Question Number 6158 by enigmeyou last updated on 16/Jun/16 $$\frac{−\mathrm{1}}{\mathrm{1048576}}=\frac{\mathrm{1}}{\mathrm{8}}×\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${find}\:{n}\:? \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $$\frac{−\mathrm{1}}{\mathrm{1048576}}=\frac{\mathrm{1}}{\mathrm{8}}×\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$\frac{−\mathrm{8}}{\mathrm{1048576}}=\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}}…
Question Number 6157 by Rasheed Soomro last updated on 16/Jun/16 Answered by Yozzii last updated on 16/Jun/16 $$\mid{BC}\mid=\mid{CD}\mid=\mid{DA}\mid=\mid{AB}\mid={x}\:{units}>\mathrm{0} \\ $$$${Form}\:{the}\:{triangle}\:\bigtriangleup{GBH}\:{where}\:\angle{GBH}=\mathrm{90}° \\ $$$${and}\angle{GHB}=\theta\:\left(\mathrm{0}<\theta<\mathrm{90}°\right).\:{Let}\:\mid{GH}\mid={r}>\mathrm{0}.\:{Since} \\ $$$${H}\:{is}\:{the}\:{centre}\:{of}\:{arc}\:{GE},\:{then}\:\mid{EH}\mid=\mid{GH}\mid={r}. \\…
Question Number 71693 by TawaTawa last updated on 18/Oct/19 Answered by MJS last updated on 19/Oct/19 $$\mathrm{the}\:\mathrm{discs}\:\mathrm{are}\:\mathrm{cylinders} \\ $$$${V}=\pi{r}^{\mathrm{2}} {h} \\ $$$${V}=\pi×\mathrm{1}^{\mathrm{2}} ×\mathrm{1}+\pi×\mathrm{2}^{\mathrm{2}} ×\mathrm{1}+\pi×\mathrm{3}^{\mathrm{2}} ×\mathrm{1}=\mathrm{14}\pi…
Question Number 137225 by yutytfjh67ihd last updated on 31/Mar/21 $${Q}\mathrm{137026} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 6154 by Rasheed Soomro last updated on 16/Jun/16 $${Prove}\:{or}\:{disprove} \\ $$$$\left(\frac{\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}}}{\mathrm{5}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \geqslant\:\:\boldsymbol{{abc}}\:\:\:\forall\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0} \\ $$ Commented by Yozzii last updated on 18/Jun/16 $$\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\left(\frac{\mathrm{2}}{{c}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\sqrt{\mathrm{2}{ab}+\mathrm{2}{bc}+{ca}}}\leqslant\mathrm{1} \\…
Question Number 137226 by SLVR last updated on 31/Mar/21 Answered by MJS_new last updated on 31/Mar/21 $${f}\left({x}\right)=\mathrm{2ln}\:{x} \\ $$$${x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{11}{x}−\mathrm{6}=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right) \\ $$$$\mathrm{the}\:\mathrm{curves}\:\mathrm{intersect}\:\mathrm{at}\:{x}=\mathrm{1} \\ $$$$\underset{\mathrm{0}}…
Question Number 6152 by sanusihammed last updated on 16/Jun/16 $${Differentiate}\:\:\:{s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\:\:\:\:{from}\:{the}\:{first}\:{principle} \\ $$$$ \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $${s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\:…
Question Number 6148 by saiful last updated on 16/Jun/16 $${if}\:\alpha,\beta\:{and}\gamma\:{is}\:{corner}\:{in}\:\Delta{abc}.\:{indication}\:{that}\:{cos}\left(\beta+\gamma\right)=−{cos}\alpha \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $${cos}\left(\beta+\gamma\right)=−{cos}\alpha \\ $$$$−−−−−−−−−− \\ $$$$\because\alpha,\beta,\gamma\:{are}\:{angles}\:{of}\:{a}\:{triangle}…
Question Number 137217 by yutytfjh67ihd last updated on 31/Mar/21 $${Q}\mathrm{137014} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 6146 by Ninik last updated on 16/Jun/16 $${Determine}\:{the}\:{value}\:{of}\:\mathrm{sin}\:\left(\frac{\Pi}{\mathrm{3}}+{p}\right)\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{6}}+{p}\right)−\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{3}}+{p}\right)\mathrm{sin}\:\left(\frac{\Pi}{\mathrm{6}}+{p}\right) \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+{p}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+{p}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+{p}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+{p}\right) \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta=\mathrm{sin}\:\left(\alpha−\beta\right) \\ $$$$=\mathrm{sin}\:\left(\left(\frac{\pi}{\mathrm{3}}+{p}\right)−\left(\frac{\pi}{\mathrm{6}}+{p}\right)\right)…