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3 years ago

This is on graphing cosine and sine functions!

Mathematics

1 answer:

yawa3891 [41]3 years ago

4 0

Check the picture below.

5 is 1*5, or 5 times greater than 1.

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Collected 20 cans this week, 25 cans next week what is the percent of increase

BartSMP [9]

Hello!!

25 - 20 = 5

5/20 = 0.25

0.25 * 100 = 25%

Good luck :)

5 0

2 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

Add 280.00 +21.40 +6.30 +.41 = 308.71, I don't know where the 7 was added though

Vitek1552 [10]

I m getting 308.11, there is an additional 0.60 that is added, maybe your source is wrong, or there is a copying error.

8 0

3 years ago

A pyramid at a museum has a height of 21 meters and a square base with side lengths of 30 meters. 21 m 30 m What is the approxim

Orlov [11]

Answer: 6,300 m^3

Step-by-step explanation:

Put numbers into calculator

5 0

2 years ago

Beth is particularly susceptible to getting strep throat. If she is infected with one streptococcus pygenes bacterium. And it ha

Neko [114]

Add a snowman amen amen Carmen caravanserais consisted

6 0

3 years ago