Question Number 10317 by Tawakalitu ayo mi last updated on 03/Feb/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}. \\ $$$$\mathrm{3}^{\mathrm{2x}} \:=\:\mathrm{18x} \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{workings}. \\ $$ Answered by mrW1 last updated on…
Question Number 141381 by loveineq last updated on 19/May/21 $$\mathrm{Let}\:\:{a},{b}\:\geqslant\:\mathrm{0}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}+{b}+\mathrm{2}\right)^{\mathrm{3}} \:\geqslant\:\frac{\mathrm{27}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 75846 by behi83417@gmail.com last updated on 18/Dec/19 $$\boldsymbol{\mathrm{sin}}^{\mathrm{5}} \boldsymbol{\mathrm{x}}+\sqrt{\mathrm{2}}\boldsymbol{\mathrm{sinx}}=\mathrm{1}\:\:\:\:\:\:\:\:,\:\:\boldsymbol{\mathrm{x}}\in\left[\mathrm{0},\mathrm{2}\boldsymbol{\pi}\right] \\ $$ Answered by MJS last updated on 18/Dec/19 $${t}=\mathrm{sin}\:{x} \\ $$$${t}^{\mathrm{5}} +\sqrt{\mathrm{2}}{t}−\mathrm{1}=\mathrm{0} \\…
Question Number 75845 by behi83417@gmail.com last updated on 18/Dec/19 $$\begin{cases}{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{yz}}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\\{\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{xz}}=\boldsymbol{\mathrm{y}}^{\mathrm{2}} }\\{\boldsymbol{\mathrm{z}}+\boldsymbol{\mathrm{xy}}=\boldsymbol{\mathrm{z}}^{\mathrm{2}} }\end{cases}\:\:\:\:\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}. \\ $$ Commented by mr W last updated on 18/Dec/19 $${x}={y}={z}=\mathrm{0} \\…
Question Number 141372 by iloveisrael last updated on 18/May/21 Answered by EDWIN88 last updated on 18/May/21 $$\left(\mathrm{1}\right)\:\mathrm{xy}\:+\:\frac{\mathrm{x}}{\mathrm{y}}\:=\:\mathrm{5} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{xy}−\frac{\mathrm{x}}{\mathrm{y}}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\begin{cases}{\mathrm{2xy}\:=\:\frac{\mathrm{26}}{\mathrm{5}}\rightarrow\mathrm{xy}=\frac{\mathrm{13}}{\mathrm{5}}}\\{\frac{\mathrm{2x}}{\mathrm{y}}\:=\:\frac{\mathrm{24}}{\mathrm{5}}\rightarrow\frac{\mathrm{x}}{\mathrm{y}}\:=\:\frac{\mathrm{12}}{\mathrm{5}}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{xy}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\:\frac{\mathrm{13}×\mathrm{12}}{\mathrm{25}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}}…
Question Number 75828 by 21042004 last updated on 18/Dec/19 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{10}^{{n}} } \\ $$ Commented by mathmax by abdo last updated on 18/Dec/19 $$\sum_{{n}=\mathrm{1}}…
Question Number 10289 by konen last updated on 02/Feb/17 $$\frac{\mathrm{9}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}}+\mathrm{3}=\frac{\mathrm{x}−\mathrm{3}}{\mathrm{3}}\Rightarrow\Sigma\mathrm{x}=? \\ $$$$ \\ $$ Answered by ridwan balatif last updated on 02/Feb/17 $$\frac{\mathrm{9}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}}+\mathrm{3}=\frac{\mathrm{x}−\mathrm{3}}{\mathrm{3}}…
Question Number 10290 by konen last updated on 02/Feb/17 $$\mathrm{3}×\mathrm{3}!+\mathrm{4}×\mathrm{4}!+…+\mathrm{12}×\mathrm{12}!=? \\ $$ Answered by ridwan balatif last updated on 02/Feb/17 $$\mathrm{3}×\mathrm{3}!+\mathrm{4}×\mathrm{4}!+…+\mathrm{12}×\mathrm{12}! \\ $$$$\left(\mathrm{4}−\mathrm{1}\right)×\mathrm{3}!+\left(\mathrm{5}−\mathrm{1}\right)×\mathrm{4}!+\left(\mathrm{6}−\mathrm{1}\right)×\mathrm{5}!+…+\left(\mathrm{13}−\mathrm{1}\right)×\mathrm{12}! \\ $$$$\mathrm{4}×\mathrm{3}!−\mathrm{1}×\mathrm{3}!+\mathrm{5}×\mathrm{4}!−\mathrm{1}×\mathrm{4}!+\mathrm{6}×\mathrm{5}!−\mathrm{1}×\mathrm{5}!+…+\mathrm{13}×\mathrm{12}!−\mathrm{1}×\mathrm{12}!…
Question Number 141356 by physicstutes last updated on 17/May/21 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:−\mathrm{1}\right) \\ $$ Answered by bemath last updated on 18/May/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{x}+\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right)=…
Question Number 141352 by mathdanisur last updated on 17/May/21 Answered by Rasheed.Sindhi last updated on 18/May/21 $${If}\:{P}\left({x}\right)\:{is}\:{divided}\:{by}\:{D}\left({x}\right)\:{giving} \\ $$$${quotient}\:{Q}\left({x}\right)\:\&\:{remainder}\:{R}\left({x}\right) \\ $$$${then} \\ $$$${P}\left({x}\right)={D}\left({x}\right).{Q}\left({x}\right)+{R}\left({x}\right) \\ $$$${x}^{\mathrm{999}}…