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yarga [219]
2 years ago
8

If you can solve this then you get the brainliest!

Mathematics
2 answers:
Law Incorporation [45]2 years ago
7 0

Answer:

x in top middle row.

x in middle left row.

x in middle right row.

x in last down middle row.

please

Alika [10]2 years ago
4 0

Start with a blank grid

\begin{array}{c|c|c} &  & \\\cline{1-3} &  & \\\cline{1-3} &  & \end{array}

Place 3 x's to clump into one corner like this

\begin{array}{c|c|c}x & x & \\\cline{1-3} x&  & \\\cline{1-3}&  & \end{array}

Then fill out the opposite corner in a similar fashion to make sure that we don't get 3 in a row/column/diagonal anywhere

\begin{array}{c|c|c}x & x & \\\cline{1-3}x &  & x\\\cline{1-3} & x & x\end{array}

Each row has exactly 2 x's and each column has exactly 2 x's. The main diagonal has 2 x's, while the off diagonal has no x's at all.

I'm not sure if this is the only configuration where this is possible (if we ignored any rotated versions of what is shown above).

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Read 2 more answers
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