Question Number 6164 by sanusihammed last updated on 16/Jun/16 $${A}\:{stone}\:{is}\:{thrown}\:{vertically}\:{upwards}\:{with}\:{velocity}\:\mathrm{20}{m}/{s}. \\ $$$${at}\:{the}\:{same}\:{time}\:,\:{and}\:\mathrm{20}{m}\:{vertically}\:{above}\:.\:{a}\:{second}\:{stone}\:{is} \\ $$$${allowed}\:{to}\:{fall}.\:{After}\:{whst}\:{time}\:{and}\:{at}\:{what}\:{height}\:{do}\:{they}\: \\ $$$${collide}\:?. \\ $$$${take}\:\:{g}\:=\:\mathrm{10}{m}/{s}^{\mathrm{2}} \\ $$ Commented by prakash jain last…
Question Number 71698 by TawaTawa last updated on 18/Oct/19 Commented by MJS last updated on 19/Oct/19 $$\mathrm{we}\:\mathrm{had}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{example}\:\mathrm{a}\:\mathrm{few}\:\mathrm{weeks}\:\mathrm{ago} \\ $$ Commented by TawaTawa last updated on…
Question Number 71696 by TawaTawa last updated on 18/Oct/19 $$\mathrm{The}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{is}\:\mathrm{measured}\:\mathrm{to}\:\mathrm{be}\:\:\mathrm{12cm}\:\mathrm{long}\:\mathrm{cofrect} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\:\mathrm{cm}.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{absolute}\:\mathrm{error} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{percentage}\:\mathrm{error}\:\mathrm{for} \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\:\:\:\:\left(\mathrm{Answer}:\:\:\mathrm{0}.\mathrm{5cm},\:\:\mathrm{4}.\mathrm{17\%}\right) \\ $$$$\left(\mathrm{b}\right)\:\:\:\:\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}.\:\:\:\:\:\:\left(\mathrm{Answer}:\:\:\:\:\mathrm{12}.\mathrm{25cm},\:\:\:\mathrm{8}.\mathrm{5\%}\right) \\ $$ Answered by MJS last updated…
Question Number 71695 by peter frank last updated on 18/Oct/19 Answered by MJS last updated on 19/Oct/19 $$\int\frac{\sqrt{\mathrm{tan}\:{x}}}{\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{\mathrm{tan}\:{x}}\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{4}}…
Question Number 6158 by enigmeyou last updated on 16/Jun/16 $$\frac{−\mathrm{1}}{\mathrm{1048576}}=\frac{\mathrm{1}}{\mathrm{8}}×\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${find}\:{n}\:? \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $$\frac{−\mathrm{1}}{\mathrm{1048576}}=\frac{\mathrm{1}}{\mathrm{8}}×\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$\frac{−\mathrm{8}}{\mathrm{1048576}}=\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}}…
Question Number 6157 by Rasheed Soomro last updated on 16/Jun/16 Answered by Yozzii last updated on 16/Jun/16 $$\mid{BC}\mid=\mid{CD}\mid=\mid{DA}\mid=\mid{AB}\mid={x}\:{units}>\mathrm{0} \\ $$$${Form}\:{the}\:{triangle}\:\bigtriangleup{GBH}\:{where}\:\angle{GBH}=\mathrm{90}° \\ $$$${and}\angle{GHB}=\theta\:\left(\mathrm{0}<\theta<\mathrm{90}°\right).\:{Let}\:\mid{GH}\mid={r}>\mathrm{0}.\:{Since} \\ $$$${H}\:{is}\:{the}\:{centre}\:{of}\:{arc}\:{GE},\:{then}\:\mid{EH}\mid=\mid{GH}\mid={r}. \\…
Question Number 71693 by TawaTawa last updated on 18/Oct/19 Answered by MJS last updated on 19/Oct/19 $$\mathrm{the}\:\mathrm{discs}\:\mathrm{are}\:\mathrm{cylinders} \\ $$$${V}=\pi{r}^{\mathrm{2}} {h} \\ $$$${V}=\pi×\mathrm{1}^{\mathrm{2}} ×\mathrm{1}+\pi×\mathrm{2}^{\mathrm{2}} ×\mathrm{1}+\pi×\mathrm{3}^{\mathrm{2}} ×\mathrm{1}=\mathrm{14}\pi…
Question Number 137225 by yutytfjh67ihd last updated on 31/Mar/21 $${Q}\mathrm{137026} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 6154 by Rasheed Soomro last updated on 16/Jun/16 $${Prove}\:{or}\:{disprove} \\ $$$$\left(\frac{\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}}}{\mathrm{5}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \geqslant\:\:\boldsymbol{{abc}}\:\:\:\forall\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0} \\ $$ Commented by Yozzii last updated on 18/Jun/16 $$\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\left(\frac{\mathrm{2}}{{c}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\sqrt{\mathrm{2}{ab}+\mathrm{2}{bc}+{ca}}}\leqslant\mathrm{1} \\…
Question Number 137226 by SLVR last updated on 31/Mar/21 Answered by MJS_new last updated on 31/Mar/21 $${f}\left({x}\right)=\mathrm{2ln}\:{x} \\ $$$${x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{11}{x}−\mathrm{6}=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right) \\ $$$$\mathrm{the}\:\mathrm{curves}\:\mathrm{intersect}\:\mathrm{at}\:{x}=\mathrm{1} \\ $$$$\underset{\mathrm{0}}…