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Mathematically, both are correct.
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Round to the nearest tenth if necessary.
Total red quadrilaterals: 2 total noble quadrilaterals: 8 unconditioned green quadrilaterals: .
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7 copies of 1 third 7 × 1 3 = 2×3 3 + 1 3 = 2 + 1 3 = 2 1 3 b.
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Count the objects in the arrays from left to right by rows and by columns.
4 + 5 = 6 + 5 = 9 + 5 = 15 8 13 7 16 12 17 16 99 14 7 13 7 16 9 16 9 sample answer: 9 + 5 because 9 is the closest act to 10, indeed you can brand a 10 and count 4 many.
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2 parallel lines of the aforesaid length drawn 3.
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