Question Number 198624 by Tinku Tara last updated on 22/Oct/23 $$\mathrm{Post}\:\mathrm{have}\:\mathrm{been}\:\mathrm{cleanup}\:\mathrm{for}\:\mathrm{users} \\ $$$$\mathrm{and}\:\mathrm{access}\:\mathrm{blocked} \\ $$ Commented by mr W last updated on 22/Oct/23 $${thanks}\:{alot}\:{sir}!\:{well}\:{done}! \\…
Question Number 198562 by cortano12 last updated on 22/Oct/23 $$\:\:\:\mathrm{f}\left(\mathrm{tan}\:\mathrm{x}\right)+\:\mathrm{2f}\left(\mathrm{cot}\:\mathrm{x}\right)\:=\:\mathrm{4x}\: \\ $$$$\:\:\:\mathrm{f}\:'\left(\mathrm{x}\right)=\:? \\ $$ Answered by dimentri last updated on 22/Oct/23 $$\:{let}\:{x}=\frac{\pi}{\mathrm{2}}−{t}\: \\ $$$$\:{f}\left(\mathrm{cot}\:{t}\right)\:+\mathrm{2}\:{f}\left(\mathrm{tan}\:{t}\right)=\:\mathrm{2}\pi−\mathrm{4}{t} \\…
Question Number 198563 by mr W last updated on 22/Oct/23 $${find}\:{all}\:{numbers}\:\left({with}\:{any}\:{number}\right. \\ $$$$\left.{of}\:{digits}\right)\:{satisfying} \\ $$$$\left(\underline{{abcd}…{xyz}}\right)×\mathrm{2}=\left(\underline{{zyx}…{dcba}}\right) \\ $$ Answered by AST last updated on 25/Oct/23 $$\mathrm{2}{z}\equiv{a}\left({mod}\:\mathrm{10}\right);{a}\leqslant\mathrm{4}\Rightarrow{a}\in\left\{\mathrm{0},\mathrm{2},\mathrm{4}\right\};{a}\geqslant{z}…
Question Number 198516 by mr W last updated on 21/Oct/23 $${To}\:{administrator}\:{Tinku}\:{Tara}\:{sir}: \\ $$$${please}\:{have}\:{a}\:{look}\:{at}\:{what}\:{the}\: \\ $$$${members}\:\underline{{Hridiana}}\:{and}\:\underline{{HomeAlone}} \\ $$$${are}\:{doing}\:{in}\:{the}\:{forum}! \\ $$$${They}\:{are}\:{rioting}! \\ $$$${Please}\:{ban}\:{them}\:{from}\:{the}\:{forum}! \\ $$ Commented by…
Question Number 198517 by hardmath last updated on 21/Oct/23 Commented by Hridiana last updated on 21/Oct/23 $${a}×\mathrm{4}+\frac{{a}}{\mathrm{2}}\:{going}\:{to}\:{a}_{\mathrm{1}} \\ $$ Commented by Hridiana last updated on…
Question Number 198519 by cortano12 last updated on 21/Oct/23 $$\:\:\mathrm{Find}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{and}\:\mathrm{d}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{abc}\:=\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{11}\right)}\\{\mathrm{abd}\:=\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\\{\mathrm{acd}\:=\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\\{\mathrm{bcd}\:=\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\end{cases} \\ $$ Answered by HomeAlone last updated on 21/Oct/23 $${ab}+{ca}+{dc}+{ad}={a}+{bd}+{cd}\:{is}\:{complicative} \\ $$ Answered…
Question Number 198511 by Hridiana last updated on 21/Oct/23 $$\mathrm{3}+{xy}\sqrt{\frac{{z}}{{xy}}} \\ $$$${xz}+{zx} \\ $$$${how}\:{to}\:{solve} \\ $$$$\mathrm{3}+\pi\left\{\frac{{x}}{\mathrm{2}}>\mathrm{0}\right\}=\left\{\frac{{zy}}{{x}+{z}}>{x}\right\} \\ $$$${operations} \\ $$$$\left\{{x}^{\mathrm{2}+\frac{{x}}{\mathrm{2}}} >\mathrm{0}\right\}\:{is}\:{the}\:{comparator} \\ $$ Commented by…
Question Number 198496 by mnjuly1970 last updated on 21/Oct/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:{A}\:{nice}\:\:{series} \\ $$$$\:\:{If}\:,\:\:\Omega\:=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\:\mathrm{1}}{{n}^{\:\mathrm{2}} \:+{n}\:−\mathrm{1}}\:=\frac{\:\pi\:{tan}\left(\:{a}\pi\:\right)}{\:{b}} \\ $$$$\:\:\:\:\:\Rightarrow\:{find}\:{the}\:{value}\:{of}\:\:\:\frac{{b}}{{a}}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$ Commented by…
Question Number 198497 by universe last updated on 21/Oct/23 $$\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\mathrm{at}} −\mathrm{e}^{−\mathrm{at}} }{\mathrm{e}^{\pi\mathrm{t}} −\mathrm{e}^{−\pi\mathrm{t}} }\:\mathrm{dt} \\ $$ Commented by Hridiana last updated on 21/Oct/23…
Question Number 198552 by HomeAlone last updated on 21/Oct/23 $$\mathrm{8}+\underbrace{\underset{\mathrm{4}+\lfloor\frac{{x}}{\:\sqrt{{x}}}\rfloor} {\mathrm{1}}}\:{to}\:{the}\:{braces} \\ $$$$\left({x}\:{may}\:{found}\:{y}\right) \\ $$$${zxyz},{xyzyzyx}_{\mathrm{1}} \:{the}\:{long}\:{multiplition} \\ $$$$\sqrt{{x}^{\mathrm{3}} } \\ $$$$\sqrt{\mathrm{2}^{\mathrm{4}^{\mathrm{4}} } } \\ $$$$\mathrm{340282366920937440}…