Question Number 120904 by pooooop last updated on 03/Nov/20 Commented by pooooop last updated on 03/Nov/20 $$????? \\ $$ Answered by MJS_new last updated on…
Question Number 55359 by gunawan last updated on 22/Feb/19 $$\mathrm{Find}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the}\:\mathrm{general}\: \\ $$$$\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{squence} \\ $$$$\mathrm{1},\:\mathrm{2},\:\mathrm{2},\:\mathrm{3},\:\mathrm{3},\:\mathrm{3},\:\mathrm{4},\:\mathrm{4},\:\mathrm{4},\mathrm{4},\:… \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 22/Feb/19 $$\mathrm{1}\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{nos}\:{of}\:{term}\rightarrow\mathrm{1} \\…
Question Number 55358 by gunawan last updated on 22/Feb/19 $$\mathrm{Determine}\:\mathrm{all}\:\mathrm{functions}\:{f}\::\:\mathbb{N}\:\rightarrow\:\mathbb{N}\: \\ $$$$\mathrm{satisfying} \\ $$$${xf}\left({y}\right)+{yf}\left({x}\right)=\left({x}+{y}\right){f}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:{x}\:\mathrm{and}\:{y} \\ $$ Terms of Service Privacy Policy…
Question Number 120882 by shaker last updated on 03/Nov/20 Answered by $@y@m last updated on 03/Nov/20 $$\mathrm{cos}^{\mathrm{3}} {x}+\mathrm{cos}\:{x}=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x} \\ $$$$\mathrm{cos}\:{x}\left(\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{1}\right)=\mathrm{sin}^{\mathrm{2}} {x} \\ $$$$\mathrm{cos}\:{x}\left(\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}}…
Question Number 186419 by mnjuly1970 last updated on 04/Feb/23 $$ \\ $$$$\:\:\mathrm{I}{f}\:,\:\sqrt[{\mathrm{3}}]{\:\mathrm{1}\:−\:{l}\overset{} {{o}g}_{\:\mathrm{2}} \left({x}\right)}\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{1}\overset{} {+}{log}_{\:\mathrm{2}} \left({x}\right)}\:−\mathrm{1}=\mathrm{0}\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:\:{x}\:=\:?\:\:\:\: \\ $$ Answered by cortano1…
Question Number 55326 by tm888 last updated on 21/Feb/19 $${find}\:\:{the}\:{minimum}\:{value} \\ $$$$\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{8}{bc}}}+\frac{{b}}{\:\sqrt{{b}^{\mathrm{2}} +\mathrm{8}{ac}}}+\frac{{c}}{\:\sqrt{{c}^{\mathrm{2}} +\mathrm{8}{ab}}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 21/Feb/19 $${trying}\:{to}\:{solve}……
Question Number 120855 by mnjuly1970 last updated on 03/Nov/20 $$\:\:\:\:\:\:\:\:\:\:\:…\:{elementary}\:\:{calculus}… \\ $$$$\:\:::\:\alpha,\beta\:{are}\:{roots}\:{of}\:\:{equation} \\ $$$$\:\:\:\:\:{of}\::\:{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{define}\:::\:{t}_{{n}} =\alpha^{{n}} −\beta^{{n}} \:\left({n}\geqslant\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{then}\:\:{evaluate}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{A}=\frac{{t}_{\mathrm{10}} −\mathrm{2}{t}_{\mathrm{8}}…
Question Number 55312 by Joel578 last updated on 21/Feb/19 $$\mathrm{Consider}\:\mathrm{the}\:\mathrm{system} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:{z}}\\{\mathrm{2}{x}\:+\:\mathrm{2}{y}\:+\:{z}\:=\:{k}}\end{cases} \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{xy}\:+\:{zk}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{system} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{is}\:… \\ $$ Answered by mr W last…
Question Number 120839 by bramlexs22 last updated on 03/Nov/20 Answered by mindispower last updated on 03/Nov/20 $${x}\in\mathbb{Z}−\left\{\mathrm{1}\right\} \\ $$$$\Rightarrow\underset{{k}=\mathrm{2}} {\overset{{x}−\mathrm{1}} {\sum}}\frac{{x}−{k}}{{x}−\mathrm{1}}=\mathrm{5} \\ $$$$\Rightarrow\left(\frac{{x}−\mathrm{2}+\mathrm{1}}{{x}−\mathrm{1}}\right)\left(\frac{{x}−\mathrm{2}}{\mathrm{2}}\right)=\mathrm{5} \\ $$$$\Rightarrow\mathrm{10}\left({x}−\mathrm{1}\right)=\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)=\mathrm{0}…
Question Number 55300 by Tawa1 last updated on 20/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 21/Feb/19 $${p}=\mathrm{2}\left(\mathrm{2}{x}+{x}\right) \\ $$$$\frac{{dp}}{{dt}}=\mathrm{6}\frac{{dx}}{{dt}} \\ $$$${A}=\mathrm{2}{x}×{x}=\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\frac{{dA}}{{dt}}=\mathrm{4}{x}×\frac{{dx}}{{dt}} \\ $$$$\mathrm{18}=\mathrm{4}×\mathrm{1}×\frac{{dx}}{{dt}}\:\left[\mathrm{2}{x}=\mathrm{2}\:{length}\:{given}\right]…