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3 years ago

PLEASE ANSWER ALL QUESTIONS WILL GIVE BRAINLY!!!!! SHOW YOUR WORK PLS

Mathematics

2 answers:

Natalija [7]3 years ago

8 0

1. 6•2/4+ 7
6• 16 + 7
96+7
103 for the final answer for the first question

kifflom [539]3 years ago

5 0

2 ^ 4 = 16

16 × 6 = 96

96 + 7 = 103

3 + 4 = 7

8 × 7 = 56

12 ^ 2 = 144

56 + 144 = 200

14 - 6 = 8

8 ^ 2 = 64

64 × 6 = 384

150 ÷ 5 = 30

30 × 6 = 180

180 + 8 = 188

hope this helps :))

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4 years ago

Calculus Problem

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

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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

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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

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