Question Number 140555 by liberty last updated on 09/May/21 $$\mathrm{The}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{A},\mathrm{B}\:,\mathrm{C}\:\mathrm{are}\: \\ $$$$\left(\mathrm{6},\mathrm{3}\right),\left(−\mathrm{3},\mathrm{5}\right)\:\mathrm{and}\:\left(\mathrm{4},−\mathrm{2}\right)\:\mathrm{respectively} \\ $$$$\mathrm{and}\:\mathrm{P}\:\mathrm{is}\:\mathrm{any}\:\mathrm{point}\:\left(\mathrm{x},\mathrm{y}\right).\:\mathrm{Find}\:\mathrm{ratio} \\ $$$$\mathrm{of}\:\mathrm{area}\:\bigtriangleup\mathrm{PBC}\:\&\:\bigtriangleup\mathrm{ABC}. \\ $$ Answered by EDWIN88 last updated on 09/May/21…
Question Number 9471 by tawakalitu last updated on 09/Dec/16 $$\mathrm{prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$$$\mathrm{a}\:+\:\left(\mathrm{a}\:+\:\mathrm{d}\right)\:+\:\left(\mathrm{a}\:+\:\mathrm{2d}\right)\:+\:…\:+\:\left[\mathrm{a}\:+\:\left(\mathrm{n}\:−\:\mathrm{1}\right)\mathrm{d}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{n}\left[\mathrm{2a}\:+\:\left(\mathrm{n}\:−\:\mathrm{1}\right)\mathrm{d}\right]\: \\ $$ Answered by mrW last updated on 10/Dec/16 $$\mathrm{for}\:\mathrm{n}=\mathrm{1}: \\ $$$$\mathrm{a}\overset{!} {=}\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\left[\mathrm{2a}+\left(\mathrm{1}−\mathrm{1}\right)\mathrm{d}\right]=\mathrm{a}…
Question Number 9470 by tawakalitu last updated on 09/Dec/16 $$\mathrm{prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}. \\ $$$$\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{2}.\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{3}.\mathrm{7}}\:+\:…\:+\:\frac{\mathrm{1}}{\left(\mathrm{2n}\:−\:\mathrm{1}\right)\left(\mathrm{2n}\:+\:\mathrm{1}\right)}\:=\:\frac{\mathrm{n}}{\mathrm{2n}\:+\:\mathrm{1}} \\ $$ Commented by mrW last updated on 10/Dec/16 $$\mathrm{please}\:\mathrm{check}\:\mathrm{the}\:\mathrm{question}! \\ $$$$\mathrm{should}\:\mathrm{it}\:\mathrm{be}\:\mathrm{following}? \\…
Question Number 9455 by RasheedSoomro last updated on 09/Dec/16 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\mathrm{3x}−\mathrm{8z}=−\mathrm{11} \\ $$$$\mathrm{2y}−\mathrm{3x}=−\mathrm{3} \\ $$$$\mathrm{y}−\mathrm{4z}=−\mathrm{7} \\ $$$${See}\:{the}\:{comment}\:{of}\:{Q}#\mathrm{9439}. \\ $$ Answered by geovane10math last updated…
Question Number 140525 by mathdanisur last updated on 09/May/21 $${x};{y};{z}>\mathrm{0}\:;\:{x}+{y}+{z}=\mathrm{3} \\ $$$${proof}\:\left({x}^{\mathrm{3}} +\mathrm{2}\right)\left({y}^{\mathrm{3}} +\mathrm{2}\right)\left({z}^{\mathrm{3}} +\mathrm{2}\right)\geqslant\mathrm{3}^{\mathrm{3}} \\ $$ Answered by mitica last updated on 09/May/21 $${jensen}…
Question Number 9449 by tawakalitu last updated on 08/Dec/16 $$\mathrm{2xy}\:=\:\mathrm{x}\:+\:\mathrm{y}\:\:\:\:\:……\:\left(\mathrm{i}\right) \\ $$$$\mathrm{6xz}\:=\:\mathrm{6z}\:−\:\mathrm{2x}\:\:\:\:…..\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{3yz}\:=\:\mathrm{3y}\:+\:\mathrm{4z}\:\:\:…….\:\left(\mathrm{iii}\right) \\ $$$$ \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{z} \\ $$ Answered by mrW last updated…
Question Number 9424 by tawakalitu last updated on 07/Dec/16 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x},\:\mathrm{y}\::\:\:\mathrm{x},\:\mathrm{y}\:\in\:\mathrm{R} \\ $$$$\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{3xy}^{\mathrm{2}} \:=\:−\:\mathrm{46}\:\:\:……..\:\left(\mathrm{i}\right) \\ $$$$\mathrm{3x}^{\mathrm{2}} \mathrm{y}\:−\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{9}\:\:\:……….\left(\mathrm{ii}\right) \\ $$ Answered by mrW last updated…
Question Number 74947 by chess1 last updated on 04/Dec/19 Answered by MJS last updated on 05/Dec/19 $$\left({x}−\mathrm{1}\right)\left(\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\pm\mathrm{1}\vee{x}=−\frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}\right)\:\vee{x}=−\frac{\mathrm{1}}{\mathrm{6}}+\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}\right)\:\vee{x}=−\frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}\right) \\ $$ Commented…
Question Number 140481 by ajfour last updated on 08/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 140477 by ajfour last updated on 08/May/21 $${If}\:\:\mathrm{sin}\:\theta+\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{sin}\:^{\mathrm{3}} \theta+…. \\ $$$$\:\:\:\:\:\:=\:\mathrm{cos}\:\theta\:\:\:\:\:{and}\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:{then} \\ $$$${find}\:\theta. \\ $$ Answered by benjo_mathlover last updated on 08/May/21…