Question Number 18004 by ajfour last updated on 13/Jul/17 Commented by ajfour last updated on 13/Jul/17 $$\mathrm{The}\:\mathrm{angular}\:\mathrm{bisector}\:\mathrm{of}\:\angle\mathrm{ABC} \\ $$$$\mathrm{intersects}\:\mathrm{side}\:\mathrm{AD}\:\mathrm{at}\:\mathrm{point}\:\mathrm{M},\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{dropped}\:\mathrm{from} \\ $$$$\mathrm{vertex}\:\mathrm{A}\:\mathrm{to}\:\mathrm{side}\:\mathrm{BC}\:\mathrm{cuts}\:\mathrm{BC}\:\mathrm{at} \\ $$$$\mathrm{point}\:\mathrm{N}\:\mathrm{so}\:\mathrm{that}\:\mathrm{BN}=\mathrm{NC}\:\mathrm{and}…
Question Number 18003 by Tinkutara last updated on 13/Jul/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}{A}\centerdot\mathrm{cos2}{A}\centerdot\mathrm{cos2}^{\mathrm{2}} {A}\:…..\:\mathrm{cos}\left(\mathrm{2}^{{n}\:−\:\mathrm{1}} {A}\right), \\ $$$$\mathrm{where}\:{A}\:\in\:{R}\:\mathrm{may}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:−\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{sin}\:\mathrm{2}^{{n}} \:{A}}{\mathrm{2}^{{n}} \:\mathrm{sin}\:{A}} \\…
Question Number 83539 by jagoll last updated on 03/Mar/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}? \\ $$ Commented by john santu last updated on 03/Mar/20 $$\mathrm{domain}\:\mathrm{x}^{\mathrm{2}} −\mathrm{1}>\mathrm{0}\:\Rightarrow\:\mathrm{x}<−\mathrm{1}\:\vee\mathrm{x}\:>\mathrm{1}…
Question Number 17999 by Tinkutara last updated on 13/Jul/17 $$\mathrm{A}\:\mathrm{large}\:\mathrm{number}\:\mathrm{of}\:\mathrm{bullets}\:\mathrm{are}\:\mathrm{fired}\:\mathrm{in} \\ $$$$\mathrm{all}\:\mathrm{direction}\:\mathrm{with}\:\mathrm{same}\:\mathrm{speed}\:{u}.\:\mathrm{The} \\ $$$$\mathrm{maximum}\:\mathrm{area}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{covered} \\ $$$$\mathrm{by}\:\mathrm{these}\:\mathrm{bullets}\:\mathrm{will}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\pi.\frac{{u}^{\mathrm{2}} }{{g}} \\ $$$$\left(\mathrm{2}\right)\:\pi.\frac{{u}^{\mathrm{4}} }{{g}^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{\pi}{\mathrm{4}}.\frac{{u}^{\mathrm{4}}…
Question Number 149068 by Tawa11 last updated on 02/Aug/21 Answered by mr W last updated on 03/Aug/21 Commented by mr W last updated on 02/Aug/21…
Question Number 17997 by 433 last updated on 13/Jul/17 $${Solve}\:{on}\:\mathbb{Z}_{\mathrm{18}} \\ $$$$\begin{cases}{\left[\mathrm{6}\right]_{\mathrm{18}} {x}+\left[\mathrm{7}\right]_{\mathrm{18}} {y}=\left[\mathrm{1}\right]_{\mathrm{18}} }\\{\left[\mathrm{2}\right]_{\mathrm{18}} {x}+\left[\mathrm{3}\right]_{\mathrm{18}} {y}=\left[\mathrm{11}\right]_{\mathrm{18}} }\end{cases} \\ $$ Terms of Service Privacy Policy…
Question Number 17995 by tawa tawa last updated on 13/Jul/17 $$\mathrm{A}\:\mathrm{pendulum}\:\mathrm{bob}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2kg}\:\mathrm{is}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{a}\:\mathrm{string}\:\:\mathrm{2m}\:\mathrm{long}\:\mathrm{and}\:\mathrm{made}\:\mathrm{to} \\ $$$$\mathrm{revolve}\:\mathrm{in}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{0}.\mathrm{8}\:,\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{string}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 149061 by jlewis last updated on 02/Aug/21 $$\mathrm{factorise}\:\:\:\:\mathrm{4}/\mathrm{5}^{\mathrm{x}} +\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 17991 by alex041103 last updated on 13/Jul/17 $${Evaluate}\:\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+…}}}} \\ $$ Commented by alex041103 last updated on 13/Jul/17 $${just}\:{for}\:{fun} \\ $$$${whoever}\:{wants}\:{can}\:{try}\:{to}\:{generalize} \\ $$$$\sqrt{{A}+{B}_{\mathrm{1}} \sqrt{{A}+{B}_{\mathrm{2}}…
Question Number 149060 by mathdanisur last updated on 02/Aug/21 $${if}\:\:\:\mathrm{4}{z}\sqrt{{z}}\:−\:\mathrm{11}\sqrt{{z}}\:=\:\mathrm{5} \\ $$$${find}\:\:\:\mathrm{2}{z}\:−\:\sqrt{{z}}\:=\:? \\ $$ Answered by bramlexs22 last updated on 02/Aug/21 $$\Rightarrow\sqrt{\mathrm{z}}\:=\:\mathrm{u}\: \\ $$$$\Rightarrow\mathrm{4z}\sqrt{\mathrm{z}}−\sqrt{\mathrm{z}}\:=\:\mathrm{5}+\mathrm{10}\sqrt{\mathrm{z}} \\…