Question Number 63689 by Rio Michael last updated on 07/Jul/19 $${Show}\:{that}\:\:{if}\:\:{a}\mid{b}\:\:{then}\:{an}\mid{bn} \\ $$ Commented by Prithwish sen last updated on 07/Jul/19 $$\mathrm{lf}\:\mathrm{a}\mid\mathrm{b} \\ $$$$\mathrm{then}\:\mathrm{b}=\mathrm{aq}\:\:\:\left(\mathrm{q}\in\mathrm{N}\:\right) \\…
Question Number 63684 by Annmuema last updated on 07/Jul/19 $$\mathrm{cot}\:\mathrm{118} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129185 by Dwaipayan Shikari last updated on 13/Jan/21 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{\mathrm{97}+{sin}^{\mathrm{2}} {x}}}\:\:{Hypergeometric}\:{form} \\ $$ Commented by Dwaipayan Shikari last updated on 13/Jan/21 $${I}\:{have}\:{found}\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{97}}}\:\:_{\mathrm{2}}…
Question Number 129145 by Dwaipayan Shikari last updated on 13/Jan/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{1}+…}}}}}} \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 129132 by BHOOPENDRA last updated on 13/Jan/21 Answered by Dwaipayan Shikari last updated on 13/Jan/21 $$\mathscr{L}^{−\mathrm{1}} \left(\frac{\mathrm{6}{as}}{\left(\mathrm{9}{s}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\right)=\frac{\mathrm{1}}{\mathrm{9}}\mathscr{L}^{−\mathrm{1}} \left(\frac{\mathrm{2}\left(\frac{{a}}{\mathrm{3}}\right){s}}{\left({s}^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{3}}\right)^{\mathrm{2}} \right)}\right)=\frac{{t}}{\mathrm{9}}{sin}\left(\frac{{at}}{\mathrm{3}}\right) \\…
Question Number 63552 by minh2001 last updated on 05/Jul/19 $${Calculate}\:\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}+\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\overset{\mathrm{1}} {\int}}{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:{dx}+\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\:{dx}+\underset{\mathrm{2}} {\overset{\mathrm{3}} {\int}}{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{5}} +\mathrm{1}\:}{dx}+…+\underset{\mathrm{78}}…
Question Number 63534 by Rio Michael last updated on 05/Jul/19 $${find}\:{the}\:{set}\:{of}\:{values}\:{of}\:{x}\:{for}\:{which}\:{y}\:{is}\:{real}\:{if}\: \\ $$$$\:{y}=\frac{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)}{{x}+\mathrm{2}}\:,\:{x}\neq−\mathrm{2},\:{x}\in\mathbb{R} \\ $$ Answered by MJS last updated on 05/Jul/19 $$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{already}\:\mathrm{given}:\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{2}\right\} \\ $$…
Question Number 63532 by Rio Michael last updated on 05/Jul/19 $${prove}\:{that}\:{there}\:{exist}\:{unique}\:{intergers}\:{p}\:{and}\:{s}\:{sucb}\:{that} \\ $$$${a}\:=\:{bp}\:+\:{s}\:{with}\:−\frac{\mid{b}\mid}{\mathrm{2}}<\:{s}\:\leqslant\frac{\mid{b}\mid}{\mathrm{2}} \\ $$$${hence}\:{find}\:{p}\:{and}\:{s}\:{given}\:{that}\:{a}=\mathrm{49}\:{and}\:{b}=\mathrm{26} \\ $$ Answered by MJS last updated on 05/Jul/19 $${a}={bp}+{s}\:\Rightarrow\:{s}={a}−{bp}…
Question Number 63517 by Rio Michael last updated on 05/Jul/19 $$\mathrm{Given}\:\mathrm{that}\:\:\mid{z}−\mathrm{6}\mid=\mathrm{2}\mid{z}+\mathrm{6}−\mathrm{9}{i}\mid, \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Use}\:\mathrm{algebra}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}, \\ $$$$\mathrm{stating}\:\mathrm{its}\:\mathrm{center}\:\mathrm{and}\:\mathrm{its}\:\mathrm{radius}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{sketch}\:\mathrm{the}\:\mathrm{locus}\:{z}\:\mathrm{on}\:\mathrm{an}\:\mathrm{argand}\:\mathrm{diagram}. \\ $$ Answered by MJS last updated on…
Question Number 129047 by Dwaipayan Shikari last updated on 12/Jan/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}.\mathrm{1}!}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}!}\right)^{\mathrm{2}} +…=\frac{\sqrt{\pi}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$ Answered by mindispower last…