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ira [324]
3 years ago
9

In Monogram and other works, Robert Rauschenberg synthesized Abstract Expressionist painting and Neo-Dada assemblage to create w

hat he called _____________ paintings.
Mathematics
1 answer:
mars1129 [50]3 years ago
7 0

Answer:White Paintings

Step-by-step explanation:

In 1951, Robert Rauschenberg painted some stretched canvanses a plain, solid white, leaving minimal roller marks. Each of his works consist of different number of panel iterations ( one to seven panels) which are collectively known as 'the white paintings'.

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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
You are opening a snow cone stand. Your cups, which are shaped like a cone, are 4" tall and have a 6" diameter. How much room is
aleksley [76]

Answer:

I'm not too sure but I think it's 37.7

Step-by-step explanation:

radius of the base times height is supposed to equal the volume but I'm bad at geometry

3 0
2 years ago
(04.01 LC)
Rina8888 [55]

The answer is 1 mile an hour

5 0
3 years ago
(8×1017)÷(2×103)Give your answer in standard form.
Alexxandr [17]
4.19004 x 10^5

sry I have to write extra stuff to send the answers :)
4 0
2 years ago
(3a2 – 5ab + b2) + (–3a2 + 2b2 + 8ab) Which of the following shows the sum of the polynomials rewritten with like terms grouped
Mila [183]
<span>[3a2 + (–3a2)] + (–5ab + 8ab) + (b2 + 2b2)</span><span>
</span>
6 0
3 years ago
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