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Goshia [24]
2 years ago
13

Which of the following are solutions to the equation below?

Mathematics
1 answer:
DENIUS [597]2 years ago
8 0
A and C could be the answer
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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}]

6 0
3 years ago
Ok heelp ummmm........
Elan Coil [88]

Answer:

c

Step-by-step explanation:

7 0
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Answer:

The answer is 9/10

Step-by-step explanation:

it's 9/10

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Find the ending balance $35,600 at 9% for 2 years
tino4ka555 [31]

Answer:

42,008

Step-by-step explanation:

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Answer please if its good then i'll give brainliest
stepladder [879]
There is no question? ...
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