Question Number 21015 by xxyy@gmail.com last updated on 10/Sep/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 21013 by aplus last updated on 10/Sep/17 Commented by aplus last updated on 10/Sep/17 $$\mathrm{hhelp}\:\mathrm{guyselp}\:\mathrm{guys} \\ $$ Answered by Tinkutara last updated on…
Question Number 21012 by Tinkutara last updated on 10/Sep/17 $$\mathrm{A}\:\mathrm{spring}\:\mathrm{with}\:\mathrm{one}\:\mathrm{end}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{mass}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{to}\:\mathrm{a}\:\mathrm{rigid}\:\mathrm{support}\:\mathrm{is} \\ $$$$\mathrm{stretched}\:\mathrm{and}\:\mathrm{released}. \\ $$$$\left({a}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{acceleration},\:\mathrm{when} \\ $$$$\mathrm{just}\:\mathrm{released}\:\mathrm{is}\:\mathrm{maximum}. \\ $$$$\left({b}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{acceleration},\:\mathrm{when} \\ $$$$\mathrm{at}\:\mathrm{equilibrium}\:\mathrm{position},\:\mathrm{is}\:\mathrm{maximum}. \\ $$$$\left({c}\right)\:\mathrm{Speed}\:\mathrm{is}\:\mathrm{maximum}\:\mathrm{when}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{at} \\…
Question Number 21009 by xxyy@gmail.com last updated on 10/Sep/17 Answered by Tinkutara last updated on 10/Sep/17 $$\mathrm{2cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{16}}×\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{3}\pi}{\mathrm{16}}×\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{5}\pi}{\mathrm{16}}×\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{7}\pi}{\mathrm{16}} \\ $$$$=\left(\mathrm{4cos}\:\frac{\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{16}}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{4cos}\:\frac{\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\mathrm{sin}\:\frac{\pi}{\mathrm{16}}\right)^{\mathrm{2}}…
Question Number 152082 by Tawa11 last updated on 25/Aug/21 Answered by Ar Brandon last updated on 25/Aug/21 $$=\underset{\mathrm{1}\leqslant\mathrm{r}\leqslant\mathrm{80}} {\sum}\mathrm{r}=\mathrm{40}\left(\mathrm{81}\right)=\mathrm{3240} \\ $$ Commented by Tawa11 last…
Question Number 21006 by Tinkutara last updated on 10/Sep/17 $$\mathrm{Let}\:{z}_{\mathrm{1}} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{be}\:\mathrm{two}\:\mathrm{distinct}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\mathrm{and}\:\mathrm{let}\:{z}\:=\:\left(\mathrm{1}\:−\:{t}\right){z}_{\mathrm{1}} \:+\:{tz}_{\mathrm{2}} \:\mathrm{for} \\ $$$$\mathrm{some}\:\mathrm{real}\:\mathrm{number}\:{t}\:\mathrm{with}\:\mathrm{0}\:<\:{t}\:<\:\mathrm{1}.\:\mathrm{If} \\ $$$$\mathrm{arg}\left({w}\right)\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{principal}\:\mathrm{argument} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{non}-\mathrm{zero}\:\mathrm{complex}\:\mathrm{number}\:{w},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mid{z}\:−\:{z}_{\mathrm{1}} \mid\:+\:\mid{z}\:−\:{z}_{\mathrm{2}}…
Question Number 86540 by jagoll last updated on 29/Mar/20 $$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{B}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\:=\: \\ $$$$\mathrm{4}\:\mathrm{cos}\:\left(\frac{\pi+\mathrm{A}}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{\pi+\mathrm{B}}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{\pi−\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\mathrm{where}\:\mathrm{A}+\mathrm{B}+\mathrm{C}\:=\:\pi \\ $$ Commented by jagoll last updated on 29/Mar/20…
Question Number 21005 by Tinkutara last updated on 10/Sep/17 $$\mathrm{If}\:{z}_{\mathrm{1}} \:=\:{a}\:+\:{ib}\:\mathrm{and}\:{z}_{\mathrm{2}} \:=\:{c}\:+\:{id}\:\mathrm{are}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\omega_{\mathrm{1}} \:=\:{a}\:+\:{ic}\:\mathrm{and}\:\omega_{\mathrm{2}} \:=\:{b}\:+\:{id}…
Question Number 86541 by peter frank last updated on 29/Mar/20 Commented by jagoll last updated on 29/Mar/20 $$\mathrm{Q}^{\mathrm{2}} \:=\:\mathrm{PR} \\ $$$$\mathrm{4Q}\:=\:\mathrm{P}+\mathrm{R}\:\Rightarrow\:\mathrm{4ar}\:=\:\mathrm{a}+\mathrm{ar}^{\mathrm{2}} \\ $$$$\mathrm{a}\left(\mathrm{r}^{\mathrm{2}} −\mathrm{4r}+\mathrm{1}\right)\:=\:\mathrm{0} \\…
Question Number 21001 by srinivasaraododda88@gmail.com last updated on 10/Sep/17 $$\begin{vmatrix}{{a}\:\mathrm{1}\:\mathrm{1}}\\{\mathrm{1}\:{b}\:\mathrm{1}}\\{\mathrm{1}\:\mathrm{1}\:{c}}\end{vmatrix}>\mathrm{0}\:{then}\:{showthat}\:{abc}>−\mathrm{8}−\mathrm{99} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com