0.49]$

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$\frac{dP(t)}{dt}=200e^{0.02t}$

$\int1.dP(t)=200e^{0.02t}dt$

$P=\frac{200}{0.02}e^{0.02t}$

$P=10,000e^{0.02t}$

$P=P_0e^{kt}$

This is initial population.

k is change in population.

So in 1995,

$P=P_0e^{kt}$

   $=10,000e^{0.02(5)}$

   $=11051$

In 2000,

$P=10,000e^{0.02(10)}$

   =12,214

Therefore, the change in the population between 1995 and 2000 = 1,163.

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3 years ago
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1 + 3i, - sqrt 10 would be the other two answers
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Answer:

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Step-by-step explanation:


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Answer:

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Step-by-step explanation:

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