find volumes of cubes: for A, L=H=w=10_5/8; for B, L=H=w=21

downandshady

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Oct 29, 2008
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The problem says

The welded tanks A and B are made from a 1/8 in steel plate. Outside measurments are given.

A 's L is 10 5/8'' H is 10 5/8'' and the w is 10 5/8''

B 's L is 21'' H is 21'' and w is 21''

find the volume of tank A. Round the answer to the nearest tenth cubic inch.

find the volume of tank B.
 
Re: I cant seem to figure it out

Hello, downandshady!

Are you waiting for some magic formula?


The welded tanks A and B are made from 18-inch steel plate.\displaystyle \text{The welded tanks }A\text{ and }B\text{ are made from }\tfrac{1}{8}\text{-inch steel plate.}

They are both cubes. Outside measurments are given.\displaystyle \text{They are both cubes. }\:\text{Outside measurments are given.}

Cube A: 1058 inches on an edge.\displaystyle \text{Cube A: }\:10\tfrac{5}{8}\text{ inches on an edge.}'

Cube B: 21 inches on an edge.\displaystyle \text{Cube B: }\:21\text{ inches on an edge.}


Find the volume of the tanks. Round to the nearest tenth cubic inch.\displaystyle \text{Find the volume of the tanks. }\:\text{Round to the nearest tenth cubic inch.}

Do to the thickness of the metal, the inner cube is a 14-inch smaller on an edge.\displaystyle \text{Do to the thickness of the metal, the inner cube is a }\tfrac{1}{4}\text{-inch smaller on an edge.}

The interior of Cube A is a cube which is 1038 inches on an edge.\displaystyle \text{The interior of Cube A is a cube which is }10\tfrac{3}{8}\text{ inches on an edge.}

\(\displaystyle \text{Its volume is: }\:(10\tfrac{3}{8})^3 \:=\:(\tfrac{83}{8})^3 \:=\:1116.771484 \;\approx\;1116.8\text{ in}^3\)


Now you try Cube B . . .

 
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