Question

Determine o resultado exato ou aproximado com uma casa decimal de cada raiz quadrada a sequir √131 e
√13​

Answer 1

$\Large\boxed{\begin{array}{l}\sf quadrados\,perfeitos\,de\,1\,a\,200:\\\sf 1,4,9,16,25,36,49,64,81,100,121,144,169,196.\\\sf \sqrt{121} < \sqrt{131} < \sqrt{144}\\\sf 11 < \sqrt{131} < 12\\\sf \sqrt{n}\approxeq=\dfrac{n+q}{2\sqrt{q}}\\\sf onde\,q\,\acute e\,o\,quadrado\,perfeito\,mais\,pr\acute oximo\,de\,n\\\sf \sqrt{131}=\dfrac{131+121}{2\sqrt{121}}=\dfrac{252}{2\cdot11}=11,4\end{array}}$$\Large\boxed{\begin{array}{l}\sf\sqrt{13}=\dfrac{13+16}{2\sqrt{16}}=\dfrac{29}{2\cdot4}=3,6\end{array}}$