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harina [27]
2 years ago
13

Answer this question, thx for the help (the 4th one is this shape O )

Mathematics
1 answer:
RideAnS [48]2 years ago
5 0

Answer:

1st one

Step-by-step explanation:

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39=1 3/10b what will be the answer show work
Lubov Fominskaja [6]
1 3/10·b=39
(1·10+3)·b/10=39
13b/10=39
13b=39·10
13b=390|:13
<u>b=30</u>

<em>Answer: 30</em>
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3 years ago
How many edges does a tetrahedron have? 6 9 12 18
Jet001 [13]
A tetrahedron is a shape made out of four triangular faces, and it has a total of <span>six edges

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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
ANSWER AND EXPLAIN PLEASE ILL MARK BRAINLIEST
Arte-miy333 [17]
Solve for x by plugging in y, then plug x back in
8 0
3 years ago
Melinda ran 8,791 meters on Monday and 2,899 meters on Tuesday. How many more meters did she run on Monday than on Tuesday?
Sati [7]

Answer:

8,791- 2,899= 5892

Step-by-step explanation:

Melinda ran 5,892 more meters on monday than tuesday

8 0
2 years ago
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