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Furkat [3]
2 years ago
15

Andrea ran 1.04 kilometers more on Wednesday than she ran on Thursday.

Mathematics
2 answers:
yanalaym [24]2 years ago
7 0

Answer:

thanks for the points little sis

Step-by-step explanation:

ivanzaharov [21]2 years ago
6 0

Answer:

DONT TAKE THIS OFF

Step-by-step explanation:

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masha68 [24]
20 percent
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6 0
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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
Help pleasee thank you so much in advance
Solnce55 [7]

Answer:

120

Step-by-step explanation:

180 subtract 60 equals 120

6 0
3 years ago
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vekshin1

Answer:

all work is shown and pictured

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3 years ago
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