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What is the sixth root of 729?
3
What is the cube root of the square root of 729?
The cube root of the square root is the sixth root. The sixth root of 729 = 3
What is one third of 3 to the sixth power?
One third of 3 to the sixth power can be calculated by first finding 3 to the sixth power, which is 3^6 = 729. Then, divide 729 by 3 to find one third of it. Therefore, one third of 3 to the sixth power is 729 divided by 3, which equals 243.
What is the sixth power of 3?
The sixth power of three is 36 or is just like saying: 3x3x3x3x3x3 which is 729
What is negative 3 to the sixth power?
3 to the negative sixth power equals 0.00137174211
What are the real-roots of the cube of 729?
The cube of 729 is calculated as (729^3). To find the real roots, we first determine the cube root of 729, which is 9, since (9^3 = 729). Therefore, the real root of the cube of 729 is 9.
What is the sixth root of 729?
3
What number to the sixth power equals 729?
To find the number that, when raised to the sixth power, equals 729, we can express this mathematically as ( x^6 = 729 ). Taking the sixth root of both sides, we find ( x = 729^{1/6} ). Since ( 729 = 3^6 ), we conclude that ( x = 3 ). Therefore, the number is 3.
What is the cube root of the square root of 729?
The cube root of the square root is the sixth root. The sixth root of 729 = 3
What are the two square roots of 729?
Square root(729) = 27
What does three to the sixth equal?
3x3x3x3x3x3=729
What is 3 raised to the sixth power?
Three raised to the sixth power equals 729.
Does every sixth-degree equation have at least two real roots?
It need not have any real roots. For example x6 + 1 = 0 has none.
What is plus 3 to the sixth power?
36 = 729
Is a 729 bullet shell real or not?
The 729 Lazzeroni is a real cartridge.
How many real sixth roots does - 1 have?
The equation (x^6 = -1) can be rewritten as (x^6 = 1 \cdot e^{i\pi}) in polar form. According to De Moivre's theorem, the sixth roots of (-1) can be found by taking the sixth root of the magnitude (which is 1) and dividing the angle (\pi) by 6, resulting in (x = e^{i(\pi/6 + 2k\pi/6)}) for (k = 0, 1, 2, 3, 4, 5). This gives us six distinct complex roots, but since the roots are complex, there are no real sixth roots of (-1). Thus, (-1) has zero real sixth roots.
What are two squre roots of 729?
√729 = 27 Therefore: 27 x 2 = 54
