Question Number 92646 by john santu last updated on 08/May/20 Commented by Tinku Tara last updated on 08/May/20 Thanks for trying out. We have now made a few changes. please download again from http://www.tinkutara.com Commented by john santu last updated…
Question Number 158175 by mathocean1 last updated on 31/Oct/21 $$\underset{{x}\rightarrow+\infty\:} {{lim}}\frac{{sinx}+{x}}{\mathrm{3}+\mathrm{2}{sinx}}=? \\ $$ Commented by cortano last updated on 01/Nov/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}+{x}}{\mathrm{3}+\mathrm{2sin}\:{x}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{sin}\:{x}}{{x}}+\mathrm{1}}{\frac{\mathrm{3}}{{x}}+\frac{\mathrm{2sin}\:{x}}{{x}}} \\ $$$$=\:\frac{\mathrm{0}+\mathrm{1}}{\mathrm{0}+\mathrm{0}}\:=\:\infty…
Question Number 92635 by otchereabdullai@gmail.com last updated on 08/May/20 Commented by mr W last updated on 08/May/20 $$\angle{BFD}=\frac{\:\:\angle{BOD}}{\mathrm{2}}=\frac{\mathrm{120}}{\mathrm{2}}=\mathrm{60}° \\ $$$$\angle{FBC}=\angle{OBC}−\angle{OBG}=\mathrm{90}−\frac{\mathrm{30}}{\mathrm{2}}=\mathrm{75}° \\ $$$$\angle{ODB}=\frac{\mathrm{180}−\mathrm{120}}{\mathrm{2}}=\mathrm{30}° \\ $$$$\angle{BCD}=\mathrm{180}−\mathrm{120}=\mathrm{60}° \\…
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Question Number 27076 by amit96 last updated on 01/Jan/18 Commented by prakash jain last updated on 05/Jan/18 $$\mathrm{Can}\:\mathrm{you}\:\mathrm{pleasw}\:\mathrm{type}\:\mathrm{your}\:\mathrm{question}\:\mathrm{or} \\ $$$$\mathrm{post}\:\mathrm{a}\:\mathrm{clearer}\:\mathrm{picture}. \\ $$ Terms of Service…
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Question Number 92594 by naka3546 last updated on 08/May/20 Commented by jagoll last updated on 08/May/20 $$\alpha+\beta+\gamma\:=\:\mathrm{180}^{\mathrm{o}} ? \\ $$ Commented by Prithwish Sen 1…
Question Number 92575 by M±th+et+s last updated on 08/May/20 $${hello}\: \\ $$$${i}\:{see}\:{a}\:{lot}\:{of}\:{questions}\:{about}\:{special} \\ $$$${functions}\:{in}\:{the}\:{forum}\:{so}\:{this}\:{book} \\ $$$${is}\:{a}\:{good}\:{book}\:{to}\:{learn}\:{special}\:{functions} \\ $$$${for}\:{anyone}\:{want}\:{to}\:{learn} \\ $$ Commented by jagoll last updated…
Question Number 158088 by Gbenga last updated on 31/Oct/21 $$\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\left({x}\right) \\ $$$${what}\:{is}\:{x}\:{note}\:{not}\:\mathrm{6} \\ $$ Commented by MJS_new last updated on 31/Oct/21 $${f}\left({n}\right)=\frac{{x}−\mathrm{6}}{\mathrm{120}}{n}^{\mathrm{5}} −\frac{{x}−\mathrm{6}}{\mathrm{8}}{n}^{\mathrm{4}} +\frac{\mathrm{17}\left({x}−\mathrm{6}\right)}{\mathrm{24}}{n}^{\mathrm{3}} −\frac{\mathrm{15}\left({x}−\mathrm{6}\right)}{\mathrm{8}}{n}^{\mathrm{2}}…
Question Number 92544 by Tinku Tara last updated on 07/May/20 $$\mathrm{A}\:\mathrm{new}\:\mathrm{version}\:\mathrm{of}\:\mathrm{application}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{few}\:\mathrm{usability}\:\mathrm{enhancement} \\ $$$$\mathrm{will}\:\:\mathrm{be}\:\mathrm{available}\:\mathrm{on}\:\mathrm{playstore} \\ $$$$\mathrm{in}\:\mathrm{next}\:\mathrm{few}\:\mathrm{days}. \\ $$$$\mathrm{Key}\:\mathrm{Changes} \\ $$$$\mathrm{a}.\:\mathrm{Continous}\:\mathrm{scrolling}\:\mathrm{on}\:\mathrm{forum}. \\ $$$$\mathrm{b}.\:\mathrm{Clicking}\:\mathrm{on}\:\mathrm{images}\:\mathrm{pop}\:\mathrm{up}\:\mathrm{image} \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{easily}\:\mathrm{scrolled}.…