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andrezito [222]
3 years ago
12

(EXTRA POINTS) Which of the following relations is a function? file in question

Mathematics
1 answer:
Taya2010 [7]3 years ago
6 0
I think it is a I am not sure though
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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab
Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

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Ann opened a new savings account with an initial deposit of $250. Which combination will result in a zero in balance in Ann's ac
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C.  (10x3)10 - (27.5x2)10 = 0
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Subtract the mixed numbers: 3 5/6 - 1 2/3 *
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Use Keep, Change, Flip
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What property is (3x10)x8=3x(10x8)
mylen [45]
(3 x 10) x 8 = 3 x (10 x 8)
this is called the associative property of multiplication.
it states : (a * b) * c = a * (b * c)
7 0
3 years ago
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The front row in a movie theater has 23 seats. If you were asked to sit in the seat that occupied the median position, in which
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Answer:

You would have to sit in the 12th seat. (12)

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