Question Number 120152 by mathocean1 last updated on 29/Oct/20 $$\mathrm{Determinate}\:\mathrm{and}\:\mathrm{construct}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:\mathrm{M} \\ $$$$\mathrm{which}\:\mathrm{have}\:\mathrm{as}\:\mathrm{affix}\:\mathrm{z}\:\:\mathrm{in}\:\mathrm{each}\:\mathrm{case}: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{arg}\left(\mathrm{i}−\mathrm{z}\right)=\mathrm{0}\left[\pi\right] \\ $$$$\left.\mathrm{2}\right)\:\mathrm{arg}\left(\mathrm{z}+\mathrm{1}−\mathrm{i}\right)=\frac{\pi}{\mathrm{6}}\left[\mathrm{2}\pi\right] \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 120154 by mathace last updated on 29/Oct/20 $${solve}\:{using}\:{LambertW}\:{function} \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} +\mathrm{17}=\mathrm{25}{x} \\ $$ Answered by mr W last updated on 29/Oct/20 $$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} =\mathrm{25}\left({x}−\frac{\mathrm{17}}{\mathrm{25}}\right)…
Question Number 120151 by mathocean1 last updated on 29/Oct/20 $$\mathrm{Represent}\:\mathrm{in}\:\mathrm{complex}\:\mathrm{plane}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points} \\ $$$$\mathrm{M}\:\mathrm{which}\:\mathrm{have}\:\mathrm{as}\:\mathrm{affix}\:\mathrm{z}\:\mathrm{such}\:\mathrm{that}\:\mid\mathrm{z}\mid=\mathrm{2}\:\mathrm{and} \\ $$$$\mathrm{arg}\left(\mathrm{z}+\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}\left[\pi\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 185684 by Shrinava last updated on 25/Jan/23 $$\mathrm{17}\:,\:\mathrm{78},\:\mathrm{143},\:\mathrm{353},\:? \\ $$$$\left.\mathrm{a}\left.\right)\left.\mathrm{3}\left.\mathrm{66}\:\:\:\:\:\mathrm{b}\right)\mathrm{0}\:\:\:\:\:\mathrm{c}\right)\mathrm{398}\:\:\:\:\:\mathrm{d}\right)\mathrm{435} \\ $$ Commented by Frix last updated on 25/Jan/23 $$\mathrm{42} \\ $$$$\left(\mathrm{Try}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}…\right) \\…
Question Number 185673 by Mingma last updated on 25/Jan/23 Answered by Frix last updated on 25/Jan/23 $$\mathrm{Due}\:\mathrm{to}\:\mathrm{symmetry}\:{x}={y}={z}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{27}} \\ $$ Terms of Service Privacy…
Question Number 54603 by behi83417@gmail.com last updated on 07/Feb/19 $${a},{b},{c}\:,{are}\:{nonnegative}\:{real}\:{numbers} \\ $$$${and}:\:\:\:{a}+{b}+{c}=\mathrm{1}\:\:. \\ $$$${show}\:{that}: \\ $$$$\:\:\:\:\:\:\mathrm{0}\leqslant\:\:\boldsymbol{\mathrm{ab}}+\boldsymbol{\mathrm{bc}}+\boldsymbol{\mathrm{ca}}−\mathrm{2}\boldsymbol{\mathrm{abc}}\:\:\:\leqslant\frac{\mathrm{7}}{\mathrm{27}}\:. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19…
Question Number 185668 by Shrinava last updated on 25/Jan/23 $$\mathrm{Find}:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{10}°}\:−\:\mathrm{4}\:\mathrm{sin}\:\mathrm{70}°\:=\:? \\ $$ Answered by Ezzat last updated on 25/Jan/23 $$=\frac{\mathrm{1}−\mathrm{4}{sin}\:\mathrm{10}°\:{sin}\mathrm{70}°\:}{{sin}\:\mathrm{10}°}\: \\ $$$$=\frac{\mathrm{1}−\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{10}°−\mathrm{70}°\right)−{cos}\left(\mathrm{10}°+\mathrm{70}°\right)\right)}{{sin}\:\mathrm{10}°} \\ $$$$=\frac{\mathrm{1}−\mathrm{2}\left({cos}\left(\mathrm{60}°\right)−{cos}\mathrm{80}°\right)}{{sin}\:\mathrm{10}°} \\…
Question Number 54600 by behi83417@gmail.com last updated on 07/Feb/19 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}:\:\boldsymbol{\mathrm{x}} \\ $$$$\left.\:\:\:\mathrm{1}\right)\:\sqrt{\mathrm{3}−\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{x}}+\mathrm{1}}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.\:\:\:\:\mathrm{2}\right)\:\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \mathrm{2}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \mathrm{3}\boldsymbol{\mathrm{x}}=\mathrm{1} \\ $$$$\left.\:\:\:\:\mathrm{3}\right)\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{p}}}+\mathrm{2}\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{1}}=\boldsymbol{\mathrm{x}}\:\:\:\:\:\left[\boldsymbol{\mathrm{p}}\in\boldsymbol{\mathrm{R}}\right] \\ $$ Commented by…
Question Number 54595 by ajfour last updated on 07/Feb/19 $$\:\:\:\:\:{x}+{y}={a} \\ $$$$\:\:\:\:\:{z}+{bx}={c} \\ $$$$\:\:\:\:\:{bz}+{xy}={d} \\ $$$$\:\:\:\:\:{Find}\:{yz}\:{in}\:{terms}\:{of}\:{a},{b},{c},{d}. \\ $$ Commented by mr W last updated on…
Question Number 185665 by mathlove last updated on 25/Jan/23 $${f}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{3}{x}+\mathrm{2}}\right)=\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}\:\:\:\:\:\:{faind}\:\:\:{f}^{−\mathrm{1}} \left(\mathrm{2}\right)=? \\ $$ Answered by cortano1 last updated on 25/Jan/23 $${f}^{−\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}\right)=\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{3}{x}+\mathrm{2}}\: \\ $$$$\Leftrightarrow\:\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}\:=\:\mathrm{2}\:;\:{x}−\mathrm{1}=\mathrm{2}{x}−\mathrm{4} \\…