Question Number 96244 by Don08q last updated on 31/May/20 $$ \\ $$$$\:\:\mathrm{The}\:\mathrm{line}\:{y}\:=\:{mx}\:\:\mathrm{meets}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\:\:{y}\:=\:\left({x}\:−\:{a}\right)\left({b}\:−\:{x}\right)\:\mathrm{tangentially}\:\mathrm{where} \\ $$$$\:\:\mathrm{0}\:<\:{a}\:<\:{b}.\:\mathrm{Show}\:\mathrm{that}\:{m}\:=\:\left(\sqrt{{b}}\:−\:\sqrt{{a}}\right)^{\mathrm{2}} \\ $$$$ \\ $$ Commented by bobhans last updated…
Question Number 161764 by Tawa11 last updated on 22/Dec/21 Commented by Tawa11 last updated on 22/Dec/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{yellow}\:\mathrm{circle}. \\ $$ Commented by cortano last updated on…
Question Number 161745 by HongKing last updated on 21/Dec/21 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{3}\:\sqrt{\mathrm{e}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}!}\:\right]\mathrm{k2}^{-\boldsymbol{\mathrm{k}}} \\ $$ Commented by mr W last updated on…
Question Number 161744 by HongKing last updated on 21/Dec/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\:+\:\frac{\mathrm{y}\:-\:\mathrm{5}}{\mathrm{5}}\:=\:\frac{\mathrm{y}\:+\:\mathrm{x}}{\mathrm{y}\:+\:\mathrm{5}}\:+\:\frac{\mathrm{5}\:+\:\mathrm{y}}{\mathrm{5}\:+\:\mathrm{x}} \\ $$ Commented by Rasheed.Sindhi last updated on 22/Dec/21 $${Not}\:{unique}\:{solution}.{Because} \\ $$$${only}\:\boldsymbol{{one}}\:{equation}\:{is}\:{given}\:{in}\:\boldsymbol{{two}} \\…
Question Number 161733 by Tawa11 last updated on 21/Dec/21 Answered by FongXD last updated on 21/Dec/21 $$\:\:\:\mathrm{let}\:\mathrm{O}\:\mathrm{be}\:\mathrm{the}\:\mathrm{point}\:\mathrm{where}\:\mathrm{the}\:\mathrm{2} \:\mathrm{intersect} \\ $$$$\:\:\:\mathrm{let}\:\mathrm{T}\in\left(\mathrm{QR}\right),\:\mathrm{where}\:\left(\mathrm{PT}\right)\bot\left(\mathrm{QR}\right) \\ $$$$\bullet\:\mathrm{in}\:\mathrm{the}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{PQT} \\ $$$$\mathrm{we}\:\mathrm{have}:\:\mathrm{PQ}^{\mathrm{2}} =\mathrm{PT}^{\mathrm{2}}…
Question Number 96189 by bemath last updated on 30/May/20 $$\mathrm{find}\:\mathrm{a}\:\mathrm{common}\:\mathrm{roots}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{quadratic}\:\mathrm{eq} \\ $$$$\mathrm{24x}^{\mathrm{2}} +\left(\mathrm{p}+\mathrm{4}\right)\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{6x}^{\mathrm{2}} +\mathrm{11x}+\mathrm{p}+\mathrm{2}=\mathrm{0} \\ $$ Answered by john santu last…
Question Number 161704 by HongKing last updated on 21/Dec/21 $$\mathrm{let}\:\:\mathrm{n}\in\mathbb{N}\:\:\mathrm{fixed}\:,\:\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\left[\mathrm{x}\right]\left\{\mathrm{x}\right\}=\mathrm{nx} \\ $$ Commented by mr W last updated on 21/Dec/21 $${two}\:{solutions}: \\ $$$${x}=−\frac{\mathrm{1}}{{n}+\mathrm{1}},\:\mathrm{0}…
Question Number 96171 by bemath last updated on 30/May/20 $$\left(\mathrm{4}+\sqrt{\mathrm{15}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\:\mathrm{k}\sqrt{\mathrm{6}} \\ $$$$\mathrm{find}\:\mathrm{k}\: \\ $$ Commented by bobhans last updated on 30/May/20 $$\mathrm{let}\sqrt{\mathrm{4}+\sqrt{\mathrm{15}}}\:=\:\mathrm{u}\:\&\:\sqrt{\mathrm{4}−\sqrt{\mathrm{15}}}\:=\:\mathrm{v}\: \\…
Question Number 30627 by Tinkutara last updated on 23/Feb/18 Answered by ajfour last updated on 23/Feb/18 $${y}=\left({x}+\mathrm{2}\right)^{\mathrm{3}} \left(\mathrm{2}−{x}\right)^{\mathrm{4}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}\left({x}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{2}−{x}\right)^{\mathrm{4}} −\mathrm{4}\left({x}+\mathrm{2}\right)^{\mathrm{3}} \left(\mathrm{2}−{x}\right)^{\mathrm{3}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{0}\:{for}\:−\mathrm{2}\:<\:{x}\:<\:\mathrm{2}\:\:\:\Rightarrow…
Question Number 161675 by cortano last updated on 21/Dec/21 $$\:\:\begin{cases}{\mathrm{2}{x}−{y}−{e}^{−{x}} =\mathrm{0}}\\{−{x}+\mathrm{2}{y}−{e}^{−{y}} =\mathrm{0}}\end{cases} \\ $$$$\:\begin{cases}{{x}=?}\\{{y}=?}\end{cases} \\ $$ Answered by mr W last updated on 21/Dec/21 $${y}=\mathrm{2}{x}−{e}^{−{x}}…