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Nutka1998 [239]
3 years ago
7

What times what is 448

Mathematics
1 answer:
Andrew [12]3 years ago
5 0
The factors of 448 include: <span>1,2,4,7,8,14,16,28,32,56,64,112,224,448</span>
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The second choice is the answer: the initial value is the y value (which is 4) and the rate of change is the slope. The slope is calculated with rise/run, so the rise in this situation is that it moves down (negative) by 3 (rise=-3) and the run is how much it moves to the right (run=4). Therefore, the rate of change is rise/run which is -3/4
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2 years ago
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,

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

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

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

                    $=10,000[0.49]$

                    =4900

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

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

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

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

$P=P_0e^{kt}$

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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Step-by-step explanation:

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Step-by-step explanation:

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