The number of hexagonal faces that are p...

The number of hexagonal faces that are present in a truncated octahedron is

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Find the number of hexagonal faces that are present in a truncated octahedral.

An octahedron has

Please help Sabu decode the jail lock. Chacha Choudhary gave Sabu a formula : f_(1) = ((x)/(z)xx y), f_(2) =((f)/(v) xxu), f_(3) = ((r)/(s)xx w) Sabu can open the lock if he finds the value of 3f_(1) +f_(2) +f_(3) = key where: Number of triangular faces in a truncated tetrahedron = x Number of hexagonal faces in a truncated tetrahedron = x Number of corners in a truncated tetrahedron = z Number of square faces in a truncated octahedron = t Number of hexagonal faces in a truncated octahedron = u Number of corners in a truncated octahedron = u Number of triangular faces in a truchcated cube = w Number of octangonal faces in a truncated cube = r Number of corners in a truncated cube = s What is the KEY ?

Cis - and trans - faces are present in

The number of faces in the given figure is

RESONANCE-SOLID STATE-Exercise -3 (Part-I)

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