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Nadya [2.5K]
3 years ago
13

The diameter of a small gear is 16cm. This is 2.5cm more than 1/4 of the diameter of a larger gear. What is the diameter of a la

rger gear? round to the nearest centimeter.
Choices: 54cm, 64cm, 44cm, and 74cm.
Mathematics
1 answer:
gogolik [260]3 years ago
8 0
(16-2.5) : 1/4 = 13.5 · 4 = 54 cm
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The answer for 80,000 * 200 is 16,000,000
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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
Whoever answers right I’ll give you Brinley
Aleks [24]

Answer:

correct answer B.

Step-by-step explanation:

hoped i helped :) can u mark brainiest pls!!

5 0
3 years ago
Please answer ASAP this is due tomorrow
Kay [80]

Answer:

227.08in^2

You want to find the surface area for this problem. SA=2lw+2lh+2hw

Step-by-step explanation:

L=2.5

W=7.4

H=9.6

SA=2lw+2lh+2hw

SA=2(2.5x7.4)+2(2.5x9.6)+2(9.6x7.4)

SA=37+142.08+48

SA=227.08in^2

8 0
3 years ago
The sum of two numbers is 98. their difference is 22. write a system of equations that describes this situation. solve by elimin
larisa [96]
Y+x=98
y-x=22

I'm going to use elimination. 

y+x=98
y-x=22
-----------
2y+0x=120 SIMPLIFY 2y=120 SIMPLIFY y=60. 

Substitute that into either equation and you get y=60 and x=38

Have a nice day! :)

6 0
3 years ago
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