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xz_007 [3.2K]
2 years ago
6

A grain silo is built from two right circular cones and a right circular cylinder with internal measurements represented by the

figure above. Of the following, which is closest to the volume of the grain silo, in cubic feet?
A) 261.8
B) 785.4
C) 916.3
D) 1047.2

Mathematics
1 answer:
Murljashka [212]2 years ago
6 0

Answer:

Step-by-step explanation: The volume of the grain silo can be found by adding the volumes of all the solids of which it is composed (a cylinder and two cones). The silo is made up of a cylinder (with height 10 feet and base radius 5 feet) and two cones (each with height 5 ft and base radius 5 ft). The formulas given at the beginning of the SAT Math section:

Volume of a Cone:

V=\frac{1}{3} \pi r^2h

Volume of a Cylinder:

V=\pi r^2h

can be used to determine the total volume of the silo. Since the two cones have identical dimensions, the total volume, in cubic feet, of the silo is given by

V_{silo} =\pi (5^2)(10)+(2)(\frac{1}{3})\pi (5^2)(5)=(\frac{4}{3})(250)\pi

which is approximately equal to 1,047.2 cubic feet.

<u>The final answer is D.</u>

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Answer:

The area of rectangle is 2\frac{80}{81}\ cm^2

Step-by-step explanation:

we know that

The perimeter of a rectangle is equal to

P=2(L+W)

we have

P=7\frac{1}{3}\ cm=7+\frac{1}{3}=\frac{22}{3}\ cm

so

\frac{22}{3}=2(L+W)

\frac{11}{3}=(L+W) ----> equation A

Remember that

The length of a rectangle is twice its width.

so

L=2W ----> equation B

substitute equation B in equation A

\frac{11}{3}=(2W+W)

solve for W

W=\frac{11}{9}\ cm

Find the value of L

L=2(\frac{11}{9})

L=\frac{22}{9}\ cm

<em>Find the area of rectangle</em>

A=LW

substitute the values

A=(\frac{22}{9})(\frac{11}{9})

A=\frac{242}{81}\ cm^2

Convert to mixed number

\frac{242}{81}\ cm^2=\frac{162}{81}+\frac{80}{81}=2\frac{80}{81}\ cm^2

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Hey mate!
Let's figure this out!

The answer is false.

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