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mojhsa [17]
3 years ago
9

Calisto Publishing Co. sells graphic novels and comics to a large retailer for $7.50 per book,

Mathematics
1 answer:
Grace [21]3 years ago
3 0

Answer:

life suck and night night

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I need help please ASAP?!!!
Lady_Fox [76]

Answer:

<em>Its 43.2 </em>

<em></em>

<em>I hope this helps ^^</em>

5 0
2 years ago
Read 2 more answers
How much do I need to subtract from 67/10 to make 6
Harrizon [31]

Answer:

0.7

Step-by-step explanation:

67/10 is the same as 6.7 when you subtract the 0.7 you will remain with 6

4 0
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}

3 0
2 years ago
I'm confused about this, please help!
Fofino [41]
D angle CDE
2552 - 22.2yyy526
7 0
3 years ago
If n(A) =15 n (B) = 9 n(A∩B) =4 <br> find n(A U B)
Xelga [282]

Answer:

20

Step-by-step explanation:

n(A) only =15-4=11

n(B) only=9-4=5

n(A n B)=4

n(A U B)=11+5+4=20

5 0
3 years ago
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