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2 years ago

11

A plane is flying at the height of 5000 meter above the sea level. at a particular point, it is excatly above a submarine floati

ng 1200 meter below the sea level. what is the vertical distance between them ?

1 answer:

AnnyKZ [126]2 years ago

3 0

Answer:

3800 meters

Step-by-step explanation:

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What is the circumference of a circle with the radius of 5 cm use 3.14 for pi

Archy [21]

Answer:

31.42cm

Step-by-step explanation:

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2 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

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

Hence,

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

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2 years ago

Alonzo is saving money to buy a game. So far he has saved $30 , which is two-thirds of the total cost of the game. How much does

Andrei [34K]

Well, it should be 45 because 1/3 of the game would be 15 and 15+15 is 30 (2/3rds cost), so 15+15+15, 45, would be the full cost of the game

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masya89 [10]

You are given the opposite: 16 km
You need to find x (adjacent)
The angle measure: 28 degrees
With opposite and adjacent, use tangent
Tan(28) = 0.5317094317
Tan(28) = 16/x
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x = 30.0916
Round to nearest tenth
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