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3 years ago

Find the missing length of the triangle. If necessary, round your answer to the nearest tenth

Mathematics

2 answers:

lisov135 [29]3 years ago

8 0

Answer:

x = 30

Step-by-step explanation:

$a^{2} = c^{2} - b^{2}$

$a^{2} = 50^{2} - 40^{2}$

$a^{2} = 2500 - 1600$

$a^{2} = 900$

$\sqrt{a^{2}} = \sqrt{900}$

a = 30

Amiraneli [1.4K]3 years ago

4 0

Answer:

Step-by-step explanation:

Since this is a right triangle, we can use the Pythagorean theorem to find the missing side

a^2 +b^2 = c^2 where a and b are the sides and c is the hypotenuse

x^2 + 40^2 = 50^2

x^2 +1600 = 2500

x^2 = 2500-1600

x^2 = 900

Take the square root of each side

sqrt(x^2) = sqrt(900)

x = 30

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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

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