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Reil [10]
3 years ago
6

WILL GET BRAINLIEST!!!! I NEED IT BY TODAY!!! NO FAKE ANSWERS!!!!

Mathematics
1 answer:
alexira [117]3 years ago
4 0
B

Hope this helps:))
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What is the answer to this question
DanielleElmas [232]

Answer:

65

Step-by-step explanation:

the angle on top is the same  because the lengths of the lines that meet to "H" are the same.

7 0
3 years ago
Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
What is the slope of the line passing through the points (2, 5) and (-1, -4)? show your work
Sati [7]

Answer:

We can solve this question using the slope equation which is y2-y1/x2-x1

If you use that formula and sub in the coordinates

-4 - 5 / -1-2

-9/-3

= 3

The slope should be 3/1

5 0
3 years ago
Read 2 more answers
Help with these two math problems?
ludmilkaskok [199]
Harold incorrectly completed the assignment.
3 0
3 years ago
Use the Distributive property to write an expression equivalent 5 times ( 3+4 ).
algol13
You can set this up as 5(3+4)....then, distribute by multiplying the terms in parentheses by 5. This gives you 15+20, which equals 35.
8 0
3 years ago
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