Question Number 98077 by Lekhraj last updated on 11/Jun/20 Commented by MJS last updated on 11/Jun/20 $$\mathrm{it}\:\mathrm{only}\:\mathrm{makes}\:\mathrm{sense}\:\mathrm{if}\:{a},\:{b}\:\mathrm{are}\:\mathrm{digits}\:\mathrm{and}\:{ab} \\ $$$$\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{a}×{b}\:\mathrm{but}\:\mathrm{10}{a}+{b}: \\ $$$${a},\:{b}\:\in\mathbb{N}\wedge\mathrm{1}\leqslant{a}\leqslant\mathrm{9}\wedge\mathrm{0}\leqslant{b}\leqslant\mathrm{9} \\ $$$$\mathrm{now}\:\mathrm{try}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{pairs} \\ $$$$\mathrm{1}!+\mathrm{0}!=\mathrm{10}^{\mathrm{2}}…
Question Number 98073 by bobhans last updated on 11/Jun/20 $$\mathrm{solve}\:\sqrt{\mathrm{6}−\mathrm{x}}\:=\:\mathrm{6}−\mathrm{x}^{\mathrm{2}} \\ $$ Commented by john santu last updated on 11/Jun/20 $$\mathrm{equation}\:\mathrm{defined}\:\mathrm{in}\:\mathrm{6}−\mathrm{x}\geqslant\mathrm{0}\:\wedge\:\mathrm{6}−\mathrm{x}^{\mathrm{2}} \:\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\:\mathrm{x}\:\leqslant\:\mathrm{6}\:\wedge\:−\sqrt{\mathrm{6}}\:\leqslant\:\mathrm{x}\:\leqslant\:\sqrt{\mathrm{6}} \\…
Question Number 32532 by rahul 19 last updated on 28/Mar/18 $$\boldsymbol{{I}}{f}\:{a},{b},{c}\:{are}\:\mathrm{3}\:{positive}\:{numbers}\:{in}\:{an} \\ $$$$\boldsymbol{{A}}.\boldsymbol{{P}}\:{and}\: \\ $$$${T}=\:\frac{{a}+\mathrm{8}{b}}{\mathrm{2}{b}−{a}}+\frac{\mathrm{8}{b}+{c}}{\mathrm{2}{b}−{c}}. \\ $$$${Then}\:{the}\:{value}\:{of}\:{T}^{\:\:\mathrm{2}\:} \:{is}\:? \\ $$$${Ans}.\:{given}\:{is}\:\mathrm{361}. \\ $$ Commented by Rasheed.Sindhi…
Question Number 98067 by 995563401 last updated on 11/Jun/20 Answered by Farruxjano last updated on 11/Jun/20 Answered by 1549442205 last updated on 12/Jun/20 $$\mathrm{The}\:\mathrm{given}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to} \\…
Question Number 32524 by Bindugirish107@gmail.com last updated on 27/Mar/18 $$\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}= \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163588 by HongKing last updated on 08/Jan/22 $${In}\:\:\bigtriangleup{ABC}\:\:{prove}\:{that} \\ $$$$\frac{{a}}{{b}}\:+\:\frac{{b}}{{c}}\:+\:\frac{{c}}{{a}}\:+\:\frac{{R}^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} }\:\geqslant\:\mathrm{1}\:+\:\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:+\:\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} } \\ $$ Terms of Service…
Question Number 163587 by HongKing last updated on 08/Jan/22 Answered by MJS_new last updated on 08/Jan/22 $$\int\frac{{x}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\int\frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\frac{{a}^{\mathrm{2}}…
Question Number 163586 by HongKing last updated on 08/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163576 by cortano1 last updated on 08/Jan/22 $$\:\:{What}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{2020}} \\ $$$$\:{in}\:\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…+{x}^{\mathrm{2020}} \right)^{\mathrm{2021}} \\ $$ Answered by bobhans last updated on 08/Jan/22 $$\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}}…
Question Number 32494 by Tinkutara last updated on 26/Mar/18 Terms of Service Privacy Policy Contact: info@tinkutara.com