Question Number 158983 by HongKing last updated on 11/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 158982 by HongKing last updated on 11/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 158978 by physicstutes last updated on 11/Nov/21 Commented by Rasheed.Sindhi last updated on 11/Nov/21 $$\mathrm{160}\:{don}'{t}\:{play}\:{any}\:{game}. \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 93439 by i jagooll last updated on 13/May/20 Answered by niroj last updated on 13/May/20 $$\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{2a}−\frac{\mathrm{8b}+\mathrm{1}}{\mathrm{4b}^{\mathrm{2}} } \\ $$$$\:\:\mathrm{6b}^{\mathrm{2}} +\mathrm{5a}=? \\…
Question Number 158961 by ajfour last updated on 11/Nov/21 Commented by ajfour last updated on 11/Nov/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Find}\:{r}\:{given},\:{c}. \\ $$ Commented by 1549442205PVT last updated on…
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Question Number 158919 by quvonnn last updated on 10/Nov/21 Answered by floor(10²Eta[1]) last updated on 10/Nov/21 $$\mathrm{4x}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{4x}}}}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+…}}}}} \\ $$$$\mathrm{4x}=\sqrt{\mathrm{x}+\mathrm{4x}}=\sqrt{\mathrm{5x}} \\ $$$$\mathrm{16x}^{\mathrm{2}} =\mathrm{5x}\Rightarrow\mathrm{x}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}=\frac{\mathrm{5}}{\mathrm{16}} \\ $$$$ \\…
Question Number 158916 by HongKing last updated on 10/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 158918 by quvonnn last updated on 10/Nov/21 Commented by quvonnn last updated on 10/Nov/21 $$???\: \\ $$ Terms of Service Privacy Policy Contact:…