Question Number 12092 by agni5 last updated on 13/Apr/17 $$\mathrm{Prove}\:\mathrm{that}\:\:\mathrm{tan}\:\mathrm{70}°−\mathrm{tan}\:\mathrm{50}°+\mathrm{tan}\:\mathrm{10}°\:=\sqrt{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 143163 by slahadjb last updated on 10/Jun/21 $$\int_{\frac{\mathrm{1}}{{x}}} ^{{x}^{\mathrm{2}} } \frac{{dt}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{3}} }}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12087 by tawa last updated on 12/Apr/17 $$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{10}\:\mathrm{objects}\:\mathrm{be}\:\mathrm{split}\:\mathrm{into}\:\mathrm{two}\:\:\mathrm{groups}\:\mathrm{containing}\: \\ $$$$\mathrm{4}\:\mathrm{and}\:\mathrm{6}\:\mathrm{objects}\:\mathrm{respectively}\:? \\ $$ Answered by sandy_suhendra last updated on 12/Apr/17 $$\mathrm{10C4}\:=\:\frac{\mathrm{10}!}{\mathrm{6}!\:\mathrm{4}!}\:=\mathrm{210} \\ $$$$\mathrm{or}\:\:\mathrm{10C6}\:=\:\mathrm{210} \\…
Question Number 143156 by mohammad17 last updated on 10/Jun/21 $${lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\left(\mathrm{1}+\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +……+{n}^{\frac{\mathrm{1}}{{n}}} \right)\:\:? \\ $$ Answered by mr W last updated on 10/Jun/21 $$\mathrm{1}=\frac{{n}×\mathrm{1}}{{n}}<{S}_{{n}} <\frac{{n}×{n}^{\frac{\mathrm{1}}{{n}}}…
Question Number 12085 by chux last updated on 12/Apr/17 Answered by ajfour last updated on 12/Apr/17 $${let}\:\mathrm{6}°=\theta \\ $$$$\mathrm{sin}\:\mathrm{24}°=\mathrm{2sin}\:\mathrm{12}°\mathrm{cos}\:\mathrm{12}° \\ $$$${let}\:\mathrm{sin}\:\mathrm{12}°={x}\:\:{and}\:\:\mathrm{cos}\:\mathrm{12}°={y} \\ $$$$\mathrm{sin}\:\left(\mathrm{3}\theta+\mathrm{2}\theta\right)=\:{y}\:\mathrm{sin}\:\mathrm{18}°+{x}\:\mathrm{cos}\:\mathrm{18}° \\ $$$$\Rightarrow\:{x}\mathrm{cos}\:\mathrm{18}°+{y}\mathrm{sin}\:\mathrm{18}°\:=\mathrm{sin}\:\mathrm{5}\theta=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 12082 by chux last updated on 12/Apr/17 Answered by ajfour last updated on 12/Apr/17 $$\mathrm{t}=\frac{\mathrm{s}}{\mathrm{v}}=\frac{\mathrm{s}}{\left(\mathrm{c}/\mu\right)}=\frac{\mu\mathrm{s}}{\mathrm{c}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(\mathrm{1}.\mathrm{5}\right)\left(\mathrm{0}.\mathrm{9}×\mathrm{10}^{−\mathrm{2}} \mathrm{m}\right)}{\left(\mathrm{3}×\mathrm{10}^{\mathrm{8}} \mathrm{m}/\mathrm{s}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{45}×\mathrm{10}^{−\mathrm{12}} \mathrm{s} \\…
Question Number 77616 by BK last updated on 08/Jan/20 Commented by MJS last updated on 08/Jan/20 $$\mathrm{are}\:\mathrm{you}\:\mathrm{serious}\:\mathrm{on}\:\mathrm{this}\:\mathrm{one}? \\ $$$$\mathrm{serious}\:\begin{pmatrix}{{BK}}\\{\mathrm{qu}.\:\mathrm{77616}}\end{pmatrix}\:=\mathrm{true}\:\Rightarrow\:{BK}\neq\mathrm{serious}\forall\mathrm{qu}.\in\mathbb{N}^{\bigstar} \\ $$ Commented by BK last…
Question Number 12080 by sin (x) last updated on 12/Apr/17 Answered by ajfour last updated on 12/Apr/17 $${f}\left(−\mathrm{1},\mathrm{3}\right)=\mathrm{max}\left(−\mathrm{4},−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$${g}\left(\mathrm{4},\mathrm{4}\right)={min}\left(\mathrm{8},\:\mathrm{16}\right)\:=\mathrm{8} \\ $$$${f}\left[{f}\left(−\mathrm{1},\mathrm{3}\right),\:{g}\left(\mathrm{4},\mathrm{4}\right)\right] \\ $$$$\:\:\:=\:{f}\left(−\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{8}\right)={max}\left(−\frac{\mathrm{25}}{\mathrm{3}},\:−\frac{\mathrm{1}}{\mathrm{24}}\right) \\…
Question Number 77614 by BK last updated on 08/Jan/20 Commented by BK last updated on 08/Jan/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{A}_{\mathrm{x}} \\ $$ Commented by mr W last updated…
Question Number 143148 by mathmax by abdo last updated on 10/Jun/21 $$\mathrm{find}\:\mathrm{v}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{3k}+\mathrm{1}}\:\mathrm{interms}\:\mathrm{of}\:\mathrm{H}_{\mathrm{n}} \\ $$$$\mathrm{H}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}} \\ $$ Terms of Service…