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fenix001 [56]
3 years ago
13

the lateral area of a cylinder is 325 pi in2 the radius is 13 in find the following Volume. surface area 

Mathematics
1 answer:
Nesterboy [21]3 years ago
5 0
Volume: 2112.5π
Surface Area: 663π

First, let's find the height of the cylinder. Using the formula for lateral surface area, 2*π*r*h:

2*π*13*h = 325π
(divide π by both sides)
26h = 325
h = 12.5 in.

Next, let's solve the surface area of the cylinder using the formula π*r²*h.

π * 13² * 12.5 = π * 169 * 12.5 = 2112.5π.

Next, let's solve the surface area using the formula 2πrh + 2πr².
 
(2 * π * 13 * 12.5) + (2 * π * 13²) = 325π + 338π = 663π.
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Answer:

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Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the  starting point in the direction of 35° E of South as shown in the figure.

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Vector \overrightarrow{DD'} is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)

and the direction is \alpha east of north as shown in the figure,

\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)

From the figure,

|AB|=|OA-OB|

\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|

\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|

\Rightarrow |AB|=45.52 miles

Again, |PQ|=|OP-OQ|

\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|

\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|

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Now, from equations (i) and (ii), we have

|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01 miles, and

\tan \alpha=\frac{|46.42|}{|45.52|}

\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}

Hence, the wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} E astof North  with respect to the destination point.

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2 years ago
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Answer:

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