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Mathematics 19 Online
OpenStudy (anonymous):

Prove that [\sum_{i}^{}m_ir_i]^2=(\sum_{i}^{}m_i)(\sum_{i}^{}m_ir_i^2)-\frac{1}{2}\sum_{i,j}^{}m_im_jr_{ij}^2

OpenStudy (anonymous):

please write in blank space.

OpenStudy (anonymous):

whats that

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