Find the sum of: 137 and -354

The correct option is (B) 581.

Solution: We have given that sum of zeroes(S)= -2√3 and product of zeroes(P) = -9 so the quadratic polynomial p(x)= k(x2 - Sx +P) p(x) = k(x2 - (-2√3)x + (-9)) p(x) = k(x2 + 2√3x - 9) where  k  is...

137 + 0.05 = 137.05

If the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number is either...

We know that the sum of first n odd natural numbers is n2. (i) 49 = 72 Therefore, 49 is the sum of first 7 odd natural numbers. 49 = 1 +...

The sum of product ab+cd The product of sum (a+b)(c+d)

Correct answer is (B) 137

We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial...

Let the larger number be x and the smaller number be y.  Then, we have:  x + y = 137 ……….(i)  x - y = 43 ………(ii)  On adding (i) and (ii), we get  2x...

Let the larger number be x and the smaller number be y.  Then, we have:  x + y = 16 ……(i)  And, 1/x + 1/y = 1/3 ……(ii)  ⇒3(x + y) = xy  ⇒3 × 16 =...