64 x 65 = 4160
(c + 8)(c - 8)= c^2 - 64
64^(2/3) = ((64^(1/3)^(2)) 64^(1/3) is the 'cube root of '64' , which is '4' Hence (4^(2)) = 16 Hence 64^(2/3) = 16
(x-2)^2+y^2=64
A = LW 64 = 64 * 1 or 1 * 64 64= 32 * 2 or 2 * 32 64 = 16 * 4 or 4 * 16
As a product of its prime factors: 2*2*2*2*2*2 = 64 or in exponent 2^6 = 64
As a product of its prime factors: 2*2*2*2*2*2 = 64 or as 26 = 64
The answer to this convoluted question is as follows: The product of prime numbers, for the composite number 64, is 2*2*2*2*2*2 or 26
It is 2*2
As product of its prime factors in exponent terms: 2^6 = 64
Yes, 64 can be a product of prime numbers. All integers are either prime numbers or a product of prime numbers (called the prime factorization of a number). The prime factorization of 64 is 2 x 2 x 2 x 2 x 2 x 2.
Just the one and it is 2 but as a product of its prime factors: 2*2*2*2*2*2 = 64
64 = 2 x 2 x 2 x 2 x 2 x 2 OR 26
There are not two prime numbers whose product is 64. The prime factorization of 64 is 2 x 2 x 2 x 2 x 2 x 2, which is 2 to the 6th power. These are the only prime numbers that can be multiplied together to result in 64. There are six of them, not two. The only pair of numbers whose product is 64 that includes one prime number is 2 x 32, but only one number is prime.
64 = 2*2*2*2*2*2
There is no such set of numbers.
64 = 2 x 2 x 2 x 2 x 2 x 2 or 26