Question Number 224042 by Tawa11 last updated on 15/Aug/25 Answered by mr W last updated on 16/Aug/25 $${let}\:{a}_{{n}} ={b}_{{n}} +{k} \\ $$$${b}_{{n}+\mathrm{2}} +{k}=\mathrm{2}\left({b}_{{n}+\mathrm{1}} +{k}\right)+{b}_{{n}} +{k}−\mathrm{1}…
Question Number 224043 by ghnaseri last updated on 15/Aug/25 $${f}\left({x}\right)=\left(\sqrt{{x}−\mathrm{5}}\right)^{\mathrm{0}} \\ $$$${Dom}_{{f}} =? \\ $$ Answered by Jyrgen last updated on 16/Aug/25 $$\mathrm{0}^{\mathrm{0}} \:{is}\:{not}\:{defined} \\…
Question Number 224036 by Buck2233Henry last updated on 15/Aug/25 $$\:\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{xy}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 224033 by fantastic last updated on 15/Aug/25 $${HAPPY}\:{INDEPENDENCE} \\ $$$${DAY}! \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 224034 by MirHasibulHossain last updated on 15/Aug/25 $$\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{18}\sqrt{\mathrm{3}}\:.\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$ Answered by BaliramKumar last updated on 15/Aug/25 $${x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{11}\sqrt{\mathrm{2}}\:+\mathrm{9}\sqrt{\mathrm{3}}\:\:} \\ $$ Answered…
Question Number 224013 by efronzo1 last updated on 14/Aug/25 Commented by efronzo1 last updated on 14/Aug/25 $$\mathrm{Given}\:\mathrm{AB}\:\mathrm{is}\:\mathrm{diameter}\:,\:\angle\mathrm{AOD}=\mathrm{22}° \\ $$$$\:\mathrm{and}\:\angle\mathrm{BCD}=\:\mathrm{p}°. \\ $$$$\:\mathrm{Find}\:\mathrm{p}. \\ $$ Answered by…
Question Number 224015 by hardmath last updated on 14/Aug/25 $$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}=\mathrm{abc} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\:+\:\sqrt{\mathrm{c}}\:\leqslant\:\frac{\mathrm{3}}{\mathrm{2}}\:\sqrt{\mathrm{abc}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 224025 by Simurdiera last updated on 14/Aug/25 $${Resuelve}\:{la}\:{ecuaci}\acute {{o}n}\:{diferencial} \\ $$$$\left[\mathrm{4}{x}^{\mathrm{3}} {y}\:−\:\frac{{e}^{{xy}} }{{x}}\:+\:{y}\:\mathrm{ln}\left({x}\right)\:+\:{x}\:\sqrt[{\mathrm{3}}]{{x}\:−\:\mathrm{4}}\right]{dx}\:+\:\left[{x}^{\mathrm{4}} −\:\frac{{e}^{{xy}} }{{y}}\:+\:{x}\:\mathrm{ln}\left({x}\right)\:−\:{x}\right]{dy} \\ $$$${Help}\:…. \\ $$ Terms of Service Privacy…
Question Number 224016 by hardmath last updated on 14/Aug/25 $$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}=\mathrm{abc} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{a}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{b}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{c}}}\:\leqslant\:\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 224017 by hardmath last updated on 14/Aug/25 $$\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0} \\ $$$$\mathrm{xy}+\mathrm{yz}+\mathrm{zx}+\mathrm{2xyz}=\mathrm{1} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }\:\leqslant\:\frac{\mathrm{3}\:\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ Terms of Service Privacy…