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nikklg [1K]
3 years ago
5

2 üssü - 12 ile 25 üssü 2 nin çarpımı nedir?​

Mathematics
1 answer:
KengaRu [80]3 years ago
7 0

Answer:

2 üssü 12 demek 2'yi 12 defa yanyana yazmak demektir yani 2×2×2×2×2×2×2×2×2×2×2×2×=4096 eder.

şimdi diğerini yapalım aynı şekilde yapmış olursak 25×25 =625 eder

ŞİMDİ SONUCU BULALIM:4096×625=2,560,000

KOLAY GELSİN

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Step-by-step explanation:

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3 years ago
Joseph and Isabelle left Omyra’s house at the same time. Joseph jogged north at 8 kilometers per hour, while Isabelle rode her b
insens350 [35]

For this case the distance will be given by:

d ^ 2 = (12 * 1.5) ^ 2 + (8 * 1.5) ^ 2

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d ^ 2 = (18) ^ 2 + (12) ^ 2

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d ^ 2 = 468

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d = 21.63 Km

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3 years ago
Calculus Problem
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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).

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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}}

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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

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

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

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\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}}

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Answer:

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Step-by-step explanation:

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