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NemiM [27]
3 years ago
13

The length of a rectangular field is represented by the expression14x-3x^2+2y. The width of the field is represented by the expr

ession 5x-7x^2+7y. How much greater is the length of the field than the width?
A) 9x+4x^2-5y
B)9x-10x^2-5y
C)19x+4x^2+9y
D) 19x-10x^2+9y
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
4 0
The length of the rectangular field is 14x-3x^2+2y, and the width is 5x-7x^2+7y.

To find <span>how much greater is the length of the field than the width we need to subtract the width from the length, so we have:

</span>(14x-3x^2+2y)-(5x-7x^2+7y)=(14x-3x^2+2y)-5x+7x^2-7y.<span>

Operating with the equal degree and variable terms, this difference is equal to

</span>(14x-5x)+(7x^2-3x^2)+(2y-7y)=9x+4x^2-5y
<span>

Answer: A

</span>
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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?

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According to the question,

The rate of change of population is given as :

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

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

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

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

$P=P_0e^{kt}$

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So in 1995,

$P=P_0e^{kt}$

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In 2000,

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

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

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