Question Number 12371 by JAZAR last updated on 20/Apr/17 $${tank}\:{you} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12332 by frank ntulah last updated on 19/Apr/17 $$\mathrm{prove}\:;\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right)=\mathrm{2} \\ $$ Commented by FilupS last updated on 20/Apr/17 $$\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{undefined} \\ $$ Commented by…
Question Number 12267 by tawa last updated on 17/Apr/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sequence} \\ $$$$\left.\mathrm{1}\right)\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\:,\:\frac{\mathrm{1}}{\mathrm{15}}\:,\:\frac{\mathrm{1}}{\mathrm{35}}\:,\:\frac{\mathrm{1}}{\mathrm{63}}\:,\:\frac{\mathrm{1}}{\mathrm{99}} \\ $$$$\left.\mathrm{2}\right)\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{6}},\:\frac{\mathrm{1}}{\mathrm{12}},\:\frac{\mathrm{1}}{\mathrm{20}},\:\frac{\mathrm{1}}{\mathrm{30}} \\ $$ Answered by ajfour last updated on 17/Apr/17 $$\left.\mathrm{1}\right)\:\:{T}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}}…
Question Number 12265 by frank ntulah last updated on 17/Apr/17 $${Determinant}\:{method}\:{can}\:{be}\:{used}\:{to}\:{solve} \\ $$$${the}\:{system}\:{below}?,\:\mathrm{if}\:\mathrm{yes}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{determinant}\:\mathrm{method}\:\mathrm{and} \\ $$$$\:\mathrm{if}\:\mathrm{no}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{another}\:\mathrm{method} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$${x}+{y}−{z}=\mathrm{8} \\ $$$$\mathrm{2}{x}+{y}−\mathrm{2}{z}=\mathrm{3} \\ $$$$\left({give}\:{clear}\:{reason}\:{for}\:{your}\:{answer}\right) \\ $$…
Question Number 77739 by ajfour last updated on 09/Jan/20 Commented by ajfour last updated on 09/Jan/20 $${Find}\:\:\:\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\phi}\:\:{using}\:{a},{b},{c},\alpha. \\ $$ Commented by key of knowledge last…
Question Number 12148 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17 $${OA}={OC}={OD}={OB}={R},{OE}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${BE}\bot{ED},{FG}\bot{CD},{FH}\bot{AB},\measuredangle{OEB}=\measuredangle{OED}. \\ $$$${find}:\:\:{HE}\:\:{and}\:{EG}\:{in}\:{term}\:{of}\::\:{R} \\ $$ Commented by…
Question Number 77681 by mr W last updated on 09/Jan/20 Commented by mr W last updated on 09/Jan/20 $${The}\:{radii}\:\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{c}}\:{are}\:{given}. \\ $$$$\mathrm{1}.\:{Find}\:{radius}\:\boldsymbol{{R}}\:{of}\:{the}\:{circumcircle} \\ $$$$\mathrm{2}.\:{Find}\:{radius}\:\boldsymbol{{d}}\:{of}\:{the}\:{fourth}\:{circle} \\ $$…
Question Number 12131 by Peter last updated on 14/Apr/17 $${a}\:{cube}\:{has}\:{a}\:{rib}\:{ABCD}.{EFGH},\:{the}\:{midle}\:{point}\:{P}\:\:{on}\:{BF}\:{so}\:{that}\:{BP}\:=\:{PF}, \\ $$$${and}\:{the}\:{midle}\:{point}\:{Q}\:{on}\:{FG}\:{so}\:{that}\:{FQ}\:=\:{QG} \\ $$$${how}\:{long}\:{projection}\:{point}\:{C}\:{to}\:{APQH}\:{field}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 77655 by TawaTawa last updated on 08/Jan/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\:\:\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{25}} \:+\:\mathrm{x}^{\mathrm{49}} \:+\:\mathrm{x}^{\mathrm{81}} \:\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\:\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{1} \\ $$ Commented by jagoll last updated on 08/Jan/20…
Question Number 77651 by ajfour last updated on 08/Jan/20 Commented by ajfour last updated on 08/Jan/20 $${Wish}\:{to}\:{know}\:{s}\:{in}\:{terms}\:{of}\:{x},{R}. \\ $$$${s}={AC},\:\:{x}={AB},\:{BP}=\mathrm{1}. \\ $$ Commented by mr W…