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2 years ago

5

Which of the following is the graph of

1 answer:

VMariaS [17]2 years ago

3 0

Answer: C

Step-by-step explanation:

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An above-ground swimming pool in the shape of a cylinder has a diameter of 18 feet and a height of 4.5 feet. If the pool is fill

laila [671]

The volume of a cylinder is

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Radius = 1/2 diameter.
Radius of this pool = (1/2) (18 ft) = 9 ft

The pool is a cylinder with height of 4.5 feet.
The water in it is also a cylinder, but only 4 ft high.

Volume of the water =

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   = (pi) x (9 ft)² x (4 ft)

   = (pi) x (81 ft²) x (4 ft)

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VladimirAG [237]

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3 years ago

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2 years ago

Demand The demand function for a product is given by

stira [4]

Answer:

$x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}$

a) $x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027$

b) $x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294$

Step-by-step explanation:

For this case we have the following function:

$P= 8000 (1- \frac{5}{5 +e^{-0.002 x}})$

We can solve for x like this. First we can reorder the expression like this:

$\frac{P}{8000} = 1- \frac{5}{5+e^{-0.002x}}$

$\frac{5}{5+e^{-0.002x}} = 1 -\frac{P}{8000} = \frac{8000-P}{8000}$

$\frac{40000}{8000-P} = 5 + e^{-0.002x}$

Now we can apply natura log on both sids and we got:

$ln[\frac{40000}{8000-P} -5] = ln e^{-0.002x}$

$ln [\frac{5P}{8000-P}] = -0.002x$

And if we solve for x we got:

$x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}$

Part a

For this case we can replace P = 200 and see what we got for x like this:

$x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027$

Part b

For this case we can replace P = 800 and see what we got for x like this:

$x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294$

4 0

3 years ago