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3 years ago

Answer nowwwwwwwwwwwwwwwwwwwwwwwwwwww

Mathematics

2 answers:

krok68 [10]3 years ago

8 0

Answer:

7/6

Step-by-step explanation:

(7x/12)/(x/2)

=7/6

Kay [80]3 years ago

6 0

Answer: 7/6

Explanation: (7x/12)(x/2)

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Can someone help me with this

trasher [3.6K]

Answer:

Angle 3 is 65°, Angle 8 is 115°

Step-by-step explanation:

Angle 3 and 1- Linear Pair

Angle 8 and 1- Alternate Exterior

7 0

3 years ago

Pentagon PQRST is shown on the coordinate grid below . Given that the pentagon was rotated 180 ° clockwise and labeled P'Q'R'ST

olganol [36]

Answer:

Step-by-step explanation:

Given is a pentagon. It has 5 sides

6 0

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

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