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x-ln-1-x-0-1614-find-x-1II-I-think-we-can-use-Lambert-BOSSES-help-your-boy-

Question Number 191499 by Mastermind last updated on 24/Apr/23 $$\mathrm{x}\:+\:\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\:=\:\mathrm{0}.\mathrm{1614},\:\mathrm{find}\:\mathrm{x}?\mathrm{1II} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{Lambert} \\ $$$$ \\ $$$$\mathrm{BOSSES},\:\mathrm{help}\:\mathrm{your}\:\mathrm{boy}! \\ $$ Commented by mr W last updated on…

If-A-2c-a-c-b-B-c-a-0-and-C-1-c-a-1-b-are-three-points-then-prove-that-i-AB-2-BC-2-CA-2-c-2-1-c-1-2-ii-AB-2-BC-2-AC-2-

Question Number 191498 by MATHEMATICSAM last updated on 24/Apr/23 $$\mathrm{If}\:\mathrm{A}\left(\frac{\mathrm{2}{c}}{{a}}\:,\:\frac{{c}}{{b}}\right),\:\mathrm{B}\left(\frac{{c}}{{a}}\:,\:\mathrm{0}\right)\:\mathrm{and}\:\mathrm{C}\left(\frac{\mathrm{1}\:+\:{c}}{{a}}\:,\:\frac{\mathrm{1}}{{b}}\right) \\ $$$$\mathrm{are}\:\mathrm{three}\:\mathrm{points},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\: \\ $$$$\mathrm{i}.\:\:\frac{\left(\mathrm{AB}\right)^{\mathrm{2}} \:+\:\left(\mathrm{BC}\right)^{\mathrm{2}} }{\left(\mathrm{CA}\right)^{\mathrm{2}} }\:=\:\frac{{c}^{\mathrm{2}} \:+\:\mathrm{1}}{\left({c}\:−\:\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{ii}.\:\left(\mathrm{AB}\right)^{\mathrm{2}} \:+\:\left(\mathrm{BC}\right)^{\mathrm{2}} \:−\:\left(\mathrm{AC}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{2}{c}\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}}…

Question-60422

Question Number 60422 by Tawa1 last updated on 20/May/19 Commented by MJS last updated on 20/May/19 $$\mathrm{what}'\mathrm{s}\:\mathrm{a}\:“\mathrm{lidless}\:\mathrm{box}''? \\ $$$$\mathrm{what}\:\mathrm{are}\:\mathrm{the}\:“\mathrm{ends}''\:\mathrm{of}\:\mathrm{a}\:\mathrm{box}? \\ $$ Commented by Tawa1 last…

Question-60423

Question Number 60423 by Tawa1 last updated on 20/May/19 Answered by MJS last updated on 20/May/19 $$\begin{pmatrix}{\mathrm{20}}\\{\angle\mathrm{30}°}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{10}\sqrt{\mathrm{3}}}\\{\mathrm{10}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{8}}\\{\angle−\mathrm{50}°}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{8sin}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}\\{−\mathrm{8cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{10}\sqrt{\mathrm{3}}}\\{\mathrm{10}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{8sin}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}\\{−\mathrm{8cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}\end{pmatrix}\:\approx\begin{pmatrix}{\mathrm{22}.\mathrm{46}}\\{\mathrm{3}.\mathrm{87}}\end{pmatrix} \\ $$$$\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{this}\:\mathrm{vector}\:\mathrm{is}\:\mathrm{22}.\mathrm{79} \\ $$$$\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{this}\:\mathrm{vector}\:\mathrm{is}\:\mathrm{E}\:\mathrm{9}.\mathrm{78}°\:\mathrm{N}…

Question-60421

Question Number 60421 by Tawa1 last updated on 20/May/19 Answered by MJS last updated on 20/May/19 $$\eta=\frac{\pi}{\mathrm{8}}×\frac{{pr}^{\mathrm{4}} }{{LQ}} \\ $$$$\eta_{{min}} =\frac{\pi}{\mathrm{8}}×\frac{.\mathrm{97}{p}×\left(.\mathrm{98}{r}\right)^{\mathrm{4}} }{\mathrm{1}.\mathrm{01}{L}×\mathrm{1}.\mathrm{005}{Q}}\approx.\mathrm{881}\frac{\pi{pr}^{\mathrm{4}} }{\mathrm{8}{LQ}} \\ $$$$\eta_{{max}}…

lim-x-0-x-2-tanxsinx-grestest-integer-function-

Question Number 60410 by tanmay last updated on 20/May/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{{x}^{\mathrm{2}} }{{tanxsinx}}\right]\:\left[.\right]={grestest}\:{integer}\:{function} \\ $$ Commented by kaivan.ahmadi last updated on 20/May/19 $${if}\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}} \\ $$$${sinx}<{x}<{tgx}\Rightarrow{xsinx}<{x}^{\mathrm{2}} <{xtgx}\Rightarrow…

Compute-the-min-max-polynomial-q-1-x-to-e-x-on-interval-1-1-Anyone-

Question Number 191483 by Mastermind last updated on 24/Apr/23 $$\mathrm{Compute}\:\mathrm{the}\:\mathrm{min}−\mathrm{max}\:\mathrm{polynomial} \\ $$$$\mathrm{q}_{\mathrm{1}} ^{\ast} \left(\mathrm{x}\right)\:\mathrm{to}\:\mathrm{e}^{\mathrm{x}} \:\mathrm{on}\:\mathrm{interval}\:\left[−\mathrm{1},\:\mathrm{1}\right]. \\ $$$$ \\ $$$$\mathrm{Anyone}?? \\ $$ Terms of Service Privacy…