Question Number 27198 by mondodotto@gmail.com last updated on 03/Jan/18 Answered by mrW1 last updated on 03/Jan/18 $$\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} −{ut}+{s}=\mathrm{0} \\ $$$$\Rightarrow\:{t}=\frac{{u}\pm\sqrt{{u}^{\mathrm{2}} −\mathrm{2}{gs}}}{{g}} \\ $$ Terms of…
Question Number 27197 by mondodotto@gmail.com last updated on 03/Jan/18 Commented by prakash jain last updated on 03/Jan/18 $$\mathrm{I}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{typo} \\ $$$$\frac{\mathrm{3}}{\mathrm{5}},\frac{\mathrm{10}}{\mathrm{8}},\frac{\mathrm{16}}{\mathrm{18}},\frac{\mathrm{36}}{\mathrm{34}},\frac{\mathrm{68}}{\mathrm{70}},\frac{\mathrm{140}}{\mathrm{138}},\frac{\mathrm{276}}{\mathrm{268}} \\ $$$$\frac{\mathrm{Numerator}\:\mathrm{of}\:{n}^{{tb}} \:\mathrm{term}}{\mathrm{Deniminator}\:\mathrm{of}\:{n}^{{th}} \:\mathrm{term}} \\…
Question Number 158267 by SANOGO last updated on 01/Nov/21 $${soit}:{F}=\left\{\left({x},{y},{z}\right)\in{R}^{\mathrm{3}} /{x}−{y}−\mathrm{2}{z}\right\}\:{et} \\ $$$${G}={Vect}\left(\mathrm{0},\mathrm{1},\mathrm{1}\right) \\ $$$${determiner}\:{l}'{intersection}\:{de}\:{F}\:{et}\:{G} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 92701 by i jagooll last updated on 08/May/20 Commented by Tinku Tara last updated on 08/May/20 $$\mathrm{No}.\:\mathrm{Several}\:\mathrm{problems}\:\mathrm{have}\:\mathrm{been}\:\mathrm{reported} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{new}\:\mathrm{version}. \\ $$$$\mathrm{Please}\:\mathrm{download}\:\mathrm{the}\:\mathrm{latest}\:\mathrm{stable}\:\mathrm{version} \\ $$$$\mathrm{from}\:\mathrm{www}.\mathrm{tinkutara}.\mathrm{com}\:\mathrm{or}\:\mathrm{playstore}.…
Question Number 92682 by Tinku Tara last updated on 08/May/20 $$\mathrm{While}\:\mathrm{uninstalling}/\mathrm{reinstalling}\:\mathrm{app} \\ $$$$\mathrm{please}\:\mathrm{use}\:\mathrm{backup}/\mathrm{import}\:\mathrm{to}\:\mathrm{keep} \\ $$$$\mathrm{your}\:\mathrm{saved}\:\mathrm{equation}\:\mathrm{on}\:\mathrm{device}. \\ $$$$\mathrm{backup}:\:\mathrm{before}\:\mathrm{unintalling} \\ $$$$\mathrm{import}:\:\mathrm{after}\:\mathrm{reinstalling} \\ $$ Commented by john santu…
Question Number 158191 by ghakhan88 last updated on 31/Oct/21 $${Can}\:{we}\:{reach}\:{to}\:\frac{{m}}{\left({m}−\mathrm{1}\right){s}}\:+\:\frac{{m}+\mathrm{1}}{{ms}^{\mathrm{2}} }\:{from}\: \\ $$$$ \\ $$$$\:\:\:\:\frac{{ms}+{m}\left({m}+\mathrm{1}\right)}{{s}\left({m}−{s}\right)}\:?? \\ $$ Commented by Rasheed.Sindhi last updated on 01/Nov/21 $$\mathcal{T}{hey}'{re}\:{not}\:{equal}\:{even}!…
Question Number 158186 by mkam last updated on 31/Oct/21 Answered by puissant last updated on 31/Oct/21 $${sinx}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+…… \\ $$$$\Rightarrow\:\frac{{sinx}}{{x}}\:=\:\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{7}!}+…..…
Question Number 92646 by john santu last updated on 08/May/20 Commented by Tinku Tara last updated on 08/May/20 Thanks for trying out. We have now made a few changes. please download again from http://www.tinkutara.com Commented by john santu last updated…
Question Number 158175 by mathocean1 last updated on 31/Oct/21 $$\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}° \\…