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3 years ago

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information about two homes in a city is given below. Home A has a value of $110,000, which is expected to increase exponentiall

y by 4% each year. Home B has a value of $100,000, which is expected to increase exponentially by 6% each year. Which equation can be used to determine t, the number of years in which the expected value of home B would be twice the expected value of home A? A 110,000(1.04)t=2(100,000)(1.06)t B 100,000(1.06)t=2(110,000)(1.04)t C 110,000(0.04)t=2(100,000)(0.06)t D 100,000(0.06)t=2(110,000)(0.04)t

1 answer:

SOVA2 [1]3 years ago

6 0

Answer:

100000(1.04)^t=2[110000(1.04)^t

Step-by-step explanation:

B=2A

100000(1.04)^t=2[110000(1.04)^t

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3 years ago

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

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Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

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Differentiate the equation with respect to time t, such that

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To differentiate the product,

Let r² = u, so that

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Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

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Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

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$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

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$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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