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Author: Tinku Tara

Solve-for-real-numbers-5-1-z-1-5-1-z-1-5-2-4-1-z-1-4-1-z-1-4-

Question Number 141654 by mathsuji last updated on 21/May/21 $${Solve}\:{for}\:{real}\:{numbers} \\ $$$$\mathrm{5}\centerdot\left(\sqrt[{\mathrm{5}}]{\mathrm{1}−{z}}\:+\:\sqrt[{\mathrm{5}}]{\mathrm{1}+{z}}\:=\:\mathrm{2}+\mathrm{4}\centerdot\left(\sqrt[{\mathrm{4}}]{\mathrm{1}−{z}}\:+\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{z}}\right)\right. \\ $$ Commented by MJS_new last updated on 22/May/21 $$\mathrm{only}\:\mathrm{solution}\:\mathrm{is}\:{z}=\mathrm{0} \\ $$ Commented…

advanced-calculus-prove-that-0-cos-2pix-2-cosh-2-pix-dx-1-4-

Question Number 141649 by mnjuly1970 last updated on 21/May/21 $$\:\:\:\:\:\:\:\:\:…….{advanced}\:\:{calculus}…… \\ $$$$\:\:\:\:{prove}\:\:{that}−:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\phi:=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{cos}\left(\mathrm{2}\pi{x}^{\mathrm{2}} \right)}{{cosh}^{\mathrm{2}} \left(\pi{x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\checkmark \\ $$ Answered by ArielVyny last updated…

Question-141651

Question Number 141651 by mathsuji last updated on 21/May/21 Commented by MJS_new last updated on 22/May/21 $$\mathrm{it}'\mathrm{s}\:\mathrm{wrong}. \\ $$$${a}={b}={c}=\mathrm{0}\:\Rightarrow\:\mathrm{6}\sqrt{\mathrm{2}}>\mathrm{0} \\ $$$${a}={b}={c}=−\sqrt{\mathrm{3}}\:\Rightarrow\:\mathrm{6}\sqrt{\mathrm{2}}>−\mathrm{6}\sqrt{\mathrm{3}}>−\mathrm{9}\sqrt{\mathrm{3}} \\ $$ Commented by…

If-the-sides-of-a-triangle-are-consecutive-integers-and-the-maximum-angle-is-twice-the-minimum-determine-the-sides-of-the-triangle-

Question Number 76112 by Maclaurin Stickker last updated on 23/Dec/19 $${If}\:{the}\:{sides}\:{of}\:{a}\:{triangle}\:{are}\:{consecutive} \\ $$$${integers}\:{and}\:{the}\:{maximum}\:{angle} \\ $$$${is}\:{twice}\:{the}\:{minimum},\:{determine} \\ $$$${the}\:{sides}\:{of}\:{the}\:{triangle}. \\ $$ Answered by mr W last updated…

3-2-2-1-2-4-1-2-8-1-2-16-1-2-32-1-2-64-1-2-128-1-

Question Number 10575 by ridwan balatif last updated on 19/Feb/17 $$\mathrm{3}\left(\mathrm{2}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{4}} +\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{8}} +\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{16}} +\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{32}} +\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{64}} +\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{128}} +\mathrm{1}\right)=…? \\ $$ Commented by FilupS last updated…

Question-76108

Question Number 76108 by Maclaurin Stickker last updated on 23/Dec/19 Commented by Maclaurin Stickker last updated on 23/Dec/19 $${in}\:{the}\:{figure},\:{the}\:{circumferences} \\ $$$${have}\:{radius}\:\mathrm{8}\:{cm}\:{and}\:\mathrm{6}\:{cm}\:{and}\:{the} \\ $$$${distance}\:{between}\:{their}\:{centers}\:{is}\:\mathrm{12}\:{cm}.\: \\ $$$${If}\:{QP}={PR},\:{find}\:{QP}.…