Let's analyze Marissa's claim step-by-step by calculating the volumes of the blocks and examining the validity of each statement.
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Volume of one block:
- For the block with dimensions \( \frac{1}{2} \times 1 \times 1 \): \[ \text{Volume} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \text{ cubic units} \]
- For the block with dimensions \( \frac{1}{3} \times 1 \times 1 \): \[ \text{Volume} = \frac{1}{3} \times 1 \times 1 = \frac{1}{3} \text{ cubic units} \]
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Volume of the stacks:
- For 6 blocks of \( \frac{1}{2} \) cubes: \[ \text{Total Volume} = 6 \times \frac{1}{2} = 3 \text{ cubic units} \]
- For 9 blocks of \( \frac{1}{3} \) cubes: \[ \text{Total Volume} = 9 \times \frac{1}{3} = 3 \text{ cubic units} \]
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Evaluating the statements:
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"Six of the one-half cubes would have a volume of 3 cubic units."
- True. We calculated this volumetric total as \( 3 \) cubic units.
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"Nine of the one-third cubes would have a volume of 6 cubic units."
- False. We calculated this volumetric total as \( 3 \) cubic units, not \( 6 \).
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"Since One-half and One-third do not have common factors, stacks of one-half cubes and stacks of third-cubes could never have the same volume."
- False. Even though \( \frac{1}{2} \) and \( \frac{1}{3} \) do not have common factors, they can still produce the same total volume, as evidenced by their individual volumes at the stack totals we calculated.
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"Three of the one-third cubes will make 1 unit cube."
- True. \[ 3 \times \frac{1}{3} = 1 \text{ unit} \]
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"Two of the one-half cubes will make 1 unit cube."
- True. \[ 2 \times \frac{1}{2} = 1 \text{ unit} \]
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"Both stacks will have a volume of 3 cubic units."
- True. We calculated that both stacks indeed have a total volume of \( 3 \) cubic units.
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Final true statements are:
- Six of the one-half cubes would have a volume of 3 cubic units.
- Three of the one-third cubes will make 1 unit cube.
- Two of the one-half cubes will make 1 unit cube.
- Both stacks will have a volume of 3 cubic units.
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