The Pythagorean theorem is `a^2 + b^2 = c^2`. Solbe for a.

The mean value theorem is f(b) - f(a) = (b-a) f'( c) . Then for the function x^(2) - 2x + 3 , in [1,3/2] , the value of c:

If a and b are natural numbers and a gt b, then show that (a^(2) + b^(2)), (a^(2) - b^(2)),(2ab) is a pythagorean triplet. Find two Pythagorean triplets using any convenient values of a and b.

If a and b are natural numbers and a gt b, then show that (a^(2)+b^(2)),(a^(2)-b^(2))_,(2ab) is a Pythagorean triplet. Find two Pythagorean triplets using any convenient values of a and b.

If a and b are natural numbers and a > b, then show that (a^(2) + b^(2)), (a^(2) - b^(2)),(2ab) is a pythagorean triplet. Find two Pythagorean triplets using any convenient values of a and b.

Using factor theorem, show that a - b is the factor of a(b^2 -c^2)+ b(c^2 -a^2)+ c(a^2 - b^2) .

Using the factor theorem it is found that a+b , b+c and c+a are three factors of the determinant |[-2a ,a+b, a+c],[ b+a,-2b,b+c],[c+a ,c+b,-2c]| . The other factor in the value of the determinant is

Using the factor theorem it is found that b + c , c + a and a + b are three factors of the determine : |(-2a,a+b,a+c),(b+a,-2b,b+c),(c+a,c+b,-2c)| . The other factor in the value of the determine is :

Using the factor theorem it is found that a+b , b+c and c+a are three factors of the determinant |[-2a ,a+b, a+c],[ b+a,-2b,b+c],[c+a ,c+b,-2c]|. The other factor in the value of the determinant is

Using the factor theorem it is found that a+b , b+c and c+a are three factors of the determinant |[-2a, a+b ,a+c],[ b+a,-2b,b+c],[c+a, c+b, -2c]|dot The other factor in the value of the determinant is (a) 4 (b) 2 (c) a+b+c (d) none of these