Question

1/49x^2+81y^2-18/7xy Find the value when x=3 and y=-3

Answer 1

Answer:

$here \: given \\ \frac{1}{49} {x}^{2} + 81 {y}^{2} - \frac{18}{7}xy \\ now \: we \: have \: put \: the \: value \\ x = 3 \: \: y = - 3 \\ so \\ = \frac{1}{49} \times {3}^{2} + 81 \times { (- 3)}^{2} - \frac{18}{7} \times 3 \times ( - 3) \\ = \frac{9}{49} + 729 + \frac{162}{7} \\ \\ = \frac{9 + 35721 + 1134}{49} = \frac{36864}{49} = 752.32$