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3 years ago

Please answer this multiple choice question correctly for 30 points and brainliest!!

Mathematics

1 answer:

Ilia_Sergeevich [38]3 years ago

8 0

Answer:

B. 12n + 2(25) ≤ 100

Step-by-step explanation:

The inequality you are asked for is intended to express ...

shirt cost + pant cost (is less than or equal to) Sandy's budget

The cost of n shirts at $12 each will be 12n. The cost of 2 pants at $25 each is 2(25). Then the inequality is ...

12n + 2(25) ≤ 100 . . . . . matches choice B

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Points A and B are on the different sides of a line 1, the distance between point A and the line lis 10 in, the distance between

givi [52]

Answer: Distance between O and the line 1 is 3 inches.

Explanation:

Since we have given that

Distance between point A and the line 1 = 10 inches

Distance between point B and the line 1 = 4 inches

Total length of segment AB is given by

$10+4=14\ inches$

Since O is midpoint of the line segment AB.

so, AO is given by

$\frac{14}{2}=7\ in.$

Now, Distance between O and the line 1 is given by

$10-7=3\ in.$

Hence, Distance between O and the line 1 is 3 inches.

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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

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

Hence,

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

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