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stiks02 [169]
3 years ago
14

Please help ASAP !!!!!

Mathematics
1 answer:
pochemuha3 years ago
7 0

Answer:

(goh) (0) = 4

Step-by-step explanation:

Given that,

g(x) = 2x

h(x) = x² + 4

We need to find the value of (goh) (0).

Firstly we find (goh),

(goh) = g(h(x))

=g(x²+4)

(goh) (0) = 0²+4

=4

Hence, the required answer is 4.

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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}))

Hence,

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

3 0
2 years ago
30 increased by 55% I need help solving this, I am good in history ect, but maths kills me
Lubov Fominskaja [6]
I honestly don't know, but I think it is 63

3 0
3 years ago
Plz help 100 points
tekilochka [14]

Answer:

the answer is one line is increased by 1, and the other increased by 6

hope this helped :))

4 0
2 years ago
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7.<br> Solve for x:<br> x^4-5x^2+4 = 0
arlik [135]

Answer:

x=2

Step-by-step explanation:

8 0
3 years ago
$18 and $20 in a expression
gogolik [260]
An expression with 18 and 20.

An expression does not have an equal sign.

18+20 

Hope I helped. 
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