Question Number 21498 by Joel577 last updated on 25/Sep/17 $${a}^{−\mathrm{2}\:} +\:{b}^{\mathrm{3}} \:+\:{c}^{−\mathrm{4}} \:=\:\frac{\mathrm{433}}{\mathrm{499}} \\ $$$$\mathrm{Find}\:{a}\:+\:{b}\:+\:{c} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 87028 by peter frank last updated on 02/Apr/20 $${solve}\:{D}.{E} \\ $$$$\frac{{dy}}{{dx}}+\frac{{y}\mathrm{sec}\:^{\mathrm{3}} {y}}{{x}}={x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 21493 by agni5 last updated on 25/Sep/17 $$\mathrm{Is}\:\mathrm{always}\:\mathrm{a}\boldsymbol{\div}\mathrm{b}\:=\:\frac{\mathrm{a}}{\mathrm{b}}\:? \\ $$ Answered by Joel577 last updated on 25/Sep/17 $${Yes} \\ $$ Terms of Service…
Question Number 21471 by Joel577 last updated on 24/Sep/17 $$\mathrm{If}\:\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0},\:\mathrm{then} \\ $$$$\frac{\left({a}\:+\:{b}\right)\left({b}\:+\:{c}\right)\left({a}\:+\:{c}\right)}{{abc}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:… \\ $$ Answered by Tinkutara last updated on 24/Sep/17 $${a}+{b}=−{c} \\ $$$$\therefore\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{{abc}}=\frac{\left(−{c}\right)\left(−{a}\right)\left(−{b}\right)}{{abc}}=−\mathrm{1} \\…
Question Number 87009 by mr W last updated on 01/Apr/20 $${solve} \\ $$$$\mathrm{7}\lfloor{x}+\mathrm{3}\rfloor^{\mathrm{2}} −\mathrm{3}\lfloor{x}\rfloor+\mathrm{6}=\mathrm{5}\:{mod}\:\mathrm{11} \\ $$ Answered by MJS last updated on 01/Apr/20 $$\lfloor{x}+\mathrm{3}\rfloor=\lfloor{x}\rfloor+\mathrm{3} \\…
Question Number 152543 by liberty last updated on 29/Aug/21 $$\:\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}=\sqrt{\mathrm{x}+\mathrm{2}} \\ $$ Answered by EDWIN88 last updated on 29/Aug/21 $${it}\:{is}\:{clear}\:{for}\:{x}\geqslant−\mathrm{2} \\ $$$${case}\left(\mathrm{1}\right)\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{2}\:,{let}\:{x}=\mathrm{2cos}\:{y}\:,\mathrm{0}\leqslant{y}\leqslant\pi…
Question Number 21470 by Joel577 last updated on 24/Sep/17 $$\mathrm{2}^{{x}} \:=\:\mathrm{3}^{{y}} \:=\:\mathrm{6}^{−{z}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\left(\frac{\mathrm{2017}}{{x}}\:+\:\frac{\mathrm{2017}}{{y}}\:+\:\frac{\mathrm{2017}}{{z}}\right)^{\mathrm{2017}} \\ $$ Commented by Joel577 last updated on 24/Sep/17…
Question Number 21463 by dioph last updated on 24/Sep/17 $$\mathrm{Let}\:{A}\:\mathrm{be}\:\mathrm{the}\:\mathrm{collection}\:\mathrm{of}\:\mathrm{functions} \\ $$$${f}\::\:\left[\mathrm{0},\:\mathrm{1}\right]\:\rightarrow\:\mathbb{R}\:\mathrm{which}\:\mathrm{have}\:\mathrm{an}\:\mathrm{infinite} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{derivatives}.\:\mathrm{Let}\:{A}_{\mathrm{0}} \:\subset\:{A} \\ $$$$\mathrm{be}\:\mathrm{the}\:\mathrm{subcollection}\:\mathrm{of}\:\mathrm{those}\:\mathrm{functions} \\ $$$${f}\:\mathrm{with}\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{0}.\:\mathrm{Define}\:{D}\::\:{A}_{\mathrm{0}} \:\rightarrow\:{A} \\ $$$$\mathrm{by}\:{D}\left({f}\right)\:=\:{df}/{dx}.\:\mathrm{Use}\:\mathrm{the}\:\mathrm{mean}\:\mathrm{value} \\ $$$$\mathrm{theorem}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\:{D}\:\mathrm{is}\:\mathrm{injective}. \\…
Question Number 152498 by mathdanisur last updated on 28/Aug/21 $$\mathrm{if}\:\:\left(\sqrt{\mathrm{5}}\:+\:\mathrm{2}\right)^{\mathrm{6}} \:<\:\mathrm{x} \\ $$$$\mathrm{find}\:\:\mathrm{min}\left(\mathrm{x}\right)\:=\:? \\ $$ Commented by imjagoll last updated on 29/Aug/21 $$\mathrm{5778} \\ $$…
Question Number 21422 by Tinkutara last updated on 23/Sep/17 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{integer}\:\mathrm{values}\:\mathrm{of}\:{a}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{quadratic}\:\mathrm{expression} \\ $$$$\left({x}\:+\:{a}\right)\left({x}\:+\:\mathrm{1991}\right)\:+\:\mathrm{1}\:\mathrm{can}\:\mathrm{be}\:\mathrm{factored} \\ $$$$\mathrm{as}\:\mathrm{a}\:\mathrm{product}\:\left({x}\:+\:{b}\right)\left({x}\:+\:{c}\right)\:\mathrm{where}\:{b}\:\mathrm{and} \\ $$$${c}\:\mathrm{are}\:\mathrm{integers}. \\ $$ Commented by Tikufly last updated…