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Author: Tinku Tara

to-all-those-who-deleted-their-posts-after-they-had-been-answered-I-will-not-answer-you-anymore-this-forum-had-been-great-but-lately-it-has-been-filling-with-unpolite-people-I-m-not-a-freebie-solver-

Question Number 72232 by MJS last updated on 26/Oct/19 $$\mathrm{to}\:\mathrm{all}\:\mathrm{those}\:\mathrm{who}\:\mathrm{deleted}\:\mathrm{their}\:\mathrm{posts}\:\mathrm{after}\:\mathrm{they} \\ $$$$\mathrm{had}\:\mathrm{been}\:\mathrm{answered}:\:\mathrm{I}\:\mathrm{will}\:\mathrm{not}\:\mathrm{answer}\:\mathrm{you} \\ $$$$\mathrm{anymore} \\ $$$$\mathrm{this}\:\mathrm{forum}\:\mathrm{had}\:\mathrm{been}\:\mathrm{great}\:\mathrm{but}\:\mathrm{lately}\:\mathrm{it}\:\mathrm{has} \\ $$$$\mathrm{been}\:\mathrm{filling}\:\mathrm{with}\:\mathrm{unpolite}\:\mathrm{people} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{a}\:\mathrm{freebie}\:\mathrm{solver}\:\mathrm{for}\:\mathrm{anybody}'\mathrm{s}\:\mathrm{homework} \\ $$ Commented by mr…

Question-137770

Question Number 137770 by peter frank last updated on 06/Apr/21 Answered by Ñï= last updated on 06/Apr/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {da}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\left(\mathrm{1}+{ax}\right)\left(\mathrm{1}+{x}^{\mathrm{2}}…

If-log-2-3-a-and-log-3-7-b-express-log-42-56-in-terms-of-a-and-b-

Question Number 137764 by byaw last updated on 06/Apr/21 $$\mathrm{If}\:\mathrm{log}_{\mathrm{2}} \mathrm{3}={a}\:\mathrm{and}\:\mathrm{log}_{\mathrm{3}} \mathrm{7}={b},\:\mathrm{express} \\ $$$$\mathrm{log}_{\mathrm{42}} \mathrm{56}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a}\:\mathrm{and}\:{b} \\ $$ Answered by EnterUsername last updated on 06/Apr/21 $${log}_{\mathrm{2}}…

limit-1-1-x-1-x-

Question Number 6692 by Tawakalitu. last updated on 14/Jul/16 $${limit}\:\:\:\:\:\:\:\:\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}\:−\:\mathrm{1}} \\ $$$${x}\:\rightarrow\:\infty \\ $$ Answered by FilupSmith last updated on 14/Jul/16 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{1}+\frac{\mathrm{1}}{{x}−\mathrm{1}}=\mathrm{1}+\frac{\mathrm{1}}{\infty−\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\infty}…

limit-1-1-1-x-x-

Question Number 6689 by Tawakalitu. last updated on 13/Jul/16 $${limit}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}} \\ $$$${x}\:\dashrightarrow\:\infty \\ $$ Answered by FilupSmith last updated on 14/Jul/16 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−\infty} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\infty}…

Question-72218

Question Number 72218 by aliesam last updated on 26/Oct/19 Commented by turbo msup by abdo last updated on 26/Oct/19 $${let}\:{A}\left({x}\right)=\left(\frac{{sinx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\Rightarrow \\ $$$${A}\left({x}\right)={e}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\frac{{sinx}}{{x}}\right)}…

Question-6681

Question Number 6681 by Tawakalitu. last updated on 11/Jul/16 Answered by Rasheed Soomro last updated on 12/Jul/16 $${Join}\:\mathrm{O}\:{and}\:\mathrm{Y}. \\ $$$$\because\:\:\mathrm{OX}=\mathrm{OY}=\mathrm{OZ}\:\:\left[{Radii}\:{of}\:{same}\:{circle}\right] \\ $$$$\therefore\:\bigtriangleup\mathrm{XOY}\:\:{and}\:\:\bigtriangleup\mathrm{ZOY}\:{are}\:{issoscel}\:{triangles}. \\ $$$$\therefore\:\:\angle\mathrm{XYO}=\angle\mathrm{OXY}=\mathrm{30}\:\:\:{and}\:\:\:\:\angle\mathrm{ZYO}=\angle\mathrm{OZY}=\mathrm{20} \\…