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3 years ago

Find the surface area of a cylinder with a height of 7 m and a base radius of 6 m.

Mathematics

1 answer:

weqwewe [10]3 years ago

6 0

Answer:

489.84 m²

Step-by-step explanation:

Area of one 2d circle: πr² ⇒ π6² ⇒ 36π ≈ 113.04 (using 3.14 for pi)

Area of both 2d circles: 113.04 + 113.04 =226.08 m²

Now we have to find the width of the rectangle, which is equal to the circumference of either circle:

Width of rectangle: 2πr ⇒ 2π6 = 12π ≈ 37.68 (using 3.14 for pi)

We can find the area of the rectangle now, since the length was given

Area of rectangle: 37.68· 7= 263.76 m²

Surface Area: 263.76+226.08= 489.84m²

Hopefully this helps!

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The radius of the sphere is 33 find the surface area

patriot [66]

Answer:

A=13684.7776

Step-by-step explanation:

$= 4\pi {r}^{2}$

$4\pi {33 }^{2} = 13684.7776$

5 0

2 years ago

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

3 years ago

At a store, Susan selected a pumpkin that weighed 35.2 ounces. • Pumpkins cost $1.80 per pound. • There are 16 ounces in 1 pound

klemol [59]

Answer:

First you have to find out the number of ounces in a pound. Then, multiply that number by the price per pound

Step-by-step explanation:

4 0

2 years ago

How do you solve (x3 + 6x2 + 3x + 1 ) ÷ (x - 2)

WINSTONCH [101]

We conduct long division, cancelling each term in x³ + 6x² + 3x + 1. As so:

x²(x -2) = x³ - 2x²
Subtracting from the original equation, we get:
8x² + 3x + 1

8x(x - 2) = 8x² - 16x
Subtracting from the new equation, we get:
21x + 1

21(x - 2) = 21x - 42
Subtracting from the new equation, we get:
43

The answer is x² + 8x + 21 with a remainder of 43.

5 0

3 years ago

Please help me with this

notka56 [123]

Answer:

Option 3

Step-by-step explanation:

Download pdf

4 0

3 years ago