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How many perfect cubes of integers between 1 and 100 inclusive are odd?
There are two.
What are the imperfect cubes between 1 to 200?
The infinitely many real numbers between 1 and 200 except for 1, 8, 27, 64 and 125.
How many small cubes with edges of 10 cm can be just accommodated in a cubical box of 1 m edge?
100
What is the volume of 100 centimeter cubes?
The volume is 1 million cm3
Gerard has a large number of cubes that measure 1 cm along an edgeHow many cubes will be necessary to build a cube that is 1 m along an edge?
1003 = 106 = 1,000,000: since 100 centimeters are required to equal one meter, the larger cube is scaled up by 100 in each of its three dimensions compared to the small cubes specified.
How many perfect cubes are there between 1 to 50?
Between the two numbers there are only two.
How many small cubes with edges of 10 cm can be just accommodated in a cubical box of 1 m edge?
100
How many zeros are there in between 1-100?
101
How many numbers between 1 and 100 are palindromes?
25 numbers are between 1-100.
How many times is 1 counted 1 to 100 between?
100
Gerard has a large number of cubes that measure 1 cm along an edgeHow many cubes will be necessary to build a cube that is 1 m along an edge?
1003 = 106 = 1,000,000: since 100 centimeters are required to equal one meter, the larger cube is scaled up by 100 in each of its three dimensions compared to the small cubes specified.
How many 1 inch cubes are created when you cut a 4 inch cube into equal pieces of 1 inch cubes?
16 1-inch cubes
How many 1 in cubes would be needed to build a 4in cubes?
64
Perfect cubes between 1 and 50?
1, 8, 27
What are perfect cubes in between 1 to 200?
The values of cubes 1 to 30 between 1 and 200 are 13 (1), 23 (8), 33 (27), 43 (64), and 53 (125).
How many small cubes with edges of 10 cm can be just accommodated in a cubical box of 1 m edge explanation?
1 metre = 100 cm so 10 cubes will fit along the length, breadth and height of the box. Therefore 10*10*10 = 1000 cubes will fit.
What is the sum of the first 100 cubes?
To solve for the sum of the first n number of cubes, use (1+2+3+4+5...+n)2. So for n=100, (1+2+3+4+5...+100)2 = c (cubes) (101 * 50)2 = c 50502 = c c = 25502500
