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Please help!!!

 Feb 10, 2019
 #1
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For the challenge...

 

The side length of the cube is \(66^{1/3}=\sqrt[3]{66}\)

The volume of a square pyramid is \(a^2\frac{h}{3}\), where \(a\) is the base edge and \(h\) is the height.

Of course, they are all \(\sqrt[3]{66}\).

So...

 

\(\sqrt[3]{66}^2\times \frac{\sqrt[3]{66}}{3}=66^{\frac{2}{3}}\times \frac{66^{\frac{1}{3}}}{3}=\frac{66}{3}=22\) in^3 is the volume.

 

You are very welcome!

:P

 

(that was a lot of LaTeX for me: \sqrt[3]{66}^2\times \frac{\sqrt[3]{66}}{3}=66^{\frac{2}{3}}\times \frac{66^{\frac{1}{3}}}{3}=\frac{66}{3}=22 was what i entered)

 Feb 10, 2019
 #2
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Do you have a diagram for the previous problems... cheeky

 Feb 10, 2019
 #3
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No I don't have diagram, I think it's asking if we know what shape has those requirements.

Guest Feb 11, 2019

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