Question Number 106653 by Dwaipayan Shikari last updated on 06/Aug/20 $$\mathrm{30}+\mathrm{144}+\mathrm{420}+\mathrm{960}+\mathrm{1890}+\mathrm{3360}+…{n} \\ $$ Commented by Dwaipayan Shikari last updated on 06/Aug/20 $${y}_{\mathrm{0}} \:\:\:\:\:\:\bigtriangleup{y}_{\mathrm{0}} \:\:\:\:\bigtriangleup^{\mathrm{2}} {y}_{\mathrm{0}}…
Question Number 106637 by john santu last updated on 06/Aug/20 $$\:\:\:\:\:\:@\mathrm{JS}@ \\ $$$$\mathrm{The}\:\mathrm{quartic}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{3}} +\mathrm{14x}+\mathrm{15}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{one}\:\mathrm{root}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1}+\mathrm{2i}\:.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{three}\:\mathrm{roots}.\: \\ $$ Answered by bemath last…
Question Number 106630 by zahaku last updated on 06/Aug/20 $${Find}\:{n}\:{in}\:{this}\:{equation}: \\ $$$$\left(−\mathrm{2}\right)^{{n}} \:=\:\mathrm{4096} \\ $$ Answered by bemath last updated on 06/Aug/20 $$\left(−\mathrm{2}\right)^{\mathrm{n}} =\left(\mathrm{1024}\right)×\mathrm{4}=\left(\mathrm{2}\right)^{\mathrm{12}} =\left(−\mathrm{2}\right)^{\mathrm{12}}…
Question Number 41066 by Tawa1 last updated on 01/Aug/18 Answered by candre last updated on 03/Aug/18 $${x}\equiv\mathrm{1}\left(\mathrm{mod3}\right) \\ $$$${x}\equiv\mathrm{3}\left(\mathrm{mod5}\right) \\ $$$${x}\equiv\mathrm{3}\left(\mathrm{mod7}\right) \\ $$$${by}\:{chinese}\:{remainder}\:{theorem} \\ $$$$\mathrm{5}\centerdot\mathrm{7}{x}_{\mathrm{1}}…
Question Number 172099 by Mastermind last updated on 23/Jun/22 $${The}\:{probability}\:{that}\:{Abiola}\:{will}\:{be} \\ $$$${late}\:{to}\:{office}\:{on}\:{a}\:{given}\:{day}\:{is}\frac{\mathrm{2}}{\mathrm{5}}\:.\:{in} \\ $$$${a}\:{given}\:{week}\:{of}\:{six}\:{days},\:{find}\:{the}\: \\ $$$$\left.\mathrm{1}\right)\:{probability}\:{that}\:{he}\:{will}\:{be}\:{late}\:{of} \\ $$$${only}\:\mathrm{3}\:{days} \\ $$$$\left.\mathrm{2}\right)\:{not}\:{be}\:{late}\:{in}\:{the}\:{week} \\ $$ Answered by mr…
Question Number 41009 by Raj Singh last updated on 31/Jul/18 Commented by $@ty@m last updated on 31/Jul/18 Answered by $@ty@m last updated on 31/Jul/18 $$\angle{C}=\angle{A}\:\:\:\left({corresponding}\:{angles}\right)….\left(\mathrm{1}\right)…
Question Number 106526 by 175mohamed last updated on 05/Aug/20 Commented by 175mohamed last updated on 05/Aug/20 solve Commented by Dwaipayan Shikari last updated on 05/Aug/20…
Question Number 106503 by Dwaipayan Shikari last updated on 05/Aug/20 $$\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{18}}+\frac{\mathrm{2}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{324}}+…. \\ $$ Commented by Dwaipayan Shikari last updated on 05/Aug/20 $$\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}.\left(\mathrm{2}+\mathrm{1}\right)^{\mathrm{3}} }+….\right) \\…
Question Number 40960 by behi83417@gmail.com last updated on 30/Jul/18 Commented by MrW3 last updated on 30/Jul/18 $${a}×\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}+{b}×\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}={c}×\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}}…
Question Number 172014 by Mikenice last updated on 23/Jun/22 $${using}\:{properties}\:{of}\:{determinats} \\ $$$${prove}\:{that} \\ $$$$\left[−{yz}\:\:\:\:\:\:{y}^{\mathrm{2}} +{yz}\:\:\:\:\:\:\:{z}^{\mathrm{2}} +{yz}\right] \\ $$$$\left[{x}^{\mathrm{2}} +{xz}\:\:\:−{xz}\:\:\:\:\:\:\:\:\:\:{z}^{\mathrm{2}} +{xy}\right]\:=\left({xy}+{yz}+{zx}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left.\:{x}^{\mathrm{2}} +{xy}\:\:\:\:\:{y}^{\mathrm{2}}…