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Burka [1]
2 years ago
13

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. Only 1 try

remaining for this problem or else I get a zero. Please help if you know the answer or how to solve.

Mathematics
1 answer:
vagabundo [1.1K]2 years ago
6 0

Using the shell method, the volume is

\displaystyle 2\pi \int_0^1 (2-x) \cdot 8x^3 \, dx = 16\pi \int_0^1 (2x^3 - x^4) \, dx

Each cylindrical shell has radius 2-x (the horizontal distance from the axis of revolution to the curve y=8x^3); has height 8x^3 (the vertical distance between a point on the x-axis in 0\le x\le1 and the curve y=8x^3).

Compute the integral.

\displaystyle 16 \pi \int_0^1 (2x^3 - x^4) \, dx = 16\pi \left(\frac{x^4}2 - \frac{x^5}5\right) \bigg|_{x=0}^{x=1} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 16\pi \left(\frac12 - \frac15\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{24}5\pi = \boxed{4.8\pi}

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Amiraneli [1.4K]
Area=length*height
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How do you work this out?
Katarina [22]
Because ABCD is a rectangle, the length of CD is 12 cm.

We need to determine the length of DE.  If we can do that, then the sum of the lengths of CD and DE represents the unknown:  the length of CE.

To find the length of CE, we have to "solve" the upper triangle.

Here's an outline of what to do:

1.  Show that BC=AD and find the length.
2.  Note that angle CAD is 60 degrees.  Why?
3.  Note that angle EAD is 30 degrees.  Why?
4.  Find the length of ED
5.  Add ED and DC, that is, ED + 12 cm.  This is your answer.

Please ask questions if need be.
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In a standard normal curve, what percentile corresponds to a z-score of 2.0?
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3 years ago
2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books
insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

  • When the books are arranged in any order, the number of arrangements is 3628800
  • When the mathematics book must not be together, the number of arrangements is 2903040
  • When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
  • When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

\mathbf{Novels = 3}

\mathbf{Mathematics = 2}

\mathbf{Chemistry = 5}

<u />

<u>(a) The books in any order</u>

First, we calculate the total number of books

\mathbf{n = Novels + Mathematics + Chemistry}

\mathbf{n = 3 + 2 +  5}

\mathbf{n = 10}

The number of arrangement is n!:

So, we have:

\mathbf{n! = 10!}

\mathbf{n! = 3628800}

<u>(b) The mathematics book, not together</u>

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

\mathbf{Maths\ together = 2 \times 9!}

Using the complement rule, we have:

\mathbf{Maths\ not\ together = Total - Maths\ together}

This gives

\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}

\mathbf{Maths\ not\ together = 2903040}

<u>(c) The novels must be together and the chemistry books, together</u>

We have:

\mathbf{Novels = 3}

\mathbf{Chemistry = 5}

First, arrange the novels in:

\mathbf{Novels = 3!\ ways}

Next, arrange the chemistry books in:

\mathbf{Chemistry = 5!\ ways}

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

\mathbf{n =Mathematics + 1 + 1}

\mathbf{n =2 + 1 + 1}

\mathbf{n =4}

So, the number of arrangements is:

\mathbf{Arrangements = n! \times 3! \times 5!}

\mathbf{Arrangements = 4! \times 3! \times 5!}

\mathbf{Arrangements = 17280}

<u>(d) The mathematics must be together and the chemistry books, not together</u>

We have:

\mathbf{Mathematics = 2}

\mathbf{Novels = 3}

\mathbf{Chemistry = 5}

First, arrange the mathematics in:

\mathbf{Mathematics = 2!}

Literally, the number of chemistry and mathematics now is:

\mathbf{n =Chemistry + 1}

\mathbf{n =5 + 1}

\mathbf{n =6}

So, the number of arrangements of these books is:

\mathbf{Arrangements = n! \times 2!}

\mathbf{Arrangements = 6! \times 2!}

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

\mathbf{Arrangements = ^7P_3}

So, the total arrangement is:

\mathbf{Total = 6! \times 2!\times ^7P_3}

\mathbf{Total = 6! \times 2!\times 210}

\mathbf{Total = 302400}

Read more about permutations at:

brainly.com/question/1216161

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I need help finding area of this shape
quester [9]

Answer:

170

Step-by-step explanation:

30 x 3 then do 10 x 8

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