Question Number 130089 by ajfour last updated on 22/Jan/21 $${To}\:{solve}\:\:\:{x}^{\mathrm{3}} ={x}+{c} \\ $$$${Let}\:\:{y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−{c}\right) \\ $$$$\:\:\frac{{dy}}{{dx}}=\left({x}^{\mathrm{3}} −{x}−{c}\right)+\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−{p}\right) \\ $$$$\:\:{let}\:\:\:\frac{{dy}}{{dx}}={mx} \\ $$$$\Rightarrow\:\:\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−{p}\right)={mx} \\ $$$$\mathrm{3}{x}^{\mathrm{3}}…
Question Number 130087 by bobhans last updated on 22/Jan/21 $$\frac{{dx}}{{dy}}\:=\:{a}\:+\:\frac{\left({b}−{a}\right){y}}{{c}}\:+\:\frac{\left({b}−{a}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi{y}}{{c}}\right)}{\mathrm{2}\pi} \\ $$$${for}\:{a}>\mathrm{0}\:,\:{b}>\mathrm{0},\:{c}>\mathrm{0}\:{on}\:{x}\geqslant\mathrm{0}\: \\ $$ Answered by benjo_mathlover last updated on 22/Jan/21 $$\mathrm{dx}\:=\:\mathrm{a}\:\mathrm{dy}\:+\:\frac{\left(\mathrm{b}−\mathrm{a}\right)}{\mathrm{c}}\:\mathrm{y}\:\mathrm{dy}\:+\:\frac{\mathrm{b}−\mathrm{a}}{\mathrm{2}\pi}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi\mathrm{y}}{\mathrm{c}}\right)\:\mathrm{dy} \\ $$$$\mathrm{x}=\:\mathrm{ay}\:+\frac{\left(\mathrm{b}−\mathrm{a}\right)\mathrm{y}^{\mathrm{2}} }{\mathrm{2c}}\:−\frac{\left(\mathrm{b}−\mathrm{a}\right)\mathrm{c}}{\mathrm{4}\pi^{\mathrm{2}}…
Question Number 130084 by mohammad17 last updated on 22/Jan/21 Commented by bobhans last updated on 22/Jan/21 $$??? \\ $$ Commented by MJS_new last updated on…
Question Number 130085 by liberty last updated on 22/Jan/21 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\begin{cases}{\mathrm{a}=\mathrm{r}^{\mathrm{2}} −\mathrm{2rs}−\mathrm{s}^{\mathrm{2}} }\\{\mathrm{b}=\mathrm{r}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} }\\{\mathrm{c}=\mathrm{r}^{\mathrm{2}} +\mathrm{2rs}−\mathrm{s}^{\mathrm{2}} }\end{cases} \\ $$$$\:\mathrm{for}\:\mathrm{some}\:\mathrm{integers}\:\mathrm{r},\mathrm{s}\:\mathrm{then}\:\mathrm{a}^{\mathrm{2}} ,\mathrm{b}^{\mathrm{2}} ,\mathrm{c}^{\mathrm{2}} \\ $$$$\mathrm{are}\:\mathrm{three}\:\mathrm{square}\:\mathrm{in}\:\mathrm{AP}. \\ $$ Answered…
Question Number 130082 by benjo_mathlover last updated on 22/Jan/21 $$\mathrm{Given}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:{a}+\mathrm{sin}\:{b}\right)+\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:{a}−\mathrm{sin}\:{b}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:^{\mathrm{2}} {a}\:+\mathrm{sin}\:^{\mathrm{2}} {b}. \\ $$ Answered by liberty last updated on 22/Jan/21…
Question Number 64545 by ajfour last updated on 19/Jul/19 Commented by ajfour last updated on 19/Jul/19 Commented by aliesam last updated on 19/Jul/19 Answered by…
Question Number 64544 by LPM last updated on 19/Jul/19 Commented by LPM last updated on 19/Jul/19 $${area}\:{of}\:\:\Box{ABCD} \\ $$ Answered by MJS last updated on…
Question Number 64542 by LPM last updated on 19/Jul/19 Answered by MJS last updated on 19/Jul/19 $${b}=\mathrm{1}−{a} \\ $$$${a}^{\mathrm{2}} +\left(\mathrm{1}−{a}\right)^{\mathrm{2}} =\mathrm{2}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\:{b}=\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${a}^{\mathrm{7}} =\frac{\mathrm{71}}{\mathrm{16}}\pm\frac{\mathrm{41}\sqrt{\mathrm{3}}}{\mathrm{16}}\:\:\:{b}^{\mathrm{7}} =\frac{\mathrm{71}}{\mathrm{16}}\mp\frac{\mathrm{41}\sqrt{\mathrm{3}}}{\mathrm{16}}…
Question Number 64541 by Chi Mes Try last updated on 19/Jul/19 $${lol}….{QUESTION}\:{OF}\:\:{THE}\:{DAY} \\ $$$$ \\ $$$${SHOW}\:{FULL}\:{WORKINGS} \\ $$$$ \\ $$$$\int{x}\left(\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right){Ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\left(\mathrm{1}−{x}^{\mathrm{2}} \right){Ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{x}^{\mathrm{4}}…
Question Number 130077 by benjo_mathlover last updated on 22/Jan/21 $$\:\mathrm{If}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\:,\:\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{x}}{\mathrm{y}+\mathrm{3}}. \\ $$ Answered by liberty last updated on 22/Jan/21 Answered by…