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3 years ago

Use a graphic calculator to determine the intervals on which (x)=2x^3-3x-1 is increasing or decreasing.

Mathematics

1 answer:

WITCHER [35]3 years ago

6 0

Answer:

im not sure of the answer but hopefully this will help

Step-by-step explanation:

basically the intervals in which its increasing and decreasing depend on the leading coefficient and the vertex

since this is opening up, the intervals will be

the function is increasing (x coord of vertex, infinity)

the function is decreasing (-infinity, x-coord of vertex)

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natulia [17]

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3 years ago

Find the volume of the solid generated when R (shaded region) is revolved about the given line. x=6−3sec y, x=6, y= π 3, and

Dmitrij [34]

Answer:

$V=9\pi\sqrt{3}$

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

$V=\int\limits^b_a {\pi r^{2}} \, dy$

where:

$r=6-(6-3 sec(y))$

$r=3 sec(y)$

a=0

$b=\frac{\pi}{3}$

so the volume becomes:

$V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy$

This can be simplified to:

$V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy$

and the integral can be rewritten like this:

$V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy$

which is a standard integral so we solve it to:

$V=9\pi[tan y]\limits^\frac{\pi}{3}_0$

so we get:

$V=9\pi[tan \frac{\pi}{3} - tan 0]$

which yields:

$V=9\pi\sqrt{3}$ ]

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3 years ago

Help!!!! 20 points!!!!!!’

Gnoma [55]

Answer:

option c is the correct answer

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3 years ago

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Colt1911 [192]

Greater
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Answer:

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Step-by-step explanation:

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