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Given-a-circle-with-the-center-at-the-point-O-and-the-radius-of-the-length-R-From-a-point-A-outside-so-that-AO-2R-drawing-two-tangents-AB-and-AC-to-the-circle-B-and-C-are-the-tangency-points-Take-a

Question Number 101937 by 1549442205 last updated on 06/Jul/20 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{the}\:\mathrm{center}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\mathrm{O}\: \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{length}\:\mathrm{R}.\mathrm{From}\:\mathrm{a}\:\mathrm{point}\:\mathrm{A}\:\mathrm{outside} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{AO}=\mathrm{2R},\mathrm{drawing}\:\mathrm{two}\:\mathrm{tangents}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{AC}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\left(\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{are}\:\mathrm{the}\:\mathrm{tangency}\:\mathrm{points}\right).\mathrm{Take}\:\mathrm{a}\:\mathrm{arbitrary}\:\mathrm{point}\:\mathrm{M} \\ $$$$\mathrm{on}\:\mathrm{smaller}\:\mathrm{arc}\:\mathrm{BC}\:\left(\mathrm{M}\:\mathrm{differ}\:\mathrm{from}\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\right) \\ $$$$\mathrm{The}\:\mathrm{tangent}\:\mathrm{pass}\:\mathrm{M}\:\mathrm{cuts}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{AC}\:\mathrm{at}\:\mathrm{Pand}\:\mathrm{Q} \\ $$$$\mathrm{respectively}.\mathrm{The}\:\mathrm{segments}\:\mathrm{OP}\:\mathrm{and}\:\mathrm{OQ}\:\mathrm{cuts} \\ $$$$\mathrm{BC}\:\mathrm{at}\:\mathrm{D}\:\mathrm{and}\:\mathrm{E}\:\mathrm{respectively}. \\…

Question-167456

Question Number 167456 by Khalmohmmad last updated on 17/Mar/22 Answered by MJS_new last updated on 17/Mar/22 $$\mathrm{tan}\:\mathrm{760}\:\neq\mathrm{4} \\ $$$$\mathrm{tan}\:\mathrm{76}°\:\neq\mathrm{4} \\ $$$$\mathrm{sin}\:\mathrm{14}\:\approx.\mathrm{990607} \\ $$$$\mathrm{sin}\:\mathrm{14}°\:\approx.\mathrm{241922} \\ $$$$\mathrm{sin}^{\mathrm{2}}…

Question-36366

Question Number 36366 by chakraborty ankit last updated on 01/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18 $$\left.\mathrm{47}\right){red}=\mathrm{3}\:\:{blue}=\mathrm{4}\:\:{green}=\mathrm{2} \\ $$$${p}=\frac{\mathrm{3}{C}_{\mathrm{1}} ×\mathrm{4}{C}_{\mathrm{1}} ×\mathrm{2}{C}_{\mathrm{1}} }{\mathrm{9}{C}_{\mathrm{3}} }=\frac{\mathrm{3}×\mathrm{4}×\mathrm{2}}{\frac{\mathrm{9}×\mathrm{8}×\mathrm{7}}{\mathrm{3}×\mathrm{2}}}==\frac{\mathrm{3}×\mathrm{4}×\mathrm{2}×\mathrm{3}×\mathrm{2}}{\mathrm{9}×\mathrm{8}×\mathrm{7}} \\…

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

Question Number 101882 by bachamohamed last updated on 05/Jul/20 $$\:\:\:\int\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{x}}}+\sqrt{\boldsymbol{{x}}+\mathrm{1}}+\sqrt{\boldsymbol{{x}}+\mathrm{2}}}\boldsymbol{{dx}} \\ $$ Commented by  M±th+et+s last updated on 05/Jul/20 $${and}\:{till}\:{now}\:{i}\:{am}\:{trying}\:{with}\:{this} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{3}} −{x}^{\mathrm{2}}…

1-2-Use-Method-and-Formula-

Question Number 167419 by Bagus1003 last updated on 16/Mar/22 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!=? \\ $$$${Use}\:{Method}\:{and}\:{Formula}!!!!!!!!! \\ $$ Commented by mkam last updated on 16/Mar/22 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!\:=\:\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$ Answered…

find-d-n-dx-n-ln-cosx-

Question Number 167402 by mkam last updated on 15/Mar/22 $$\boldsymbol{{find}}\:\frac{\boldsymbol{{d}}^{\boldsymbol{{n}}} }{\boldsymbol{{dx}}^{\boldsymbol{{n}}} }\:\left[\:\boldsymbol{{ln}}\:\left(\boldsymbol{{cosx}}\right)\:\right] \\ $$ Commented by DrHZ last updated on 15/Mar/22 I think this is wrong! Commented by mkam…