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Since both objects are assumed to have the same volume, matching the formulas we'll get:
π
h = 
π
h
Having both <span>π and h at both sides of the equation, we can ignore them, so:
</span><span> =
</span>
And clearing R, according to the equation rules, we'll get:
![\sqrt[2]{3 r^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B2%5D%7B3%20r%5E%7B2%7D%20%7D%20)
= R
π
Having both <span>π and h at both sides of the equation, we can ignore them, so:
</span><span>
</span>
And clearing R, according to the equation rules, we'll get:
At x = 8, y = 4
At x = 14, y = 7
At y = 3, x = 6
Solution:
Given rule:
At x = 8,
Hence at x = 8, y = 4.
At x = 14.
Hence at x = 14, y = 7.
At y = 3
Multiply by 2 on both sides.
Hence at y = 3, x = 6.
The image of the table is attached below.
