3025+23310+?=(22)2\sqrt{30\frac25+23\frac{3}{10}+?}=(2\sqrt{2})^23052+23103+?=(22)2$$ആയാൽ? ൻ്റെ സ്ഥാനത്തുള്ള സംഖ്യ ഏത്. A10 ³/10B8⅘C10 ⁷/10D7⅖Answer: A. 10 ³/10 Read Explanation: 3025+23310+?=(22)2\sqrt{30\frac25+23\frac{3}{10}+?}=(2\sqrt{2})^23052+23103+?=(22)21525+23310+?=8\sqrt{\frac{152}{5}+\frac{233}{10}+?}=85152+10233+?=8304+23310+?=8\sqrt{\frac{304+233}{10}+?}=810304+233+?=853710+?=82=64\frac{537}{10}+?=8^2=6410537+?=82=64?=64−53710?=64-\frac{537}{10}?=64−10537=640−53710=\frac{640-537}{10}=10640−537=10310=\frac{103}{10}=10103=10310=10\frac{3}{10}=10103 Read more in App