Question Number 8273 by lepan last updated on 05/Oct/16 $${Express}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha\:{in}\:{the}\:{form}\: \\ $$$${Rsin}\left(\alpha+\beta\right)\:{where}\:{R}>\mathrm{0}\:{and}\:\mathrm{0}°<\beta<\mathrm{90}°. \\ $$$${Hence}\:{solve}\:{the}\:{equation}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha=\mathrm{2} \\ $$$${for}\:\mathrm{0}°<\alpha<\mathrm{270}°. \\ $$ Answered by prakash jain last updated on…
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Question Number 8269 by tawakalitu last updated on 04/Oct/16 Commented by tawakalitu last updated on 04/Oct/16 $$\mathrm{Find}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}.\:\mathrm{thanks}\:\mathrm{in}\:\mathrm{advance}. \\ $$ Answered by prakash jain last updated…
Question Number 139342 by BHOOPENDRA last updated on 26/Apr/21 Commented by BHOOPENDRA last updated on 26/Apr/21 $${mr}.{W}\:{sir}\:{help}\:{me}\:{out}\:{this} \\ $$ Answered by mr W last updated…
Question Number 73805 by Waseem Yaqoob last updated on 16/Nov/19 $${topic}\:{binomial}\:{theorem} \\ $$$${Evaluate} \\ $$$$\left(\mathrm{2}{x}−\mathrm{6}{y}\right)^{−\mathrm{8}} \\ $$ Commented by mr W last updated on 16/Nov/19…
Question Number 8267 by lepan last updated on 04/Oct/16 $${Show}\:{that}\:{sinA}+{sinB}=\mathrm{2}{sin}\frac{{A}+{B}}{\mathrm{2}}\:{cos}\frac{{A}−{B}}{\mathrm{2}}. \\ $$$$ \\ $$ Answered by Yozzias last updated on 04/Oct/16 $$\mathrm{Let}\:\mathrm{p}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{and}\:\mathrm{q}=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}−\mathrm{B}\right). \\ $$$$\therefore\:\mathrm{2p}=\mathrm{A}+\mathrm{B}\:\mathrm{and}\:\mathrm{2q}=\mathrm{A}−\mathrm{B}. \\…
Question Number 73800 by TawaTawa last updated on 15/Nov/19 $$\mathrm{Please}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{angles} \\ $$$$\mathrm{QR}\:\:\:=\:\:\mathrm{6}\:\:\mathrm{cm} \\ $$$$\mathrm{RS}\:\:\:=\:\:\mathrm{7}\:\mathrm{cm} \\ $$$$\mathrm{PS}\:\:=\:\:\mathrm{4}\:\mathrm{cm} \\ $$ Commented by mr W last updated on…
Question Number 139332 by bemath last updated on 26/Apr/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{50}} \:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{expression}\:\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{1000}} \:+\mathrm{2x}\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{999}} + \\ $$$$\mathrm{3x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{998}} +…+\mathrm{1001x}^{\mathrm{1000}} \\ $$ Answered by mr W…
Question Number 8262 by 666225 last updated on 04/Oct/16 $$\mid{x}−\mathrm{1}\mid\:<\:\mathrm{2}\:\Rightarrow\:\mid{x}−\mathrm{3}\mid \\ $$ Commented by 123456 last updated on 04/Oct/16 $$\mid{x}−\mathrm{1}\mid<\mathrm{2}\Leftrightarrow−\mathrm{2}<{x}−\mathrm{1}<\mathrm{2} \\ $$$$−\mathrm{2}−\mathrm{2}<{x}−\mathrm{3}<\mathrm{2}−\mathrm{2} \\ $$$$−\mathrm{4}<{x}−\mathrm{3}<\mathrm{0} \\…
Question Number 8259 by 314159 last updated on 04/Oct/16 Answered by Yozzias last updated on 04/Oct/16 $$\mathrm{Let}\:\mathrm{u}\left(\mathrm{x}\right)=\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} \:\:\left(\mathrm{n}\in\mathbb{N},\mathrm{n}\geqslant\mathrm{2}\right) \\ $$$$\mathrm{2}^{\mathrm{n}} −\mathrm{1}=\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} +\mathrm{C}_{\mathrm{3}} +\mathrm{C}_{\mathrm{4}} +…+\mathrm{C}_{\mathrm{n}}…