ik that angle one would be the same as 3 and 2 and 4 are also the same so if u find an answer for 2 or 4 they would be the same but is there like a total of the entire thing

7 0

3 years ago

Read 2 more answers

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,

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

3 0

3 years ago

What is the solutions to the equation w/2w-3=4/w

labwork [276]

Answer:

{2, 6}

Step-by-step explanation:

w/2w-3=4/w is ambiguous. Did you mean

w 4

---------- = ------ ? If so, please enclose "2w - 3" inside parentheses.

2w - 3 w

Cross multiplying, we get w² = 8w - 12.

Putting this into standard form, we get:

w² - 8w + 12 = 0, which factors into (w - 2)(w - 6) = 0.

The solutions are found by setting each factor = 0 separately and solving the resulting equations for w: {2, 6}.

7 0

3 years ago

Which expression is equivalent to -18a^-2b^5/-12a^-4b^-6

crimeas [40]

Answer:

c. 3a^2b^11/2

Step-by-step explanation:

The applicable rule of exponents is ...

(a^b)/(a^c) = a^(b-c)

$\dfrac{-18a^{-2}b^5}{-12a^{-4}b^{-6}}=\dfrac{-18}{-12}a^{-2-(-4)}b^{5-(-6)}=\boxed{\dfrac{3a^2b^{11}}{2}}$

3 0

2 years ago