There is no pair of perfect squares that sums to 21. And the question is pointless if it is not about perfect squares because in that case there are infinitely many answers.There is no pair of perfect squares that sums to 21. And the question is pointless if it is not about perfect squares because in that case there are infinitely many answers.There is no pair of perfect squares that sums to 21. And the question is pointless if it is not about perfect squares because in that case there are infinitely many answers.There is no pair of perfect squares that sums to 21. And the question is pointless if it is not about perfect squares because in that case there are infinitely many answers.
Difference between the sum of the squares and the square of the sums of n numbers?Read more:Difference_between_the_sum_of_the_squares_and_the_square_of_the_sums_of_n_numbers
When comparing the sums of squares of normal variates.
Yes.
No.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theorem
There is no pair of perfect squares that sums to 21. And the question is pointless if it is not about perfect squares because in that case there are infinitely many answers.There is no pair of perfect squares that sums to 21. And the question is pointless if it is not about perfect squares because in that case there are infinitely many answers.There is no pair of perfect squares that sums to 21. And the question is pointless if it is not about perfect squares because in that case there are infinitely many answers.There is no pair of perfect squares that sums to 21. And the question is pointless if it is not about perfect squares because in that case there are infinitely many answers.
Difference between the sum of the squares and the square of the sums of n numbers?Read more:Difference_between_the_sum_of_the_squares_and_the_square_of_the_sums_of_n_numbers
When comparing the sums of squares of normal variates.
For an array of numbers, it is the square of the sums divided by the sum of the squares.
Yes.
Carlos J. Moreno has written: 'Sums of squares of integers'
No.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theorem
The sums of squares of sqrt(59) and sqrt(2) is 61.
Only in squares and rectangles. In a rhombus, the consecutive angle is supplementary (sums to 180 degrees).
Pythagoras is, of course, best remembered for the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sums of the squares of the other two sides.
There is no single formula.It is necessary to calculate the total sum of squares and the regression sum of squares. These are used to calculate the residual sum of squares. The next step is to use the appropriate degrees of freedom to calculate the mean regression sum of squares and the mean residual sum of squares.The ratio of these two is distributed as Fisher's F statistics with the degrees of freedom which were used to obtain the average sums of squares. The ratio is compared with published values of the F-statistic since there is no simple analytical form for the integral.
The answer will depend on what kind of sums.