There are five odd numbers on an odd number cube?
the cube of 1 is 1 as is any operation done against 1 3x3x3= 27 so 27 is the cube of the odd number 3
The cube root of -1 ("minus 1" or "negative 1") is -1. In fact all odd-numbered roots (i.e., the fifth root, the seventh root, the ninth root, etc.) of -1 are -1.
Half of them.
Yes, it is.
1728 is even so it cannot have an odd cube root.
Any odd root, for example the cube root, of -3.24 is a real number.
There are five odd numbers on an odd number cube?
Assuming you know that your number is a perfect square, the square root of an even number is even, and the square root of an odd number is odd.
the cube of 1 is 1 as is any operation done against 1 3x3x3= 27 so 27 is the cube of the odd number 3
The cube root of -1 ("minus 1" or "negative 1") is -1. In fact all odd-numbered roots (i.e., the fifth root, the seventh root, the ninth root, etc.) of -1 are -1.
Two odd numbers always sum to an even number. Always. Two even numbers always sum to an even number, and an odd number and an even number always sum to an odd number.
3/6=1/2 on a number cube it depends on how many sides of a spinner there are if its an even number always 1/2 if its an odd number of sides it cant be simplified
Half of them.
The cube root of 2 is irrational. The proof that the square root of 2 is irrational may be used, with only slight modification. Assume the cube root of 2 is rational. Then, it may be written as a/b, where a and b are integers with no common factors. (This is possible for all nonzero rational numbers). Since a/b is the cube root of 2, its cube must equal 2. That is, (a/b)3 = 2 a3/b3 = 2 a3 = 2b3. The right side is even, so the left side must be even also, that is, a3 is even. Since a3 is even, a is also even (because the cube of an odd number is always odd). Since a is even, there exists an integer c such that a = 2c. Now, (2c)3 = 2b3 8c3 = 2b3 4c3 = b3. The left side is now even, so the right side must be even too. The product of two odd numbers is always odd, so b3 cannot be odd; it must be even. Therefore b is even as well. Since a and b are both even, the fraction a/b is not in lowest terms, thus contradicting our initial assumption. Since the initial assumption cannot have been true, it must be false, and the cube root of 2 is irrational.
Yes, it is.
No, it always has an EVEN answer