Question

raiz quadrada de 2490

Answer 1

$\mathsf{F\acute ormula\!\!:\,\sqrt n\approxeq \dfrac{n+Q}{2\sqrt Q}}$

$\mathsf{\sqrt{2490}\approxeq \dfrac{ 2490+2500}{2\sqrt{2500}}=\dfrac{4990}{2\cdot50}}$

$\boxed{\boxed{\mathsf{\sqrt{2490}\approxeq\dfrac{4990}{100}= 49{,}9 }}}$