Question Number 151961 by Tawa11 last updated on 24/Aug/21 Commented by Tawa11 last updated on 24/Aug/21 $$\mathrm{The}\:\mathrm{pencils}\:\mathrm{has}\:\mathrm{diameter}\:\:\:\mathrm{7mm}. \\ $$$$ \\ $$$$\mathrm{Question} \\ $$Seven pencils are…
Question Number 86426 by M±th+et£s last updated on 28/Mar/20 $${solve}\:{in}\:{R} \\ $$$${x}^{\mathrm{3}} −\mathrm{5}=\left[{x}\right] \\ $$ Answered by mr W last updated on 28/Mar/20 $${let}\:{x}={n}+\delta \\…
Question Number 20891 by Tinkutara last updated on 06/Sep/17 $$\mathrm{The}\:\mathrm{Figure}\:\mathrm{shows}\:\mathrm{a}\:\mathrm{system}\:\mathrm{consisting} \\ $$$$\mathrm{of}\:\left({i}\right)\:\mathrm{a}\:\mathrm{ring}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{radius}\:\mathrm{3}{R}\:\mathrm{rolling} \\ $$$$\mathrm{clockwise}\:\mathrm{without}\:\mathrm{slipping}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{with}\:\mathrm{angular}\:\mathrm{speed} \\ $$$$\omega\:\mathrm{and}\:\left({ii}\right)\:\mathrm{an}\:\mathrm{inner}\:\mathrm{disc}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{2}{R} \\ $$$$\mathrm{rotating}\:\mathrm{anti}-\mathrm{clockwise}\:\mathrm{with}\:\mathrm{angular} \\ $$$$\mathrm{speed}\:\omega/\mathrm{2}.\:\mathrm{The}\:\mathrm{ring}\:\mathrm{and}\:\mathrm{disc}\:\mathrm{are} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\mathrm{frictionless}\:\mathrm{ball}\:\mathrm{bearing}. \\…
Question Number 151956 by mathdanisur last updated on 24/Aug/21 $$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2k}^{\mathrm{2}} }\:\mathrm{arctan}\:\frac{\mathrm{2k}^{\mathrm{2}} \:-\:\mathrm{1}}{\mathrm{2k}^{\mathrm{2}} }\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20886 by tammi last updated on 06/Sep/17 $${if}\:\mathrm{sin}\:{x}={m}\mathrm{sin}\:{y} \\ $$$${so}\:{proof}\:{that} \\ $$$$\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)=\frac{{m}−\mathrm{1}}{{m}+\mathrm{1}}\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}\right) \\ $$ Answered by ajfour last updated on 06/Sep/17 $${m}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{y}}\:\:\:\:\Rightarrow\:\:\frac{{m}−\mathrm{1}}{{m}+\mathrm{1}}=\frac{\mathrm{sin}\:{x}−\mathrm{sin}\:{y}}{\mathrm{sin}\:{x}+\mathrm{sin}\:{y}} \\…
Question Number 151959 by john_santu last updated on 24/Aug/21 Answered by iloveisrael last updated on 24/Aug/21 $$\left(\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} +\left(\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{1}\right)\left(\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}−\left(\mathrm{sin}\:{x}\mathrm{cos}\:{x}\right)^{\mathrm{2}}…
Question Number 20885 by tammi last updated on 06/Sep/17 $${if}\:\left(\theta−\varphi\right){subtle}\:{and}\:\:\:\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi= \\ $$$$\left(\mathrm{cos}\:\varphi−\mathrm{cos}\:\theta\right)\sqrt{\mathrm{3}} \\ $$$${so}\:{proof}\:\mathrm{sin}\:\mathrm{3}\theta+\mathrm{sin}\:\mathrm{3}\varphi=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151958 by mnjuly1970 last updated on 24/Aug/21 $${f}\:\left(\:{x}\:\right)\:=\:{a}\:−\sqrt{\frac{{x}}{\mathrm{1}+{x}}\:}\:\:\:,\:{D}_{\:{f}} \::\:\left[\:\mathrm{0},\:\infty\right) \\ $$$$,\:{a}\geqslant\:\mathrm{1}\:\:\:,\:\:{h}\:\left({x}\:\right):=\sqrt{\frac{\:{f}^{\:−\mathrm{1}} \left({a}−{ax}\:\right)}{{f}^{\:−\mathrm{1}} \left(\:{a}−\:\mathrm{2}{x}\:\right)}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{D}_{\:{h}} \:=\:?\:\:\:\left(\:\:\:{D}\::=\:{Domain}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$ Terms of Service…
Question Number 20884 by tammi last updated on 06/Sep/17 $$\mathrm{2cos}\:\frac{\pi}{\mathrm{3}}\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{13}}=\mathrm{0} \\ $$ Answered by ajfour last updated on 06/Sep/17 $${must}\:{start}\:{with}\:\:\:\:\mathrm{2cos}\:\frac{\pi}{\mathrm{13}}…,\:{Then} \\ $$$$\:{L}.{H}.{S}.\:=\:\mathrm{cos}\:\left(\frac{\mathrm{10}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{13}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{13}}\right) \\…
Question Number 20882 by soufiane zarik last updated on 06/Sep/17 Answered by $@ty@m last updated on 08/Sep/17 Commented by soufiane zarik last updated on 11/Sep/17…