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Question-88881

Question Number 88881 by Power last updated on 13/Apr/20 Answered by mahdi last updated on 13/Apr/20 $$\mathrm{A}=\sqrt{\mathrm{4}+\sqrt{\mathrm{4}+\sqrt{\mathrm{4}−\sqrt{\mathrm{4}+\sqrt{\mathrm{4}+\sqrt{\mathrm{4}−…}}}}}} \\ $$$$\mathrm{A}=\sqrt{\mathrm{4}+\sqrt{\mathrm{4}+\sqrt{\mathrm{4}−\mathrm{A}}}}\Rightarrow\mathrm{A}^{\mathrm{2}} =\mathrm{4}+\sqrt{\mathrm{4}+\sqrt{\mathrm{4}−\mathrm{A}}} \\ $$$$\mathrm{A}\backsimeq\mathrm{2}.\mathrm{5070} \\ $$ Terms…

Given-a-10-digit-number-X-1345789026-How-many-10-digit-number-that-can-be-made-using-every-digit-from-X-with-condition-If-a-number-n-is-located-in-k-th-position-of-X-then-the-new-created-numb

Question Number 88876 by Joel578 last updated on 13/Apr/20 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{10}−\mathrm{digit}\:\mathrm{number}\:{X}\:=\:\mathrm{1345789026} \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{10}−\mathrm{digit}\:\mathrm{number}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made} \\ $$$$\mathrm{using}\:\mathrm{every}\:\mathrm{digit}\:\mathrm{from}\:{X},\:\mathrm{with}\:\mathrm{condition}: \\ $$$$\mathrm{If}\:\mathrm{a}\:\mathrm{number}\:{n}\:\:\mathrm{is}\:\mathrm{located}\:\mathrm{in}\:{k}^{{th}} \:\mathrm{position}\:\mathrm{of}\:{X},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{new}\:\mathrm{created}\:\mathrm{number}\:\mathrm{must}\:\mathrm{not}\:\mathrm{contain} \\ $$$$\mathrm{number}\:{n}\:\mathrm{in}\:{k}^{{th}} \:\mathrm{position} \\ $$$$ \\…

nice-calculus-prove-that-I-0-1-e-x-sin-2-x-x-3-2-2pi-1-2-1-m-n-

Question Number 154409 by mnjuly1970 last updated on 18/Sep/21 $$ \\ $$$$\:\:{nice}\:{calculus}.. \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\: \\ $$$$\:\mathrm{I}:=\int_{\mathrm{0}} ^{\:\infty} \frac{\:\left(\mathrm{1}+{e}^{\:−{x}} \:\right){sin}^{\:\mathrm{2}} \left({x}\right)}{{x}^{\:\frac{\mathrm{3}}{\mathrm{2}}} }\:=\sqrt{\mathrm{2}\pi}\:\left(\:\mathrm{1}+\:\sqrt{\sqrt{\mathrm{2}}\:−\:\mathrm{1}}\:\right) \\ $$$$\:{m}.{n}…

3-x-2-5-gt-x-1-

Question Number 88873 by jagoll last updated on 13/Apr/20 $$\mathrm{3}+\sqrt{{x}^{\mathrm{2}} −\mathrm{5}}\:>\:\mid{x}−\mathrm{1}\mid\: \\ $$ Commented by john santu last updated on 13/Apr/20 $$\Rightarrow\left(\mathrm{3}+\sqrt{{x}^{\mathrm{2}} −\mathrm{5}}\right)^{\mathrm{2}} \:>\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\…