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2 years ago

Find the value of x given m (I need a lot of help)

Mathematics

2 answers:

FrozenT [24]2 years ago

8 0

Answer:

x=7

Step-by-step explanation:

m angle SAR = m angle SQR (angles lying in the same segment of a circle are always equal)

that's why

7x+7=9x-7

9x-7x= 7+7

2x= 14

x=7

LuckyWell [14K]2 years ago

4 0

Answer:

x=7

Step-by-step explanation:

Given that measure of angle SAR= measure of angle SQR

7x+7=9x-7

-2x=-14

x=7

CHECK:

7(7)+7=56

9(7)-7=56

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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

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Answer:

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Step-by-step explanation:

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

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Apply substitution

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

Then

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

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

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