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find-all-irational-numbers-x-such-that-x-2-20x-20-and-x-3-2020x-1-both-is-a-rasional-number-

Question Number 114965 by joki last updated on 22/Sep/20 $${find}\:{all}\:{irational}\:{numbers}\:{x}\:{such}\:{that}\:{x} \\ $$$$\mathrm{2}+\mathrm{20}{x}+\mathrm{20}\:{and}\:{x}^{\mathrm{3}\:} −\mathrm{2020}{x}+\mathrm{1}\:\:\:{both}\:{is}\:{a} \\ $$$${rasional}\:{number}. \\ $$ Answered by MJS_new last updated on 22/Sep/20 $$\left(\mathrm{1}\right)\:{x}^{\mathrm{2}}…

note-the-tringle-ABC-is-not-isosceles-with-the-elevtions-of-AA1-BB1-and-CC1-suppose-BA-amd-CA-respectively-point-at-BB1-and-CC1-so-that-A1BA-is-pependiculer-to-BB1-and-A1CA-perpendiculer-to-CC1-th

Question Number 114950 by joki last updated on 22/Sep/20 $${note}\:{the}\:{tringle}\:{ABC}\:\:{is}\:{not}\:{isosceles}\:{with}\: \\ $$$${the}\:{elevtions}\:{of}\:{AA}\mathrm{1},{BB}\mathrm{1},\:{and}\:{CC}\mathrm{1}.{suppose} \\ $$$${BA}\:\:{amd}\:{CA}\:{respectively}\:{point}\:{at}\:{BB}\mathrm{1}\:{and} \\ $$$${CC}\mathrm{1}\:{so}\:{that}\:{A}\mathrm{1}{BA}\:{is}\:{pependiculer}\:{to}\:{BB}\mathrm{1} \\ $$$${and}\:{A}\mathrm{1}{CA}\:{perpendiculer}\:{to}\:{CC}\mathrm{1}.{the}\:{read}\: \\ $$$${and}\:{BC}\:{lines}\:{intersect}\:{at}\:{the}\:{TA}\:{point}. \\ $$$${define}\:{in}\:{the}\:{same}\:{way}\:\:{TB}\:{and}\:{TC}\:{poits}. \\ $$$${prove}\:{that}\:{TA},{TB},{and}\:{TC}\:{are}\:{collinear}. \\…

find-four-consecutive-multiples-of-5-such-that-twice-the-sum-of-the-two-greatest-integer-exceed-five-times-the-least-by-5-

Question Number 114863 by mathdave last updated on 21/Sep/20 $${find}\:{four}\:{consecutive}\:{multiples}\:{of}\:\mathrm{5} \\ $$$${such}\:{that}\:{twice}\:{the}\:{sum}\:{of}\:{the}\:{two} \\ $$$${greatest}\:{integer}\:{exceed}\:{five}\:{times}\:{the} \\ $$$${least}\:{by}\:\mathrm{5} \\ $$ Answered by PRITHWISH SEN 2 last updated…

we-know-that-e-pii-1-ln-e-pii-ln-1-pii-ln-1-How-good-is-this-prove-

Question Number 114857 by Rio Michael last updated on 21/Sep/20 $$\mathrm{we}\:\mathrm{know}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:{e}^{\pi{i}} \:=\:−\mathrm{1}\: \\ $$$$\Rightarrow\:\mathrm{ln}\:\left({e}^{\pi{i}} \right)\:=\:\mathrm{ln}\left(−\mathrm{1}\right) \\ $$$$\:\:\pi{i}\:=\:\mathrm{ln}\:\left(−\mathrm{1}\right).\: \\ $$$$\mathrm{How}\:\mathrm{good}\:\mathrm{is}\:\mathrm{this}\:\mathrm{prove}? \\ $$ Commented by…

Question-114840

Question Number 114840 by zakirullah last updated on 21/Sep/20 Answered by Aziztisffola last updated on 21/Sep/20 $$\:\left(\mathrm{i}\right)\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\:\mathrm{z}+\overset{−} {\mathrm{z}}=\mathrm{x}+\mathrm{iy}+\mathrm{x}−\mathrm{iy}=\mathrm{2x}=\mathrm{2Re}\left(\mathrm{z}\right) \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{z}−\overset{−} {\mathrm{z}}=\mathrm{x}+\mathrm{iy}−\mathrm{x}+\mathrm{iy}=\mathrm{2iy}=\mathrm{2iIm}\left(\mathrm{z}\right) \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{z}\overset{−}…