6.4.10 CYU
Flashcard
•
Mathematics
•
6th Grade
•
Practice Problem
•
Hard
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1.
FLASHCARD QUESTION
Front
What is the volume of a cube with edge length 4 inches?
Back
The volume of a cube is calculated using the formula V = side^3. For a cube with edge length 4 inches, V = 4^3 = 64 cubic inches.
2.
FLASHCARD QUESTION
Front
How many little cubes with edge length \( \frac{1}{2} \) inch fit into a larger cube with edge length 4 inches?
Back
To find the number of little cubes, calculate the volume of the larger cube (4^3 = 64 cubic inches) and the volume of one little cube (\( \left(\frac{1}{2}\right)^3 = \frac{1}{8} \) cubic inches). Then, divide the volume of the larger cube by the volume of one little cube: \( 64 \div \frac{1}{8} = 64 \times 8 = 512 \).
3.
FLASHCARD QUESTION
Front
What expression can be used to find how many cubes with edge length of \( \frac{1}{3} \) unit fit in a prism that is 5 units by 5 units by 8 units?
Back
The expression is \( (5 \cdot 3) \cdot (5 \cdot 3) \cdot (8 \cdot 3) \). This calculates the number of cubes along each dimension.
4.
FLASHCARD QUESTION
Front
What is \( \frac{5}{6} \div \frac{1}{2} \)?
Back
To divide fractions, multiply by the reciprocal: \( \frac{5}{6} \times \frac{2}{1} = \frac{10}{6} = 1\frac{2}{3} \).
5.
FLASHCARD QUESTION
Front
What is \( \frac{8}{9} \div \frac{2}{3} \)?
Back
To divide fractions, multiply by the reciprocal: \( \frac{8}{9} \times \frac{3}{2} = \frac{24}{18} = 1\frac{1}{3} \).
6.
FLASHCARD QUESTION
Front
What is \( \frac{7}{8} \div \frac{1}{4} \)?
Back
To divide fractions, multiply by the reciprocal: \( \frac{7}{8} \times \frac{4}{1} = \frac{28}{8} = 3\frac{1}{2} \).
7.
FLASHCARD QUESTION
Front
Define volume in the context of geometry.
Back
Volume is the amount of space occupied by a three-dimensional object, measured in cubic units.
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