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Question Number 7630 by sandy_suhendra last updated on 06/Sep/16 Commented by Chantria Math last updated on 06/Sep/16 $${There}\:{are}\:{some}\:{false}\:{in}\:{this} \\ $$$${problem},\: \\ $$$$ \\ $$$$ \\…
Question Number 138703 by mathocean1 last updated on 16/Apr/21 $${Calculate} \\ $$$$\underset{−\frac{\pi}{\mathrm{6}}} {\overset{\mathrm{0}} {\int}}\frac{{cos}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{2}{sinx}}{dx} \\ $$ Commented by SanyamJoshi last updated on 17/Apr/21 $${Calculate}…
Question Number 138702 by peter frank last updated on 16/Apr/21 Answered by mr W last updated on 17/Apr/21 $${y}=\mathrm{sinh}\:{x}+{k}\:\mathrm{cosh}\:{x} \\ $$$$\:\:=\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\:\left(\frac{\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}}\mathrm{sinh}\:{x}+\frac{{k}}{\:\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{cosh}\:{x}\right) \\…
Question Number 7628 by Tawakalitu. last updated on 06/Sep/16 $${By}\:{the}\:{use}\:{of}\:{substitution}\:\:{x}\:=\:\mu^{\mathrm{2}} ,\:{show}\:{that} \\ $$$${the}\:{legendary}\:{equation}\:, \\ $$$$\left(\mathrm{1}\:−\:\mu^{\mathrm{2}} \right){y}''\:−\:\mathrm{2}\mu{y}'\:+\:{n}\left({n}\:+\:\mathrm{1}\right){y}\:=\:\mathrm{0},\: \\ $$$${where}\:{n}\:{is}\:{a}\:{constant}\:{change}\:{to}\:{hyper}\:{geometric} \\ $$$${equation}\:.\:{hence}\:{obtain}\:{the}\:{solution}\:{to}\:{the}\: \\ $$$${resulting}\:{hyper}\:{geometric}\:{differential}\:{equation}\: \\ $$$${by}\:{way}\:{of}\:{comparison}. \\…
Question Number 138696 by mr W last updated on 16/Apr/21 Commented by mr W last updated on 16/Apr/21 $${a}\:{small}\:{mass}\:{is}\:{released}\:{on}\:{the} \\ $$$${interior}\:{surface}\:{of}\:{a}\:{hollow}\:{cone} \\ $$$${at}\:{the}\:{position}\:{as}\:{shown}.\:{find}\:{the} \\ $$$${time}\:{the}\:{mass}\:{needs}\:{to}\:{reach}\:{the}\:{tip}…
Question Number 73160 by Aditya789 last updated on 07/Nov/19 $$\mathrm{1}+{tanAtan}\frac{{A}}{\mathrm{2}}={tanAcot}\frac{{A}}{\mathrm{2}}−\mathrm{1}={secA} \\ $$ Answered by $@ty@m123 last updated on 07/Nov/19 $$\mathrm{1}+\mathrm{tan}\:{A}\mathrm{tan}\:\frac{{A}}{\mathrm{2}} \\ $$$$=\mathrm{1}+\frac{\mathrm{2tan}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}\:.\mathrm{tan}\:\frac{{A}}{\mathrm{2}} \\ $$$$=\mathrm{1}+\frac{\mathrm{2tan}^{\mathrm{2}}…
Question Number 138698 by mathlove last updated on 16/Apr/21 Answered by nimnim last updated on 16/Apr/21 $$\frac{{a}}{{b}}=\mathrm{1}.\mathrm{454545}…..=\frac{\mathrm{145}−\mathrm{1}}{\mathrm{99}}=\frac{\mathrm{144}}{\mathrm{99}}=\frac{\mathrm{16}}{\mathrm{11}} \\ $$$$\Rightarrow{a}=\mathrm{16},\:{b}=\mathrm{11} \\ $$$$\therefore\:{a}+{b}=\mathrm{16}+\mathrm{11}=\mathrm{27} \\ $$ Commented by…
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Question Number 7621 by 314159 last updated on 06/Sep/16 $${Prove}\:{that}\: \\ $$$$\pi=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$ Answered by prakash jain last updated on 26/Sep/16…