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Author: Tinku Tara

Find-the-cordinates-of-the-two-points-on-the-curve-y-4-x-2-whose-tangents-pass-through-the-point-1-7-

Question Number 5614 by Rasheed Soomro last updated on 22/May/16 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{cordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{points} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{4}−\mathrm{x}^{\mathrm{2}} \:\:\mathrm{whose}\:\mathrm{tangents} \\ $$$$\mathrm{pass}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\left(−\mathrm{1},\:\mathrm{7}\right)\:. \\ $$ Commented by Rasheed Soomro last updated on…

Question-71149

Question Number 71149 by naka3546 last updated on 12/Oct/19 Commented by MJS last updated on 17/Oct/19 $$\mathrm{if}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is}\:\mathrm{regular}\:\mathrm{it}'\mathrm{s}\:\mathrm{symmetric} \\ $$$${BE}\:\mathrm{is}\:\mathrm{the}\:\mathrm{symmetry}\:\mathrm{axis} \\ $$$${M}\in\left({BE}\right) \\ $$$${A}={C}'\:\wedge\:{F}={D}' \\ $$$$\Rightarrow\:\left({AF}\right)=\left({CD}\right)'\:\mathrm{and}\:\mathrm{similar}\:\mathrm{for}\:\mathrm{all}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{lines}…

sin-12-13-and-is-in-the-2nd-quadrent-prove-cos-5-13-

Question Number 71146 by sadimuhmud 136 last updated on 12/Oct/19 $$\mathrm{sin}\:\alpha=\frac{\mathrm{12}}{\mathrm{13}}\:\:\mathrm{and}\:\alpha\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{2nd}\:\mathrm{quadrent}. \\ $$$$\mathrm{prove}\:\mathrm{cos}\:\alpha=−\frac{\mathrm{5}}{\mathrm{13}} \\ $$ Commented by Rio Michael last updated on 12/Oct/19 $${sin}\alpha\:=\:\frac{\mathrm{12}}{\mathrm{13}} \\…

Question-136680

Question Number 136680 by aupo14 last updated on 24/Mar/21 Answered by Olaf last updated on 24/Mar/21 $$\mathrm{F}\left({x}\right)\:=\:\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Let}\:{x}\:=\:\mathrm{3sh}{u} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\int\frac{\mathrm{3ch}{u}}{\left(\mathrm{9sh}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{2}} }\:{du}…

x-floor-function-x-ceiling-function-I-have-seen-somewhere-either-x-or-x-i-dont-remember-which-What-is-it-are-they-

Question Number 5609 by FilupSmith last updated on 22/May/16 $$\lfloor{x}\rfloor\:=\:\mathrm{floor}\:\mathrm{function} \\ $$$$\lceil{x}\rceil\:=\:\mathrm{ceiling}\:\mathrm{function} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{seen}\:\mathrm{somewhere}\:\mathrm{either}: \\ $$$$\lfloor{x}\rceil\:\:\mathrm{or}\:\:\lceil{x}\rfloor\:\:\:\left(\mathrm{i}\:\mathrm{dont}\:\mathrm{remember}\:\mathrm{which}\right) \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{it}/\mathrm{are}\:\mathrm{they}? \\ $$ Terms of Service…

Im-C-0-1-2-z-dz-pi-2-

Question Number 136682 by snipers237 last updated on 24/Mar/21 $$\:{Im}\left(\int_{{C}^{+} \left(\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right)} \overset{−} {{z}dz}\:\right)=\:\frac{\pi}{\mathrm{2}}\:\: \\ $$ Answered by Olaf last updated on 24/Mar/21 $$\Omega\:=\:\mathrm{Im}\int_{\mathrm{C}^{+} \left(\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right)} \overset{−}…

find-x-x-1-x-2-dx-

Question Number 71142 by mathmax by abdo last updated on 12/Oct/19 $${find}\:\int\sqrt{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx} \\ $$ Commented by MJS last updated on 12/Oct/19 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this} \\ $$ Terms…

arcsin-x-arcsin-1-x-t-sin-x-sin-1-x-t-sin-cos-cos-sin-sin-t-x-1-1-x-1-x-1-x-x-1-x-1-t-pi-2-

Question Number 136679 by mnjuly1970 last updated on 24/Mar/21 $$\:\:\:\:\:\:\:\:\:{arcsin}\sqrt{{x}}\:+{arcsin}\left(\sqrt{\mathrm{1}−{x}}\:\right)={t} \\ $$$$\:\:\:\:\:\:{sin}\left(\alpha\right)=\sqrt{{x}}\:\:\:\:{sin}\left(\beta\right)=\sqrt{\mathrm{1}−{x}}\: \\ $$$$\:\:\:\:\alpha+\beta={t} \\ $$$$\:\:\:\:{sin}\left(\alpha\right){cos}\left(\beta\right)+{cos}\left(\alpha\right){sin}\left(\beta\right)={sin}\left({t}\right) \\ $$$$\:\:\sqrt{{x}\:}\:.\sqrt{\mathrm{1}−\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}\:.\sqrt{\mathrm{1}−{x}}\: \\ $$$$={x}+\mathrm{1}−{x}=\mathrm{1} \\ $$$$\:\:\:{t}=\frac{\pi}{\mathrm{2}}… \\ $$$$\:\:\: \\…