Question Number 158425 by HongKing last updated on 04/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 158424 by HongKing last updated on 04/Nov/21 Answered by ghimisi last updated on 04/Nov/21 $${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}+\mathrm{3}\sqrt[{\mathrm{3}}]{{y}}+\mathrm{3}\sqrt[{\mathrm{3}}]{{z}}\:\overset{{am}−{gm}} {\geqslant}\mathrm{12}\sqrt[{\mathrm{12}}]{\mathrm{4}{x}^{\mathrm{4}} {y}^{\mathrm{4}} {z}^{\mathrm{4}} }=\mathrm{12}\Rightarrow \\…
Question Number 92885 by unknown last updated on 09/May/20 Commented by unknown last updated on 09/May/20 $${a}+{b}+{c}=\mathrm{3}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{A} \\ $$ Commented by mr W last updated…
Question Number 92880 by fath035990 last updated on 09/May/20 $$\mathrm{solve}\:\mathrm{8}\varkappa+\mathrm{4}=\mathrm{3}\left(\varkappa−\mathrm{1}\right)+\mathrm{7} \\ $$ Answered by niroj last updated on 09/May/20 $$\:\:\:\mathrm{8}\varkappa+\mathrm{4}=\mathrm{3}\left(\varkappa−\mathrm{1}\right)+\mathrm{7} \\ $$$$\:\:\mathrm{or},\:\mathrm{8}\varkappa+\mathrm{4}=\mathrm{3}\varkappa−\mathrm{3}+\mathrm{7} \\ $$$$\:\:\mathrm{or},\:\:\mathrm{8}\varkappa−\mathrm{3}\varkappa=\:\mathrm{4}−\mathrm{4}\:\:\: \\…
Question Number 158410 by tebohlouis last updated on 03/Nov/21 Answered by MJS_new last updated on 03/Nov/21 $$\mathrm{2}{x}+\mathrm{3}\geqslant\mathrm{0}\wedge\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)>\mathrm{0}\:\vee\:\mathrm{2}{x}+\mathrm{3}\leqslant\mathrm{0}\wedge\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)<\mathrm{0} \\ $$$$ \\ $$$$\mathrm{2}{x}+\mathrm{3}\geqslant\mathrm{0}\:\Rightarrow\:{x}\geqslant−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)>\mathrm{0}\:\Rightarrow\:{x}<−\mathrm{1}\vee{x}>\mathrm{2} \\ $$$$\Rightarrow\:−\frac{\mathrm{3}}{\mathrm{2}}\leqslant{x}<−\mathrm{1}\vee{x}>\mathrm{2}…
Question Number 158383 by HongKing last updated on 03/Nov/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{4}} \left(\boldsymbol{\mathrm{x}}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{10}}\:=\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)} \\ $$$$ \\ $$ Answered by MJS_new last updated on 03/Nov/21…
Question Number 158378 by Rasheed.Sindhi last updated on 03/Nov/21 $${How}\:{to}\:{graph}\:{order}\:{pair}\:\left(\mathrm{3}+\mathrm{5}{i}\:,\:\mathrm{4}−\mathrm{2}{i}\right)? \\ $$ Answered by MJS_new last updated on 03/Nov/21 $$\mathrm{you}\:\mathrm{would}\:\mathrm{need}\:\mathrm{a}\:\mathrm{4}−\mathrm{dimensional}\:\mathrm{coordinate} \\ $$$$\mathrm{system}… \\ $$ Commented…
Question Number 27300 by mondodotto@gmail.com last updated on 04/Jan/18 Commented by mondodotto@gmail.com last updated on 04/Jan/18 $$\mathrm{any}\:\mathrm{one}\:\mathrm{to}\:\mathrm{help}\:\mathrm{this}\:\mathrm{please} \\ $$ Answered by bsayani309@gmail.com last updated on…
Question Number 92839 by i jagooll last updated on 09/May/20 $$\mathrm{let}\:\mathrm{a}\:\mathrm{is}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\: \\ $$$$\mathrm{that}\:\mathrm{a}^{\mathrm{10}} \:+\:\mathrm{a}^{\mathrm{5}} \:+\mathrm{1}\:=\:\mathrm{0}. \\ $$$$\mathrm{find}\:\mathrm{a}^{\mathrm{2005}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2005}} }\:? \\ $$ Answered by john santu…
Question Number 92835 by unknown last updated on 09/May/20 Commented by unknown last updated on 09/May/20 $$\mathrm{let}\:{a},{b},{c}\:\mathrm{is}\:\mathrm{real}\:\mathrm{positive}\:\mathrm{number}.\:{a}+{b}+{c}=\mathrm{3} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:: \\ $$$${A}=\frac{\mathrm{2}−{a}^{\mathrm{3}} }{{a}}+\frac{\mathrm{2}−{b}^{\mathrm{3}} }{{b}}+\frac{\mathrm{2}−{c}^{\mathrm{3}} }{{c}} \\…