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3 years ago

Can anyone please help me with this attachment?

Mathematics

1 answer:

Vesna [10]3 years ago

6 0

Answer:

Perimeter of the sector = (36 + 8π) cm.

Step-by-step explanation:

Formula to find the of an arc of the circle is,

Length of the arc = $\frac{\theta}{360}(2\pi r)$

Here, θ = central angle formed by the arc

r = radius of the circle

Therefore, length of the arc given in the picture = $\frac{80}{360}(2\pi) (18)$

= 8π cm

Perimeter of the sector = Sum of all sides of the sector

= Length of radius + length of the arc + length of radius

= 18 + 8π + 18

= (36 + 8π) cm

Perimeter of the sector is (36 + 8π) cm.

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3 years ago

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Step-by-step explanation:

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3 years ago

What is the point slope equation of the line that passes through (9,-1) and has a slope of 5

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2 years ago

Read 2 more answers

The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any meth

Lubov Fominskaja [6]

Answer:

$\frac{128\pi}{3}$ cubic units

Step-by-step explanation:

We are given that a region bounded by a curves $x=(y-7)^2$

x=4 and about y=5

We have to find the value of volume of the resulting solid

To find the volume of resulting solid we are using cylinder method

Substitute the values of x then we get

$(y-7)^2=4$

$y-7=\pm2$

y-7=2 and y-7=-2

y=7+2=9 and y=-2+7=5

Radius of cylinder =r=y-5

and height =h= $4-(y-7)^2=4-y^2-49+14 y=-y^2+14 y -45$

Using cylinder method and integrate along y -axis from y=5 to y=9

Volume = $\int_{a}^{b}2\pi r h dy=\int_{5}^{9}2\pi(y-5)(-y^2+14 y -45) dy$

volume= $2\pi\int_{5}^{9}(-y^3+19y^2-115y+225)dy$

Volume = $2\pi[-\frac{y^4}{4}+19\frac{y^3}{3}-115\frac{y^2}{2}+225y]^9_5$

Volume = $2\pi[-\frac{6561}{4}+\frac{625}{4}+19(\frac{729-125}{3})-115(\frac{81-25}{2})+225(9-5)]$

Volume= $2\pi(-1484+\frac{11476}{3}-3220+900)$

Volume = $2\pi(\frac{11476}{3}-3804)$

Volume = $2\pi(\frac{11476-11412}{3})=\frac{128\pi}{3}$

Hence, volume of resulting solid = $\frac{128\pi}{3}$ cubic units

3 0

3 years ago