Question Number 87737 by M±th+et£s last updated on 05/Apr/20 $${solve} \\ $$$${sin}\left(\frac{\pi}{\left[\frac{\left[{x}\right]}{\mathrm{4}}\right]}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by mahdi last updated on 06/Apr/20 $$\mathrm{u}=\left[\frac{\left[\mathrm{x}\right]}{\mathrm{4}}\right]\Rightarrow−\mathrm{1}\leqslant\frac{\mathrm{1}}{\mathrm{u}}\leqslant\mathrm{1}\Rightarrow−\pi\leqslant\frac{\pi}{\mathrm{u}}\leqslant\pi\:\:\:\left\{\mathrm{u}\neq\mathrm{0}\Rightarrow\mathrm{x}\notin\left[\mathrm{0},\mathrm{4}\right)\right\} \\ $$$$\mathrm{sin}\left(\frac{\pi}{\mathrm{u}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\begin{cases}{\frac{\pi}{\mathrm{u}}=\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi}\\{\frac{\pi}{\mathrm{u}}=\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2k}\pi}\end{cases} \\…
Question Number 87733 by TawaTawa1 last updated on 05/Apr/20 Commented by mahdi last updated on 05/Apr/20 $$\mathrm{27y}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} }=\left(\mathrm{3y}+\frac{\mathrm{1}}{\mathrm{y}}\right)\left(\mathrm{9y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }−\mathrm{3}\right)= \\ $$$$\left(\mathrm{3y}+\frac{\mathrm{1}}{\mathrm{y}}\right)\left(\mathrm{3}−\mathrm{3}\right)=\mathrm{0} \\ $$…
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Question Number 87724 by mr W last updated on 05/Apr/20 $${solve}\:{the}\:{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{cos}}\:\lfloor\boldsymbol{{x}}\rfloor\right)=\mathrm{1} \\ $$ Answered by mahdi last updated on 05/Apr/20 $$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\left[\mathrm{x}\right]\right)=\mathrm{1}\Rightarrow\mathrm{cos}\left[\mathrm{x}\right]=\mathrm{sin}\left(\mathrm{1}+\mathrm{2k}\pi\right)…
Question Number 153263 by mnjuly1970 last updated on 06/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 153256 by Lekhraj last updated on 06/Sep/21 Commented by MJS_new last updated on 06/Sep/21 $${x}<{y} \\ $$$${x}=\frac{\mathrm{25}}{\mathrm{9}{t}}\wedge{y}=\frac{\mathrm{25}{t}}{\mathrm{9}} \\ $$$${t}=\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\:\Rightarrow\:\frac{{x}}{{y}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\left(\mathrm{for}\:{y}<{x}\:\mathrm{we}\:\mathrm{get}\:\frac{{x}}{{y}}=\frac{\mathrm{5}}{\mathrm{3}}\right) \\ $$…
Question Number 22177 by Tinkutara last updated on 12/Oct/17 $$\frac{{C}_{\mathrm{0}} }{\mathrm{2}}\:−\:\frac{{C}_{\mathrm{1}} }{\mathrm{3}}\:+\:\frac{{C}_{\mathrm{2}} }{\mathrm{4}}\:−\:\frac{{C}_{\mathrm{3}} }{\mathrm{5}}\:+\:………. \\ $$ Answered by ajfour last updated on 12/Oct/17 $${x}\left(\mathrm{1}−{x}\right)^{{n}} ={C}_{\mathrm{0}}…
Question Number 153245 by liberty last updated on 06/Sep/21 $$\:\:\begin{cases}{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} {y}=\mathrm{30}}\\{{y}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} =\mathrm{10}}\end{cases} \\ $$$$\:\left({x},{y}\right)=? \\ $$ Answered by Rasheed.Sindhi last updated on 06/Sep/21…
Question Number 22154 by Joel577 last updated on 12/Oct/17 $$\mathrm{Given}\:{a},{b},{c}\:\mathrm{real}\:\mathrm{and}\:\mathrm{positive}\:\mathrm{numbers},\:\mathrm{and} \\ $$$${a}\:+\:{b}\:+\:{c}\:=\:\mathrm{1} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\:\frac{{a}\:+\:{b}}{{abc}} \\ $$ Answered by ajfour last updated on 12/Oct/17 $${let}\:{a}={cx}\:\:\:\:{and}\:\:{b}={cy} \\…
Question Number 153208 by mathdanisur last updated on 05/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com