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3 years ago

See image attached for the question. HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Greeley [361]3 years ago

3 0

Answer:

B, C

Step-by-step explanation:

that's the answerrr

krok68 [10]3 years ago

3 0

Answer:

2/8 and 5/20 are Equivalent Fractions

B and C options are right ones

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How many does 95 fit into 1660

nadezda [96]

17.4736843 or 17 remainder 45

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3 years ago

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I need help quick please help me thnak you

quester [9]

Answer:

28 numbers

Step-by-step explanation:

There is a distance of 81 between -0.5 and 80.5. Divide this number by 3, since you are only counting every third number and you will get 27, but you need to count the first number, which has been excluded. Add one more to get the answer, 28.

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3 years ago

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

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